Radical Expressions — Free Algebra 1 Review Games.

\(6 \times 6 = 36\), so \(\sqrt{36} = 6\).

\(7 \times 7 = 49\), so \(\sqrt{49} = 7\).

\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).

\(10 \times 10 = 100\), so \(\sqrt{100} = 10\).

\(\sqrt{4}\sqrt{x^2} = 2x\) (assuming \(x \geq 0\)).

\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).

Combine like radicals: \((3+5)\sqrt{2} = 8\sqrt{2}\).

\(\sqrt{3}\sqrt{12} = \sqrt{36} = 6\).

\(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\).

Multiply top and bottom by \(\sqrt{5}\): \(\frac{\sqrt{5}}{5}\).

Square both sides: \(x = 49\).

\(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\). Sum \(= 5\sqrt{2}\).

Square both sides: \(2x+3=25\), \(2x=22\), \(x=11\). Check: \(\sqrt{25}=5\).

\((\sqrt{3})^4 = ((\sqrt{3})^2)^2 = 3^2 = 9\).

Multiply by conjugate: \(\frac{3(2+\sqrt{3})}{(2)^2-(\sqrt{3})^2} = \frac{3(2+\sqrt{3})}{4-3} = 6+3\sqrt{3}\).

Since \(11 \times 11 = 121\), we have \(\sqrt{121} = 11\). A common mistake is confusing this with \(\sqrt{144} = 12\).

\(\sqrt{x^6} = x^{6/2} = x^3\). The square root operation divides the exponent by 2. A common error is dividing by 3 instead of 2, which would give \(x^2\).

64 = 8^2, so it is a perfect square. The numbers 50, 60, and 72 do not have whole-number square roots and are therefore not perfect squares.

\(\sqrt{\frac{9}{25}} = \frac{\sqrt{9}}{\sqrt{25}} = \frac{3}{5}\). The square root applies to both numerator and denominator separately. A common error is forgetting to take the square root of the denominator.

\(\sqrt{4} = 2\) and \(\sqrt{9} = 3\), so the product is \(2 \times 3 = 6\). Equivalently, \(\sqrt{4} \times \sqrt{9} = \sqrt{36} = 6\). Adding under the radical to get \(\sqrt{13}\) is a common error that misapplies the product rule.

\(0.01 = (0.1)^2\), so \(\sqrt{0.01} = 0.1\). A common mistake is dividing by 10 to get \(0.001\), but taking a square root requires finding the number that, when squared, returns the original value.

By the definition of a square root, \((\sqrt{x})^2 = x\). Equivalently, \(\sqrt{x} \times \sqrt{x} = \sqrt{x \times x} = \sqrt{x^2} = x\) for \(x \geq 0\). A common error is thinking the result is \(2x\) or \(2\sqrt{x}\) by treating multiplication as addition.

Factor 45 as \(9 \times 5\), where 9 is a perfect square. Then \(\sqrt{45} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}\). Writing \(5\sqrt{3}\) is a common error from misidentifying which factor is the perfect square.

Simplify each radical first: \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{27} = 3\sqrt{3}\). Then add like terms: \(2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}\). Adding radicands to get \(\sqrt{39}\) is incorrect -- the sum rule does not apply to addition under the radical.

\((3\sqrt{2})^2 = 3^2 \times (\sqrt{2})^2 = 9 \times 2 = 18\). A common mistake is squaring only the coefficient to get \(9\sqrt{2}\), forgetting to also square the \(\sqrt{2}\) factor. Squaring just the 3 and just the 2 without the radical gives \(36\), another frequent error.

Multiply numerator and denominator by \(\sqrt{7}\): \(\frac{2 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{2\sqrt{7}}{7}\). Choice A divides only the numerator by \(\sqrt{7}\). Choice B multiplies only the numerator, leaving the denominator unchanged.

Find the largest perfect square factor of 48: \(48 = 16 \times 3\). Then \(\sqrt{48} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}\). Using a smaller perfect square like 4 gives \(2\sqrt{12}\), which is not fully simplified because \(\sqrt{12}\) can be reduced further.

Multiply coefficients and radicands separately: \((3 \times 2) \times (\sqrt{5} \times \sqrt{5}) = 6 \times 5 = 30\). A common error is writing \(6\sqrt{5}\), which means the student correctly multiplied the coefficients but forgot that \(\sqrt{5} \times \sqrt{5} = 5\).

\(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\). Choice A uses 4 as the coefficient instead of \(\sqrt{4} = 2\). Choice B would equal \(\sqrt{18}\), and Choice C would equal \(\sqrt{36}\).

This is a difference of squares pattern: \((a + b)(a - b) = a^2 - b^2\). With \(a = \sqrt{2}\) and \(b = \sqrt{3}\), the result is \((\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1\). Adding the radicands to get 5 ignores the minus sign in the second factor.

Factor inside the radical: \(18x^2 = 9 \times 2 \times x^2\). Then \(\sqrt{18x^2} = \sqrt{9} \times \sqrt{x^2} \times \sqrt{2} = 3 \times x \times \sqrt{2} = 3x\sqrt{2}\). Note that \(\sqrt{9} = 3\), not 9, which makes choice A a common error.

\(\sqrt{16x^4} = \sqrt{16} \times \sqrt{x^4} = 4 \times x^2 = 4x^2\). A common mistake is writing \(4x\) by incorrectly computing \(\sqrt{x^4} = x\) instead of \(x^2\). Remember that \(\sqrt{x^4} = x^{4/2} = x^2\).

Multiply numerator and denominator by \(\sqrt{2}\): \(\frac{5\sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{5\sqrt{2}}{2}\). A key error leading to choice C is thinking \((\sqrt{2})^2 = 4\), when in fact \((\sqrt{2})^2 = 2\).

Square both sides: \(5 - x = 9\). Solving gives \(-x = 4\), so \(x = -4\). Check: \(\sqrt{5 - (-4)} = \sqrt{9} = 3\). The solution is negative, which surprises many students, but a negative \(x\) simply makes the expression inside the radical larger.

Multiply the coefficients: \(2 \times 3 = 6\). Multiply the radicands: \(\sqrt{3} \times \sqrt{6} = \sqrt{18} = 3\sqrt{2}\). Combine: \(6 \times 3\sqrt{2} = 18\sqrt{2}\). Choice A, \(6\sqrt{18}\), equals the same value but is not fully simplified since \(\sqrt{18} = 3\sqrt{2}\).

Square both sides: \(x + 1 = (x - 1)^2 = x^2 - 2x + 1\), giving \(x^2 - 3x = 0\), so \(x = 0\) or \(x = 3\). Checking \(x = 0\): \(\sqrt{1} = 1\), but \(0 - 1 = -1\), so \(x = 0\) is an extraneous solution. Checking \(x = 3\): \(\sqrt{4} = 2 = 3 - 1\). Only \(x = 3\) is valid.

Multiply by the conjugate \((\sqrt{5} - \sqrt{3})\) over itself. Numerator: \(4(\sqrt{5} - \sqrt{3})\). Denominator: \((\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2\). Result: \(\frac{4(\sqrt{5} - \sqrt{3})}{2} = 2(\sqrt{5} - \sqrt{3})\). Using the same sign as the conjugate (choice C) fails to eliminate the radicals from the denominator.

Using the product rule for radicals, \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b} = 3 \times 5 = 15\). Choice A adds the two square root values (\(3 + 5 = 8\)), which confuses the product of radicals with their sum.

Take square roots of both sides: \(\sqrt{x} - 2 = \pm 3\). Case 1: \(\sqrt{x} = 5\), giving \(x = 25\). Case 2: \(\sqrt{x} = -1\), which has no solution since square roots are non-negative. Therefore \(x = 25\) only. Students who choose \(x = 1\) and \(x = 25\) fail to check that \(\sqrt{1} = 1\) leads to \(\sqrt{x} = -1\), which is impossible.

Square both sides: \(x^2 + 5 = 9\), so \(x^2 = 4\), giving \(x = \pm 2\). Both solutions are valid: \(\sqrt{4 + 5} = \sqrt{9} = 3\). Unlike an equation such as \(\sqrt{x} = c\) where \(x\) must be non-negative, here \(x\) appears as \(x^2\), producing a symmetric equation with two real solutions.

\(\sqrt{81} = 9\) because \(9 \times 9 = 81\). The number 27 is the cube root of 729, not the square root of 81. \(\sqrt{9} = 3\), which is the square root of 9, not 81.

