Radical Functions — Free Algebra 2 Review Games.

\(9^2 = 81\), so \(\sqrt{81} = 9\).

3^3 = 27, so the cube root of 27 is 3.

\(x^{1/2} = \sqrt{x}\) by definition of rational exponents.

\(x^{a/b}\) means the \(b\)-th root of \(x^a\).

Square root requires non-negative input: \(x \geq 0\).

Cube root of 8 = 2, then 2^2 = 4.

\(\sqrt{x^6} = x^3\) (assuming \(x \geq 0\)).

Square both sides: \(x + 5 = 9\), \(x = 4\).

The cube root of 27 is 3.

\(x - 4 \geq 0\), so \(x \geq 4\).

Square: \(2x+1=x^2-2x+1\), \(x^2-4x=0\), \(x(x-4)=0\). Check \(x=4\): \(\sqrt{9}=3=4-1\). Check \(x=0\): \(\sqrt{1}=1\) but \(0-1=-1\). Only \(x=4\) works.

Multiply exponents: \((3/4)\cdot(4/3) = 1\), so \(x^1 = x\).

Cube both sides: x - 1 = -8, x = -7.

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

Isolate one radical: \(\sqrt{x} = 5