\(\sqrt{36x^2} = \sqrt{36} \times \sqrt{x^2} = 6 \times x = 6x\). Since \(x \geq 0\), \(\sqrt{x^2} = x\). The choice \(6x^2\) incorrectly doubles the exponent, and \(18x\) confuses the square root of 36 with half of 36.

\(\sqrt{11}\) is in simplest form because 11 is prime and has no perfect square factors. In contrast: \(\sqrt{12} = 2\sqrt{3}\), \(\sqrt{50} = 5\sqrt{2}\), and \(\sqrt{45} = 3\sqrt{5}\) can all be simplified further.

Using the quotient property: \(\sqrt{4/9} = \sqrt{4} / \sqrt{9} = 2/3\). Choice \(2/9\) incorrectly divides 4 by 9 without taking square roots. Choice \(4/3\) takes the square root of the denominator only.

\(\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}\). \(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\), not \(5\sqrt{3}\). Neither \(\sqrt{15}\) nor \(\sqrt{30}\) has perfect square factors that produce a coefficient of 5.

\(\sqrt{x^{10}} = x^{10/2} = x^5\). The rule is \(\sqrt{x^n} = x^{n/2}\). Since \(x \geq 0\), no absolute value is needed. \(x^2\) applies the wrong exponent, and \(x^{1/10}\) confuses square root with tenth root.

Like radicals have the same radicand. \(5\sqrt{7}\) and \(3\sqrt{7}\) both contain \(\sqrt{7}\), so they are like radicals and can be combined. \(\sqrt{4} = 2\) and \(\sqrt{9} = 3\) are rational numbers, not radical expressions.

Squaring a square root cancels the radical: \((\sqrt{6})^2 = \sqrt{6} \times \sqrt{6} = 6\). This follows from the definition of square roots. Choice 36 would require \((\sqrt{6})^4\), and choice 12 is a common error from doubling the radicand instead of squaring the expression.

Like radicals are combined by adding coefficients while keeping the radicand unchanged: \(3\sqrt{2} + 5\sqrt{2} = (3 + 5)\sqrt{2} = 8\sqrt{2}\). Choice \(8\sqrt{4} = 16\) is incorrect. Choice \(15\sqrt{2}\) results from multiplying the coefficients instead of adding them.

Factor 75 as \(25 \times 3\), where 25 is a perfect square: \(\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}\). Choice \(3\sqrt{5}\) is incorrect because \(9 \times 5 = 45\), not 75. \(\sqrt{25 \times 3}\) does not equal \(25\sqrt{3}\).

Using the quotient rule: \(\sqrt{50} / \sqrt{2} = \sqrt{50/2} = \sqrt{25} = 5\). Choice \(5\sqrt{2}\) is a common error — \(\sqrt{25} = 5\), not \(5\sqrt{2}\). Choice \(\sqrt{48}\) incorrectly subtracts radicands instead of dividing, and 25 forgets to take the square root.

Using the product rule: \(\sqrt{8} \cdot \sqrt{2} = \sqrt{8 \cdot 2} = \sqrt{16} = 4\). Choice \(\sqrt{10}\) incorrectly adds the radicands (\(8 + 2\)) instead of multiplying them. Choice \(2\sqrt{4} = 4\) is equivalent but not fully simplified.

Simplify each radical first: \(\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\) and \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\). Then add like radicals: \(5\sqrt{2} + 2\sqrt{2} = 7\sqrt{2}\). Adding the radicands directly to get \(\sqrt{58}\) is a very common error.

\(\sqrt{20x^2} = \sqrt{4 \cdot 5 \cdot x^2} = \sqrt{4} \cdot \sqrt{x^2} \cdot \sqrt{5} = 2 \cdot x \cdot \sqrt{5} = 2x\sqrt{5}\). Since \(x > 0\), \(\sqrt{x^2} = x\). Choice \(4x\sqrt{5}\) incorrectly uses \(\sqrt{4} = 4\) rather than \(2\).

Multiply numerator and denominator by \(\sqrt{5}\): \(\dfrac{3 \cdot \sqrt{5}}{2 \cdot \sqrt{5} \cdot \sqrt{5}} = \dfrac{3\sqrt{5}}{2 \cdot 5} = \dfrac{3\sqrt{5}}{10}\). Choice \(\dfrac{3\sqrt{5}}{2}\) forgets to square \(\sqrt{5}\) in the denominator. Choice \(\dfrac{3\sqrt{5}}{5}\) ignores the coefficient \(2\) already in the denominator.

Using \((a + b)^2 = a^2 + 2ab + b^2\): \((\sqrt{5})^2 + 2\sqrt{5}\sqrt{3} + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}\). Choice \(8\) forgets the middle term \(2\sqrt{15}\). Choice \(15 + 2\sqrt{15}\) incorrectly multiplies \(5\) and \(3\) instead of adding them.

First simplify \(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\). Then subtract like radicals: \(4\sqrt{5} - 2\sqrt{5} = 2\sqrt{5}\). The key step is recognizing \(\sqrt{20}\) must be simplified before subtracting. Choice \(3\sqrt{5}\) comes from incorrectly treating \(\sqrt{20}\) as \(\sqrt{5}\).

Square both sides: \(2x + 3 = 25\). Solve: \(2x = 22\), so \(x = 11\). Verify: \(\sqrt{2(11) + 3} = \sqrt{25} = 5\). Choice \(14\) results from solving \(2x = 25 + 3\) rather than \(25 - 3\). Choice \(22\) is the value of \(2x\) before dividing by \(2\).

Isolate the radical: \(\sqrt{x - 3} = 4\). Square both sides: \(x - 3 = 16\), so \(x = 19\). Verify: \(\sqrt{19 - 3} + 2 = \sqrt{16} + 2 = 4 + 2 = 6\). Choice \(7\) solves \(x - 3 = 4\) without squaring first. Choice \(35\) squares \(6\) directly instead of the isolated radical \(4\).

Multiply by the conjugate \((\sqrt{6} + \sqrt{2})\): the denominator becomes \((\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4\). The numerator becomes \(8(\sqrt{6} + \sqrt{2})\). Dividing: \(\dfrac{8(\sqrt{6} + \sqrt{2})}{4} = 2(\sqrt{6} + \sqrt{2})\). Choice B uses the wrong sign in the conjugate, which leaves an irrational denominator.

Square both sides: \(3x + 4 = x^2\). Rearrange: \(x^2 - 3x - 4 = 0\), which factors as \((x - 4)(x + 1) = 0\), giving \(x = 4\) or \(x = -1\). Check \(x = 4\): \(\sqrt{16} = 4\). Check \(x = -1\): \(\sqrt{1} = 1\), but the right side is \(-1\). Since a square root cannot equal a negative number, \(x = -1\) is extraneous. Only \(x = 4\) satisfies the original equation.

Simplify each term: \(3\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}\); \(\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}\); \(2\sqrt{3}\) stays as is. Combine: \(6\sqrt{3} - 4\sqrt{3} + 2\sqrt{3} = 4\sqrt{3}\). A common error is computing \(\sqrt{12} = \sqrt{3}\), giving \(3\sqrt{3}\) instead of \(6\sqrt{3}\) for the first term.

Factor out the largest perfect-square from each variable: \(a^3 = a^2 \cdot a\) and \(b^4 = (b^2)^2\). Then \(\sqrt{a^3 b^4} = \sqrt{a^2} \cdot \sqrt{a} \cdot \sqrt{b^4} = a \cdot \sqrt{a} \cdot b^2 = ab^2\sqrt{a}\). Choice \(a^2b^2\) would require \(a^4 b^4\) under the radical, not \(a^3 b^4\).

Using the product rule: \(\sqrt{16x} = \sqrt{16} \cdot \sqrt{x} = 4 \cdot k = 4k\). Since \(\sqrt{x} = k\) by the given condition. Choice \(16k\) incorrectly uses \(16\) itself instead of \(\sqrt{16} = 4\). Choice \(4k^2\) wrongly squares \(k\) when it should stay as \(k\), since \(\sqrt{x} = k\) not \(x = k\).

When two square root expressions are equal, square both sides to get \(2x - 1 = x + 3\). Solving: \(x = 4\). Verify both radicands are non-negative: \(2(4) - 1 = 7\) and \(4 + 3 = 7\). Check: \(\sqrt{7} = \sqrt{7}\). Choice \(x = 2\) comes from incorrectly solving \(x = 3 + 1\) instead of \(2x - x = 3 + 1\).

\(144 = 12 \times 12\), so \(\sqrt{144} = 12\). Dividing \(144\) by \(2\) gives \(72\), but square roots require finding what number multiplied by itself equals the radicand, not dividing by \(2\).

Using the quotient property: \(\sqrt{\dfrac{1}{9}} = \dfrac{\sqrt{1}}{\sqrt{9}} = \dfrac{1}{3}\). A common error is taking \(\sqrt{9} = 9\) without evaluating the square root, or flipping the fraction to get \(3\).

A radical is simplified when its radicand has no perfect square factors greater than \(1\). \(\sqrt{7}\) satisfies this because \(7\) is prime. The others are not simplified: \(\sqrt{12} = 2\sqrt{3}\), \(\sqrt{50} = 5\sqrt{2}\), and \(\sqrt{45} = 3\sqrt{5}\).

By the product property of radicals, \(\sqrt{4} \cdot \sqrt{25} = \sqrt{4 \cdot 25} = \sqrt{100} = 10\). Alternatively, \(\sqrt{4} = 2\) and \(\sqrt{25} = 5\), so \(2 \cdot 5 = 10\). A common error is adding the radicands: \(\sqrt{4 + 25} = \sqrt{29}\).

\(\sqrt{x^4} = (x^4)^{1/2} = x^{4/2} = x^2\). The square root halves the exponent. Since \(x \geq 0\), no absolute value is needed. A common mistake is leaving the exponent as \(4\) or thinking the answer is \(2x^2\) by confusing the exponent rule.

36 = 6^2, 49 = 7^2, and 64 = 8^2 are all perfect squares. 45 is not a perfect square because no integer multiplied by itself equals 45. Since 6^2 = 36 and 7^2 = 49, the value 45 falls between two consecutive perfect squares.

First evaluate each radical: \(\sqrt{9} = 3\) and \(\sqrt{4} = 2\). Then compute: \(2 \cdot 3 + 2 = 6 + 2 = 8\). A common error is combining the radicands before simplifying, treating it as \(\sqrt{9 + 4} = \sqrt{13}\), which incorrectly treats unlike radicals as addable under one root.

Find the largest perfect square factor of \(72\): \(72 = 36 \cdot 2\). So \(\sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}\). To check the distractors: \(8\sqrt{2}\) would require \(64 \cdot 2 = 128\), not \(72\); \(4\sqrt{3}\) gives \(16 \cdot 3 = 48\), not \(72\).

Like radicals are combined by adding their coefficients, just like like terms in algebra: \(3\sqrt{7} + 5\sqrt{7} = (3 + 5)\sqrt{7} = 8\sqrt{7}\). A common error is multiplying the coefficients to get \(15\sqrt{7}\), or adding the radicands to get \(8\sqrt{14}\).

First simplify \(\sqrt{45}\): \(45 = 9 \cdot 5\), so \(\sqrt{45} = 3\sqrt{5}\). Then subtract the like radicals: \(3\sqrt{5} - 1\sqrt{5} = 2\sqrt{5}\). A common mistake is subtracting directly under the radical: \(\sqrt{45 - 5} = \sqrt{40}\), which violates the rules for radical subtraction.

Multiply coefficients together and radicands together: \((2 \cdot 5)(\sqrt{3} \cdot \sqrt{3}) = 10 \cdot 3 = 30\). Because \(\sqrt{3} \cdot \sqrt{3} = 3\), the radical disappears and the result is rational. A common error is keeping the radical to get \(10\sqrt{3}\).

Simplify each term: \(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\) and \(\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}\). Then add the like radicals: \(5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}\). A common error is adding directly under the radical: \(\sqrt{75 + 27} = \sqrt{102}\), which is incorrect.

Multiply numerator and denominator by \(\sqrt{3}\): \(\frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3}\). The choice \(\frac{3}{\sqrt{3}}\) still has an irrational denominator and is not rationalized. The choice \(\frac{1}{3}\) incorrectly drops the radical from the numerator.

Factor the radicand: \(98x^2 = 49 \cdot 2 \cdot x^2\). Extract the perfect square factors: \(\sqrt{49} = 7\) and \(\sqrt{x^2} = x\). So \(\sqrt{98x^2} = 7x\sqrt{2}\). A common error is doubling 7 to get 14 (confusing \(\sqrt{49}\) with \(49/2\)), or incorrectly leaving \(x^2\) inside the radical.

First apply the product property: \(\sqrt{6} \cdot \sqrt{10} = \sqrt{60}\). Then simplify: \(60 = 4 \cdot 15\), so \(\sqrt{60} = \sqrt{4} \cdot \sqrt{15} = 2\sqrt{15}\). Leaving the answer as \(\sqrt{60}\) is not fully simplified because 60 has a perfect square factor of 4.

\((3\sqrt{2})^2 = 3^2 \cdot (\sqrt{2})^2 = 9 \cdot 2 = 18\). Both the coefficient and the radical factor must be squared. A common error is squaring only the coefficient and keeping the radical to get \(9\sqrt{2}\), or multiplying the factors instead of squaring to get \(3 \cdot 2 = 6\).

Recognize that \(x^2 + 4x + 4 = (x + 2)^2\), a perfect square trinomial. So \(\sqrt{(x + 2)^2} = x + 2\), since \(x \geq 0\) means \(x + 2 > 0\) and no absolute value is needed. A common error is trying to distribute the radical across each term, treating \(\sqrt{x^2 + 4x + 4}\) as \(\sqrt{x^2} + \sqrt{4x} + \sqrt{4}\).

To isolate x, square both sides: \((\sqrt{x})^2 = 7^2\), so \(x = 49\). Verify: \(\sqrt{49} = 7\). A common error is doubling 7 to get 14, which confuses the square root operation with dividing by 2.

Square both sides: \(x + 5 = (x - 1)^2 = x^2 - 2x + 1\). Rearranging gives \(x^2 - 3x - 4 = 0\), which factors as \((x - 4)(x + 1) = 0\), so \(x = 4\) or \(x = -1\). Check \(x = 4\): \(\sqrt{9} = 3\) and \(4 - 1 = 3\). Valid. Check \(x = -1\): \(\sqrt{4} = 2\) but \(-1 - 1 = -2\). Extraneous. Only \(x = 4\) satisfies the original equation.

This is a difference of squares pattern: \((a + b)(a - b) = a^2 - b^2\), where \(a = \sqrt{2}\) and \(b = \sqrt{6}\). The result is \((\sqrt{2})^2 - (\sqrt{6})^2 = 2 - 6 = -4\). A common mistake is assuming the product of two radicals must be positive, or incorrectly FOILing and leaving unsimplified radical terms.

Multiply numerator and denominator by the conjugate \((3 - \sqrt{5})\). Numerator: \(4(3 - \sqrt{5}) = 12 - 4\sqrt{5}\). Denominator: \(3^2 - (\sqrt{5})^2 = 9 - 5 = 4\). Result: \(\frac{12 - 4\sqrt{5}}{4} = 3 - \sqrt{5}\). The distractor \(\frac{12 - 4\sqrt{5}}{14}\) comes from incorrectly computing \(3 + 5 = 14\) in the denominator instead of using the difference of squares.

By the product property of radicals: \(\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} = 5 \cdot 3 = 15\). Since \(a = 25\) and \(b = 9\), we have \(a \cdot b = 225\), and \(\sqrt{225} = 15\). Note that 225 is the value of \(a \cdot b\) itself, not \(\sqrt{a \cdot b}\). The value 8 is the sum \(\sqrt{a} + \sqrt{b}\), not the square root of the product.

Isolate the radical first: subtract 3 from both sides to get \(2\sqrt{x - 1} = 8\), then divide by 2 to get \(\sqrt{x - 1} = 4\). Square both sides: \(x - 1 = 16\), so \(x = 17\). Check: \(2\sqrt{16} + 3 = 8 + 3 = 11\). A common error is squaring both sides before isolating the radical, which leads to an incorrect equation.

Combine using the product and quotient properties: \(\frac{\sqrt{12} \cdot \sqrt{6}}{\sqrt{2}} = \sqrt{\frac{12 \cdot 6}{2}} = \sqrt{36} = 6\). Alternatively, simplify \(\frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}\) first, then \(\sqrt{12} \cdot \sqrt{3} = \sqrt{36} = 6\). The distractor 36 is the value under the radical before taking the square root, a common step-skipping error.

Apply \((a - b)^2 = a^2 - 2ab + b^2\) where \(a = 2\sqrt{5}\) and \(b = \sqrt{3}\). Then \(a^2 = 4 \cdot 5 = 20\), \(2ab = 2(2\sqrt{5})(\sqrt{3}) = 4\sqrt{15}\), and \(b^2 = 3\). The result is \(20 - 4\sqrt{15} + 3 = 23 - 4\sqrt{15}\). The choice 17 omits the middle term entirely, and \(23 + 4\sqrt{15}\) has the wrong sign on the cross term.

\(\sqrt{49} = 7\) because \(7 \cdot 7 = 49\). A common error is to confuse this with \(\sqrt{9} = 3\), or to write \(\sqrt{7}\) by treating the radical as a variable rather than evaluating it.

\(\sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2}\). The square root distributes over division. A common error is to divide 25 by 4 first to get 6.25, whose square root is not a clean value.

\(\sqrt{5}\) is in simplest form because 5 has no perfect square factors other than 1. By contrast, \(\sqrt{8} = 2\sqrt{2}\), \(\sqrt{12} = 2\sqrt{3}\), and \(\sqrt{18} = 3\sqrt{2}\) all contain extractable perfect square factors.

\(\sqrt{x^4} = x^{4/2} = x^2\). The square root halves the exponent. Since \(x \geq 0\), no absolute value is needed. A common error is to write x by halving the base coefficient 4 instead of halving the exponent.

Like radicals are combined exactly like like terms: \((3 + 2)\sqrt{5} = 5\sqrt{5}\). Choice A, \(5\sqrt{10}\), would result from multiplying the two terms, not adding them.

\(\sqrt{36x^2} = \sqrt{36} \cdot \sqrt{x^2} = 6 \cdot x = 6x\). Since \(x \geq 0\), \(\sqrt{x^2} = x\) without needing absolute value. A common error is to forget to take the square root of \(x^2\) and write \(6x^2\) instead.

\(\sqrt{121} = 11\) because \(11 \cdot 11 = 121\). The value 60.5 is a distractor from dividing 121 by 2 rather than taking the square root. Memorizing perfect squares through \(12^2 = 144\) makes these quick to evaluate.

\(4\sqrt{3} - 1\sqrt{3} = (4 - 1)\sqrt{3} = 3\sqrt{3}\). Treat \(\sqrt{3}\) as a common factor just like a variable. A common mistake is to subtract the radicands (\(3 - 3 = 0\)) and conclude the answer is 0.

First simplify each term: \(2\sqrt{8} = 2 \cdot 2\sqrt{2} = 4\sqrt{2}\) and \(\sqrt{32} = 4\sqrt{2}\). So \(4\sqrt{2} - 4\sqrt{2} = 0\). A common error is to simplify only one of the two terms and not recognize they become equal.

\(\sqrt{3} \cdot \sqrt{27} = \sqrt{3 \cdot 27} = \sqrt{81} = 9\). Alternatively, \(\sqrt{27} = 3\sqrt{3}\), so \(\sqrt{3} \cdot 3\sqrt{3} = 3(\sqrt{3})^2 = 3 \cdot 3 = 9\). A common error is to stop at \(\sqrt{81}\) without evaluating it.

Multiply numerator and denominator by \(\sqrt{2}\): \(\frac{6\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}\). Choice A forgets to divide by 2 after rationalizing. Choice C equals \(2\sqrt{3}\), which is a different value.

Square both sides: \(2x + 1 = 25\). Subtract 1: \(2x = 24\). Divide: \(x = 12\). Check: \(\sqrt{2 \cdot 12 + 1} = \sqrt{25} = 5\). A common error is to square only the 5 and write \(2x + 1 = 25\), then incorrectly divide 25 by 2 without subtracting 1 first.

\(\sqrt{18} = 3\sqrt{2}\) and \(\sqrt{8} = 2\sqrt{2}\). Adding like radicals: \(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\). You cannot add the radicands directly: \(\sqrt{18} + \sqrt{8}\) does not equal \(\sqrt{26}\).

\(\sqrt{x^4 y^2} = \sqrt{x^4} \sqrt{y^2} = x^2 y\). The square root halves each exponent: \(4/2 = 2\) and \(2/2 = 1\). Since both variables are non-negative, no absolute value is needed. Choice B, \(xy\), results from halving 4 to 2 incorrectly as 1.

This is a difference-of-squares pattern: \((a + b)(a - b) = a^2 - b^2\). With \(a = \sqrt{5}\) and \(b = 2\): \((\sqrt{5})^2 - 2^2 = 5 - 4 = 1\). Choice B (9) is the result of \((\sqrt{5} + 2)^2\), which uses the wrong formula.

First, \(3\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}\). Then subtract: \(6\sqrt{3} - 1\sqrt{3} = 5\sqrt{3}\). A common error is to attempt to subtract before simplifying \(\sqrt{12}\), which leads to combining unlike radicals.

First isolate the radical: \(\sqrt{x} = 7 - 3 = 4\). Then square both sides: \(x = 16\). Check: \(\sqrt{16} + 3 = 4 + 3 = 7\). A common error is to square both 7 and 3 individually before subtracting, yielding \(x = 49 - 9 = 40\).

Multiply by the conjugate \((\sqrt{7} + \sqrt{3})\) over itself. The denominator becomes \((\sqrt{7})^2 - (\sqrt{3})^2 = 7 - 3 = 4\). The numerator becomes \(2(\sqrt{7} + \sqrt{3})\). Dividing by 4 gives \((\sqrt{7} + \sqrt{3}) / 2\). Choice C omits the division by 4.

Square both sides: \(x + 6 = x^2\). Rearrange: \(x^2 - x - 6 = 0\), which factors as \((x - 3)(x + 2) = 0\), giving \(x = 3\) or \(x = -2\). Check \(x = 3\): \(\sqrt{9} = 3\). Check \(x = -2\): \(\sqrt{4} = 2\), not \(-2\). Because a square root is always non-negative, \(x = -2\) is extraneous. Only \(x = 3\) is valid.

Combine under one radical: \(\sqrt{50x^3 / 2x} = \sqrt{25x^2} = 5x\). Since \(x > 0\), \(\sqrt{x^2} = x\). A common error is to expand \(\sqrt{50x^3} = 5x\sqrt{2x}\) and then divide, missing the cancellation of \(\sqrt{2x}\) in the denominator.

Use \((a + b)^2 = a^2 + 2ab + b^2\): \((\sqrt{3})^2 + 2\sqrt{3}\sqrt{5} + (\sqrt{5})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}\). Choice A (8) drops the cross-term entirely. Choice D halves the middle term incorrectly.

Distribute: \(2\sqrt{6}\sqrt{3} + \sqrt{2}\sqrt{3} = 2\sqrt{18} + \sqrt{6}\). Since \(\sqrt{18} = 3\sqrt{2}\), this becomes \(6\sqrt{2} + \sqrt{6}\). Choice C incorrectly writes \(\sqrt{6}\sqrt{3} = \sqrt{6}\) rather than \(\sqrt{18} = 3\sqrt{2}\).

Add the two equations: \(2\sqrt{a} = 6\), so \(\sqrt{a} = 3\). Subtract the second from the first: \(2\sqrt{b} = 4\), so \(\sqrt{b} = 2\). Then \(\sqrt{ab} = \sqrt{a}\sqrt{b} = 3 \cdot 2 = 6\). Choice A (4) is the value of \(b\) itself, not \(\sqrt{ab}\).

Square both sides: \(5x - 4 = 4x\). Subtract \(4x\): \(x = 4\). Check: \(\sqrt{5 \cdot 4 - 4} = \sqrt{16} = 4\) and \(2\sqrt{4} = 4\). Both sides equal 4, confirming \(x = 4\). Choice D (\(x = 1\)) fails the check: \(\sqrt{1} = 1\) on the left but \(2\sqrt{1} = 2\) on the right.

Isolate one radical: \(\sqrt{x + 4} = 1 + \sqrt{x - 1}\). Square both sides: \(x + 4 = 1 + 2\sqrt{x - 1} + (x - 1)\). Simplify: \(4 = 2\sqrt{x - 1}\). Divide by 2: \(\sqrt{x - 1} = 2\). Square: \(x - 1 = 4\), so \(x = 5\). Check: \(\sqrt{9} - \sqrt{4} = 3 - 2 = 1\). Correct.

\(\sqrt{49} = 7\) because \(7 \times 7 = 49\). The square root asks: what number multiplied by itself equals 49? The answer is 7. Choice 24.5 is 49 divided by 2, which is not the square root. Choice \(\sqrt{7}\) would be the fourth root of 49, not the square root.

\(\sqrt{81} = 9\) because \(9 \times 9 = 81\). Choice 27 is the cube root of 81 (since \(3^3 = 27\) and \(27 \times 3 = 81\)), not the square root. Choice 3 is the fourth root of 81, since \(3^4 = 81\). Only 9 satisfies \(9^2 = 81\).

\(\sqrt{7}\) is in simplest form because 7 is prime and has no perfect square factors other than 1. The others simplify further: \(\sqrt{12} = 2\sqrt{3}\), \(\sqrt{18} = 3\sqrt{2}\), and \(\sqrt{50} = 5\sqrt{2}\). A radical is in simplest form when the radicand has no perfect square factor greater than 1.

\(\sqrt{16/25} = \sqrt{16}/\sqrt{25} = 4/5\). The square root rule applies to numerator and denominator separately. \(\sqrt{16} = 4\) and \(\sqrt{25} = 5\), giving 4/5. Choice 2/5 equals \(\sqrt{4/25}\), not \(\sqrt{16/25}\). Choice 4/25 applies \(\sqrt{}\) to the numerator only and forgets to take \(\sqrt{}\) of the denominator.

\(\sqrt{4} \sqrt{9} = 2 \times 3 = 6\). You can also apply the product rule: \(\sqrt{4} \sqrt{9} = \sqrt{4 \times 9} = \sqrt{36} = 6\). Choice 36 = \(6^2\) is the square of the correct answer. Choice \(\sqrt{13}\) is wrong because radicals under separate roots cannot be added: \(\sqrt{4} + \sqrt{9}\) does not equal \(\sqrt{4 + 9}\).

\(\sqrt{x^6} = (x^6)^{1/2} = x^{6/2} = x^3\). Taking the square root divides the exponent by 2. Since \(x > 0\), \(\sqrt{x^6} = x^3\) without needing absolute value. Choice \(x^2\) would result from \(\sqrt{x^4}\). Choice \(x^{1/3}\) confuses the square root with the cube root.

\(\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \sqrt{5} = 3\sqrt{5}\). The largest perfect square factor of 45 is 9. Choice \(9\sqrt{5}\) incorrectly takes \(\sqrt{9} = 9\) instead of 3. Choice \(5\sqrt{9}\) is not fully simplified since \(\sqrt{9} = 3\), giving 15. Choice \(5\sqrt{3}\) is wrong because \(5^2 \times 3 = 75\), not 45.

\(\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \sqrt{2} = 6\sqrt{2}\). The largest perfect square factor of 72 is 36. Choice \(8\sqrt{3}\) is wrong because \(8^2 \times 3 = 192\), not 72. Choice \(36\sqrt{2}\) is wrong because \(36^2 \times 2 = 2592\), not 72. Always extract the largest perfect square to reach simplest form in one step.

Using the product rule: \(\sqrt{3} \sqrt{27} = \sqrt{3 \times 27} = \sqrt{81} = 9\). Choice \(3\sqrt{3}\) is the simplified form of \(\sqrt{27}\) alone, not the product. Choice \(3\sqrt{9}\) equals 9 numerically but is not fully simplified since \(\sqrt{9} = 3\) should be evaluated. Choice 27 would result from incorrectly computing \(\sqrt{3} \sqrt{27} = \sqrt{3} \sqrt{27}\) and treating it as \(3 \times 9\).

Simplify each term first: \(2\sqrt{48} = 2 \sqrt{16 \times 3} = 2 \times 4\sqrt{3} = 8\sqrt{3}\), and \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\). Then combine like radicals: \(8\sqrt{3} + 5\sqrt{3} = 13\sqrt{3}\). Choice \(11\sqrt{3}\) results from computing \(\sqrt{48} = 3\sqrt{3}\) (wrong) instead of \(4\sqrt{3}\). Choice \(2\sqrt{123}\) incorrectly adds radicands: \(48 + 75 = 123\), but you cannot combine unlike radicals under one square root.

Multiply numerator and denominator by \(\sqrt{5}\): \((4/\sqrt{5}) \times (\sqrt{5}/\sqrt{5}) = 4\sqrt{5} / (\sqrt{5} \sqrt{5}) = 4\sqrt{5}/5\). The denominator becomes \(\sqrt{5}^2 = 5\). Choice 4/5 incorrectly removes the radical without rationalizing. Choice \(4\sqrt{5}\) keeps the multiplication in the numerator but forgets that the denominator also becomes 5.

Square both sides: \(x - 3 = 4^2 = 16\), so \(x = 19\). Check: \(\sqrt{19 - 3} = \sqrt{16} = 4\). Choice \(x = 7\) results from adding 3 to 4 without first squaring: \(4 + 3 = 7\). Choice \(x = 13\) comes from a similar error: computing \(4^2 = 16\) but then only adding 3 to get \(16 - 3 = 13\) using wrong algebra. Always square both sides before isolating \(x\).

\((\sqrt{5})^4 = (5^{1/2})^4 = 5^{4/2} = 5^2 = 25\). Alternatively, group the exponents: \(((\sqrt{5})^2)^2 = 5^2 = 25\). Choice \(625 = 5^4\) ignores the square root and treats \(\sqrt{5}\) as if it were \(5\). Choice \(5\) stops after squaring once: \((\sqrt{5})^2 = 5\), but the exponent is \(4\), not \(2\).

\(\sqrt{2} \cdot \sqrt{8} = \sqrt{2 \cdot 8} = \sqrt{16} = 4\). Choice \(16 = 4^2\) is the square of the answer, not the answer itself. Choice \(2\sqrt{2}\) equals \(\sqrt{8}\) alone, not the product of \(\sqrt{2}\) and \(\sqrt{8}\). Choice \(\sqrt{10}\) results from incorrectly adding the radicands: \(\sqrt{2 + 8} = \sqrt{10}\), but multiplication means the radicands multiply, not add.

Like radicals are combined by subtracting their coefficients: \(5\sqrt{3} - 2\sqrt{3} = (5 - 2)\sqrt{3} = 3\sqrt{3}\). This mirrors combining like terms: \(5x - 2x = 3x\). Choice \(3\sqrt{6}\) results from mistakenly adding radicands when combining terms: treating it as \(\sqrt{3 + 3} = \sqrt{6}\). Choice \(3\) drops the \(\sqrt{3}\) factor entirely.

Square both sides: \(3x - 2 = 16\). Add 2 to both sides: \(3x = 18\). Divide by 3: \(x = 6\). Check: \(\sqrt{3(6) - 2} = \sqrt{16} = 4\). Choice \(x = 4\) results from setting \(3x - 2 = 4\) without squaring the right side first. Choice \(x = 2\) comes from computing \(3x = 4 + 2 = 6\) using \(4\) instead of \(16\).

\(\sqrt{50x^2} = \sqrt{25 \cdot 2 \cdot x^2} = \sqrt{25} \cdot \sqrt{x^2} \cdot \sqrt{2} = 5 \cdot x \cdot \sqrt{2} = 5x\sqrt{2}\). Since \(x > 0\), \(\sqrt{x^2} = x\). Choice \(5x\) drops the remaining \(\sqrt{2}\) factor. Choice \(25x\sqrt{2}\) takes \(\sqrt{25} = 25\) instead of \(5\). Choice \(5\sqrt{2x}\) fails to extract \(x^2\) fully from under the radical.

\(\sqrt{6} \cdot \sqrt{15} = \sqrt{6 \cdot 15} = \sqrt{90} = \sqrt{9 \cdot 10} = \sqrt{9} \cdot \sqrt{10} = 3\sqrt{10}\). Choice \(\sqrt{21}\) incorrectly adds the radicands: \(\sqrt{6 + 15} = \sqrt{21}\), but the product rule multiplies them. Choice \(9\sqrt{10}\) takes \(\sqrt{9} = 9\) instead of \(3\). Always factor out the largest perfect square from the product.

Multiply numerator and denominator by the conjugate \((3 + \sqrt{2})\). Numerator: \((3 + \sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}\). Denominator: \(3^2 - (\sqrt{2})^2 = 9 - 2 = 7\). Result: \((11 + 6\sqrt{2})/7\). Choice \((11 - 6\sqrt{2})/7\) has a sign error in the cross term. Choice \((11 + 2\sqrt{2})/7\) uses \(2 \cdot 1 \cdot \sqrt{2} = 2\sqrt{2}\) for the middle term instead of \(2 \cdot 3 \cdot \sqrt{2} = 6\sqrt{2}\).

Factor out perfect squares: \(\sqrt{12x^5} = \sqrt{4 \cdot 3 \cdot x^4 \cdot x} = \sqrt{4} \cdot \sqrt{x^4} \cdot \sqrt{3x} = 2 \cdot x^2 \cdot \sqrt{3x} = 2x^2\sqrt{3x}\). Choice \(4x^2\sqrt{3x}\) incorrectly takes \(\sqrt{4} = 4\) instead of \(2\). Choice \(2x\sqrt{3x^3}\) extracts only \(x^2\) (not \(x^4\)) from \(x^5\), leaving too much inside the radical.

Square both sides: \(2x - 1 = (x - 2)^2 = x^2 - 4x + 4\), which gives \(x^2 - 6x + 5 = 0\), factoring as \((x - 1)(x - 5) = 0\), so \(x = 1\) or \(x = 5\). Check \(x = 1\): \(\sqrt{1} = 1\) but \(x - 2 = -1\). Since a square root cannot equal a negative number, \(x = 1\) is extraneous. Check \(x = 5\): \(\sqrt{9} = 3\) and \(x - 2 = 3\). Valid. Only \(x = 5\) is a solution.

Simplify each term individually: \(\sqrt{8} = 2\sqrt{2}\), \(\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}\), \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\). Then combine like radicals: \(2\sqrt{2} + 4\sqrt{2} - 3\sqrt{2} = 3\sqrt{2}\). Choice \(\sqrt{22}\) comes from adding and subtracting radicands directly: \(8 + 32 - 18 = 22\), which is not valid. Choice \(5\sqrt{2}\) forgets to subtract the \(3\sqrt{2}\) term from \(\sqrt{18}\).

Use the formula \((a - b)^2 = a^2 - 2ab + b^2\): \((\sqrt{6})^2 - 2 \cdot \sqrt{6} \cdot \sqrt{2} + (\sqrt{2})^2 = 6 - 2\sqrt{12} + 2 = 8 - 2\sqrt{12}\). Then simplify \(\sqrt{12} = 2\sqrt{3}\): \(8 - 2 \cdot 2\sqrt{3} = 8 - 4\sqrt{3}\). Choice \(4\) applies the difference-of-squares pattern \((a-b)(a+b)\), not \((a-b)^2\). Choice \(8 - 2\sqrt{3}\) skips the simplification of \(\sqrt{12}\) to \(2\sqrt{3}\).

Multiply by the conjugate \((\sqrt{5} + \sqrt{3}) / (\sqrt{5} + \sqrt{3})\). Numerator: \((\sqrt{5} + \sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}\). Denominator: \((\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2\). Result: \((8 + 2\sqrt{15})/2 = 4 + \sqrt{15}\). Choice \((8 + 2\sqrt{15})/2\) is correct but not fully simplified. Choice \(4 - \sqrt{15}\) has a sign error in the numerator cross term.

Isolate the radical: \(\sqrt{3x + 4} = x - 2\). For the square root to be valid, \(x - 2\) must be non-negative, so \(x \geq 2\). Square both sides: \(3x + 4 = x^2 - 4x + 4\), giving \(x^2 - 7x = 0\), so \(x(x - 7) = 0\), meaning \(x = 0\) or \(x = 7\). Check \(x = 0\): \(\sqrt{4} + 2 = 4\) but \(0\) does not equal \(4\) — extraneous. Check \(x = 7\): \(\sqrt{25} + 2 = 7\). Valid. Only \(x = 7\) is a solution.

\(13 \cdot 13 = 169\), so \(\sqrt{169} = 13\). Note that \(11^2 = 121\) and \(12^2 = 144\), both less than \(169\), making \(13\) the correct answer.

81 = 9 * 9 = 9^2, making it a perfect square with an integer square root. The numbers 45, 56, and 90 have no integer square roots.

\(\sqrt{x^6} = x^{6/2} = x^3\). A common mistake is choosing \(x^2\), which would correspond to \(\sqrt{x^4}\), or choosing \(6x\), which incorrectly treats the exponent as a coefficient.

\(\sqrt{16/49} = \sqrt{16}/\sqrt{49} = 4/7\). The square root of a fraction equals the square root of the numerator divided by the square root of the denominator.

Like radicals are added by combining coefficients: \(6 + 5 = 11\), keeping the radicand \(\sqrt{2}\) unchanged. Choosing \(11\sqrt{4}\) is wrong because adding like radicals does not change the radicand.

Subtract the coefficients: \(12 - 4 = 8\), keeping \(\sqrt{7}\). The expression \(8\sqrt{14}\) is wrong because subtracting like radical terms does not change the radicand from \(7\) to \(14\).

\(\sqrt{4} = 2\) and \(\sqrt{16} = 4\), so the product is \(2 \cdot 4 = 8\). Alternatively, \(\sqrt{4 \cdot 16} = \sqrt{64} = 8\). Choosing \(64\) mistakes the product of the original radicands (\(4 \cdot 16\)) for the final answer without taking the square root.

\(0.5 \cdot 0.5 = 0.25\), so \(\sqrt{0.25} = 0.5\). Equivalently, \(0.25 = 1/4\), and \(\sqrt{1/4} = 1/2 = 0.5\). The other choices are not equal to \(0.25\) when squared.

Factor out the largest perfect square: \(72 = 36 \cdot 2\), so \(\sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}\). The choice \(8\sqrt{3}\) is incorrect because \(8^2 \cdot 3 = 192\), not \(72\).

First simplify \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\). Then add like radicals: \(2\sqrt{3} + 2\sqrt{3} = 4\sqrt{3}\). The error \(\sqrt{15}\) comes from treating the expression as a product instead of a sum.

Multiply numerator and denominator by \(\sqrt{2}\): \((8 \cdot \sqrt{2}) / (\sqrt{2} \cdot \sqrt{2}) = 8\sqrt{2}/2 = 4\sqrt{2}\). Choosing \(8\sqrt{2}\) forgets to simplify by dividing by the resulting denominator of \(2\).

\(\sqrt{3} \cdot \sqrt{27} = \sqrt{3 \cdot 27} = \sqrt{81} = 9\). A common error is stopping at \(\sqrt{81}\) without evaluating it, or choosing \(81\) by squaring the product of \(3\) and \(27\) incorrectly.

Square both sides: \(x + 1 = 36\), so \(x = 35\). Check: \(\sqrt{35 + 1} = \sqrt{36} = 6\), which is correct. Choosing \(x = 36\) is a common mistake that results from forgetting to subtract \(1\) after squaring both sides.

\(\sqrt{75} = \sqrt{25 * 3} = 5\sqrt{3}\) and \(\sqrt{27} = \sqrt{9 * 3} = 3\sqrt{3}\). Subtracting like radicals: \(5\sqrt{3} - 3\sqrt{3} = 2\sqrt{3}\). Choosing \(4\sqrt{3}\) or \(\sqrt{48}\) would result from adding rather than subtracting.

Rewrite as \(\sqrt{9x^2 * 2x} = \sqrt{9x^2} * \sqrt{2x} = 3x\sqrt{2x}\). The choice \(3x^2\sqrt{2}\) is incorrect because \(x^3\) is not a perfect square, so only \(x^2\) can be pulled outside leaving one factor of \(x\) inside.

\((\sqrt{3})^6 = (3^{1/2})^6 = 3^{6/2} = 3^3 = 27\). Choosing 9 applies only the exponent 2 instead of 3, and choosing \(81 = 3^4\) uses the wrong resulting exponent.

\(\sqrt{45} = \sqrt{9 * 5} = 3\sqrt{5}\) and \(\sqrt{20} = \sqrt{4 * 5} = 2\sqrt{5}\). Adding like radicals: \(3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}\). The error \(\sqrt{65}\) comes from adding the radicands (\(45 + 20 = 65\)) without first simplifying.

Multiply by the conjugate \((\sqrt{5} + 1)/(\sqrt{5} + 1)\). Numerator: \(4(\sqrt{5} + 1)\). Denominator: \((\sqrt{5})^2 - 1^2 = 5 - 1 = 4\). Result: \(4(\sqrt{5} + 1)/4 = \sqrt{5} + 1\). The choice \(4(\sqrt{5} - 1)\) incorrectly applies the wrong sign in the conjugate result.

Square both sides: \(x + 6 = x^2\), so \(x^2 - x - 6 = 0\), which factors as \((x - 3)(x + 2) = 0\) giving \(x = 3\) or \(x = -2\). Check \(x = 3\): \(\sqrt{9} = 3\), which is valid. Check \(x = -2\): \(\sqrt{4} = 2\), but 2 is not equal to \(-2\), so \(x = -2\) is an extraneous solution.

Apply \((a + b)^2 = a^2 + 2ab + b^2\): \((\sqrt{3})^2 + 2\sqrt{3}\sqrt{5} + (\sqrt{5})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}\). Choosing just 8 ignores the middle cross term \(2\sqrt{15}\), which is a common binomial expansion error.

\(\sqrt{200} = \sqrt{100 * 2} = 10\sqrt{2}\), \(\sqrt{8} = \sqrt{4 * 2} = 2\sqrt{2}\), and \(\sqrt{50} = \sqrt{25 * 2} = 5\sqrt{2}\). Combining: \(10\sqrt{2} - 2\sqrt{2} + 5\sqrt{2} = 13\sqrt{2}\). Choosing \(15\sqrt{2}\) results from adding all three instead of subtracting the middle term.

Multiply by the conjugate \((\sqrt{3} + 1)/(\sqrt{3} + 1)\). Numerator: \((2 + \sqrt{3})(\sqrt{3} + 1) = 2\sqrt{3} + 2 + 3 + \sqrt{3} = 5 + 3\sqrt{3}\). Denominator: \((\sqrt{3})^2 - 1^2 = 3 - 1 = 2\). Final result: \((5 + 3\sqrt{3})/2\).

Isolate the radical: \(\sqrt{3x - 5} = x - 1\). Square both sides: \(3x - 5 = x^2 - 2x + 1\), so \(x^2 - 5x + 6 = 0\) and \((x - 2)(x - 3) = 0\). Check \(x = 2\): \(\sqrt{1} + 1 = 1 + 1 = 2\), which is valid. Check \(x = 3\): \(\sqrt{4} + 1 = 2 + 1 = 3\), which is also valid. Both solutions check out.

Simplify the numerator: \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{27} = 3\sqrt{3}\), so the numerator is \(5\sqrt{3}\). Dividing: \(5\sqrt{3} / \sqrt{3} = 5\). Choosing \(5\sqrt{3}\) forgets to carry out the final division, and \(\sqrt{13}\) incorrectly adds the radicands.

Square both sides: \(x - 2 = 2x - 7\), so \(5 = x\). Check: \(\sqrt{5 - 2} = \sqrt{3}\) and \(\sqrt{2*5 - 7} = \sqrt{3}\), which confirms equality. Also verify the domain: both radicands must be non-negative, requiring \(x\) at least 3.5. Since \(x = 5\) satisfies this, the solution is valid.

\(\sqrt{64} = 8\) because \(8^2 = 64\). A common error is to divide 64 by 2 and get 32, but the square root asks 'what number times itself equals 64?' — that number is 8, not 32.

\(\sqrt{121} = 11\) because \(11^2 = 121\). Since \(10^2 = 100\) and \(12^2 = 144\), the only integer whose square equals 121 is 11.

Using the quotient property of radicals: \(\sqrt{25/4} = \sqrt{25} / \sqrt{4} = 5/2\). A common mistake is to take \(\sqrt{25} = 5\) but forget to take \(\sqrt{4} = 2\), leaving 5/4 instead of the correct 5/2.

A radical is in simplest form when the radicand has no perfect square factors other than 1. \(\sqrt{8} = 2\sqrt{2}\), \(\sqrt{50} = 5\sqrt{2}\), and \(\sqrt{12} = 2\sqrt{3}\) — all can be simplified further. \(\sqrt{13}\) is already in simplest form because 13 is prime and has no perfect square factors.

\(\sqrt{9} = 3\) and \(\sqrt{16} = 4\), so \(\sqrt{9} + \sqrt{16} = 3 + 4 = 7\). A very common error is to combine radicands first: \(\sqrt{9 + 16} = \sqrt{25} = 5\). This is incorrect — each square root must be evaluated separately before adding.

\(\sqrt{4} * \sqrt{9} = 2 * 3 = 6\). Equivalently, by the product property: \(\sqrt{4 * 9} = \sqrt{36} = 6\). The choice 36 is wrong because that would be \((2 * 3)^2\), not the product itself. The choice \(\sqrt{13}\) mistakenly adds the radicands.

\(\sqrt{x^6} = x^{6/2} = x^3\). Taking a square root means dividing the exponent by 2. Since \(x > 0\), no absolute value is needed. The choice \(x^2\) would be the answer for \(\sqrt{x^4}\), not \(\sqrt{x^6}\). The choice \(3x\) incorrectly treats the exponent as a coefficient.

Factor out the largest perfect square: \(48 = 16 * 3\), so \(\sqrt{48} = \sqrt{16} * \sqrt{3} = 4\sqrt{3}\). The choice \(3\sqrt{4} = 3 * 2 = 6\), which is much less than \(\sqrt{48} \approx 6.93\). The choice \(8\sqrt{3}\) would correspond to \(\sqrt{192}\), not \(\sqrt{48}\).

First simplify \(\sqrt{12} = \sqrt{4 * 3} = 2\sqrt{3}\). Then add like radicals: \(2\sqrt{3} + 1\sqrt{3} = 3\sqrt{3}\). The choice \(\sqrt{15}\) is wrong because you cannot combine radicands by adding them — \(\sqrt{12 + 3} = \sqrt{15}\) is a common but incorrect shortcut.

Factor out the largest perfect square: \(72 = 36 * 2\), so \(\sqrt{72} = \sqrt{36} * \sqrt{2} = 6\sqrt{2}\). The choice \(4\sqrt{3} \approx 6.93\) while \(\sqrt{72} \approx 8.49\), so they are not equal. The choice \(8\sqrt{2}\) corresponds to \(\sqrt{128}\), not \(\sqrt{72}\).

By the product property: \(\sqrt{5} * \sqrt{45} = \sqrt{5 * 45} = \sqrt{225} = 15\). Alternatively, \(\sqrt{45} = 3\sqrt{5}\), so \(\sqrt{5} * 3\sqrt{5} = 3 * (\sqrt{5})^2 = 3 * 5 = 15\). The choice \(3\sqrt{5} \approx 6.7\), which is not equal to 15.

To undo the square root, square both sides: \(x = 7^2 = 49\). Verify: \(\sqrt{49} = 7\) ✓. A common mistake is to confuse the equation \(\sqrt{x} = 7\) with \(x = 7\) — but the equation says the square root of \(x\) is 7, so \(x\) itself must be 49.

Multiply numerator and denominator by \(\sqrt{5}\): \((5 * \sqrt{5}) / (\sqrt{5} * \sqrt{5}) = 5\sqrt{5} / 5 = \sqrt{5}\). The answer is simply \(\sqrt{5}\). The choice 5 would require \(5/\sqrt{5} = 5\), which implies \(\sqrt{5} = 1\) — clearly false.

Factor out the largest perfect square: \(98 = 49 \times 2\), so \(\sqrt{98} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2}\). The choice \(9\sqrt{2} \approx 12.73\), while the correct value \(7\sqrt{2} \approx 9.90\). The choice \(49\sqrt{2}\) mistakenly uses \(49\) as a coefficient without taking its square root.

Like radicals are combined by adding their coefficients, just like combining like terms: \(2\sqrt{5} + 3\sqrt{5} = (2 + 3)\sqrt{5} = 5\sqrt{5}\). The choice \(5\sqrt{10}\) incorrectly multiplies the radicands instead of adding the coefficients. Only the coefficients change; the radicand stays the same.

Split into separate square roots: \(\sqrt{x^2} \times \sqrt{y^4} = x \times y^2\). For \(x^2\): \(\sqrt{x^2} = x^{2/2} = x\). For \(y^4\): \(\sqrt{y^4} = y^{4/2} = y^2\). The choice \(x^2 y^2\) incorrectly leaves the exponent on \(x\) unchanged instead of halving it.

Square both sides: \(x - 3 = 16\). Add 3 to both sides: \(x = 19\). Verify: \(\sqrt{19 - 3} = \sqrt{16} = 4\) ✓. The choice \(7\) comes from the arithmetic error \(3 + 4 = 7\) instead of squaring \(4\) first. The choice \(13\) comes from incorrectly computing \(16 - 3\) instead of \(16 + 3\).

First simplify \(\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}\). Then add like radicals: \(4\sqrt{2} + 1\sqrt{2} = 5\sqrt{2}\). The choice \(\sqrt{34}\) results from incorrectly adding under the radical: \(\sqrt{32 + 2} = \sqrt{34}\), which is the wrong procedure for addition of radicals.

Apply \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 2\sqrt{3}\) and \(b = \sqrt{5}\). \((2\sqrt{3})^2 = 4 \times 3 = 12\); \(2(2\sqrt{3})(\sqrt{5}) = 4\sqrt{15}\); \((\sqrt{5})^2 = 5\). Result: \(12 - 4\sqrt{15} + 5 = 17 - 4\sqrt{15}\). The choice \(17 + 4\sqrt{15}\) forgets the negative from \((a - b)^2\). The choice \(7\) forgets the middle term entirely.

Multiply by the conjugate \((\sqrt{6} + \sqrt{3})\) over itself. Numerator: \(\sqrt{3}(\sqrt{6} + \sqrt{3}) = \sqrt{18} + 3 = 3\sqrt{2} + 3\). Denominator: \((\sqrt{6})^2 - (\sqrt{3})^2 = 6 - 3 = 3\). Simplify: \(\dfrac{3\sqrt{2} + 3}{3} = \sqrt{2} + 1\). The choice \(\sqrt{2} - 1\) would result from a sign error when distributing in the numerator.

Square both sides: \(x + 5 = (x - 1)^2 = x^2 - 2x + 1\). Rearranging gives \(x^2 - 3x - 4 = 0\), which factors as \((x - 4)(x + 1) = 0\), so \(x = 4\) or \(x = -1\). Check \(x = 4\): \(\sqrt{9} = 3 = 4 - 1\) ✓. Check \(x = -1\): \(\sqrt{4} = 2\), but \(-1 - 1 = -2\), and a square root cannot equal a negative number, so \(x = -1\) is extraneous. Only \(x = 4\) is valid.

Rewrite as \(\sqrt{25 \times 3 \times x^2 \times x \times y^2}\). Extract perfect squares: \(\sqrt{25} = 5\), \(\sqrt{x^2} = x\), \(\sqrt{y^2} = y\). The remaining factor under the radical is \(3x\). Final result: \(5xy\sqrt{3x}\). The choice \(5x^2y\sqrt{3}\) incorrectly pulls out \(x^2\) from \(x^3\), leaving no \(x\) under the radical, which is wrong.

Isolate one radical: \(\sqrt{x + 7} = 1 + \sqrt{x}\). Square both sides: \(x + 7 = 1 + 2\sqrt{x} + x\). The \(x\) terms cancel, giving \(6 = 2\sqrt{x}\), so \(\sqrt{x} = 3\) and \(x = 9\). Verify: \(\sqrt{16} - \sqrt{9} = 4 - 3 = 1\) ✓. The choice \(x = 4\) gives \(\sqrt{11} - 2 \approx 1.32\), which does not equal \(1\).

This is a difference of squares: \((a + b)(a - b) = a^2 - b^2\), with \(a = \sqrt{2}\) and \(b = \sqrt{6}\). Result: \((\sqrt{2})^2 - (\sqrt{6})^2 = 2 - 6 = -4\). The choice \(4\) has the wrong sign. Note that \(-\sqrt{4} = -2\), which is not the same as \(-4\), so that choice is also incorrect.

Isolate the radical: subtract 3 from both sides to get \(2\sqrt{x - 1} = 6\), then divide by 2 to get \(\sqrt{x - 1} = 3\). Square both sides: \(x - 1 = 9\), so \(x = 10\). Verify: \(2\sqrt{10 - 1} + 3 = 2\sqrt{9} + 3 = 2(3) + 3 = 9\) ✓. The choice \(x = 13\) comes from forgetting to divide by 2 before squaring, giving \(x - 1 = 36\) and \(x = 37\).

\(\sqrt{144} = 12\) because \(12^2 = 144\). A common mistake is to halve the radicand, giving \(72\), but dividing by 2 is not how square roots work. You must find the number that multiplies by itself to equal \(144\).

Apply the product rule: \(\sqrt{49x^2} = \sqrt{49} \times \sqrt{x^2} = 7 \times x = 7x\). Since \(x > 0\), \(\sqrt{x^2} = x\) (no absolute value needed). A common error is forgetting to take the square root of the coefficient, leaving \(49x\) instead of \(7x\).

A radical is in simplest form when the radicand has no perfect square factors other than 1. \(\sqrt{50} = 5\sqrt{2}\), \(\sqrt{18} = 3\sqrt{2}\), and \(\sqrt{8} = 2\sqrt{2}\) can all be simplified further. Since \(13\) is prime, it has no perfect square factors, so \(\sqrt{13}\) is already fully simplified.

Use the quotient rule for radicals: \(\sqrt{50} / \sqrt{2} = \sqrt{50/2} = \sqrt{25} = 5\). A common mistake is to subtract radicands instead of dividing them, giving \(\sqrt{50 - 2} = \sqrt{48}\), which is incorrect. Another error is forgetting to take the final square root and stopping at \(25\).

First find \(x\): if \(\sqrt{x} = 4\), then \(x = 16\). Now substitute: \(\sqrt{16 + 9} = \sqrt{25} = 5\). A very common error is to add the separate square roots: \(\sqrt{x} + \sqrt{9} = 4 + 3 = 7\). However, \(\sqrt{a + b}\) does not equal \(\sqrt{a} + \sqrt{b}\); you must add inside the radical first, then take the root.

First simplify \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\). Then \(3\sqrt{12} = 3 \times 2\sqrt{3} = 6\sqrt{3}\). Now the like terms can be combined: \(6\sqrt{3} - 1\sqrt{3} = 5\sqrt{3}\). A common error is subtracting the radicands directly and getting \(2\sqrt{9}\), which ignores the coefficient rule for simplifying.

Square both sides: \((\sqrt{3x})^2 = 6^2\) gives \(3x = 36\). Divide both sides by 3 to get \(x = 12\). Verify: \(\sqrt{3 \times 12} = \sqrt{36} = 6\). A common error is squaring only the variable part and treating the 3 as a coefficient outside the radical, leading to \(x = 36\).

Multiply numerator and denominator by the conjugate \((3 + \sqrt{3})\): numerator becomes \(6(3 + \sqrt{3}) = 18 + 6\sqrt{3}\); denominator becomes \((3)^2 - (\sqrt{3})^2 = 9 - 3 = 6\). So the result is \(\dfrac{18 + 6\sqrt{3}}{6} = 3 + \sqrt{3}\). A common error is stopping after multiplying and leaving the unsimplified form \(18 + 6\sqrt{3}\) without dividing through by 6.

Square both sides: \(x + 3 = (x - 3)^2 = x^2 - 6x + 9\). Rearranging: \(0 = x^2 - 7x + 6 = (x - 6)(x - 1)\), giving \(x = 6\) or \(x = 1\). Now check both. For \(x = 6\): \(\sqrt{9} = 3\) and \(6 - 3 = 3\). Valid. For \(x = 1\): \(\sqrt{4} = 2\) but \(1 - 3 = -2\). Since a principal square root cannot equal a negative number, \(x = 1\) is extraneous. Only \(x = 6\) is a valid solution.

Use \((a + b)^2 = a^2 + 2ab + b^2\): \((\sqrt{6})^2 + 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2 = 6 + 2\sqrt{12} + 2 = 8 + 2\sqrt{12}\). Now simplify \(\sqrt{12} = 2\sqrt{3}\), so \(2\sqrt{12} = 4\sqrt{3}\). The final answer is \(8 + 4\sqrt{3}\). The most common error is computing \((\sqrt{6} + \sqrt{2})^2\) as simply \(6 + 2 = 8\), which omits the crucial middle cross term \(2ab\).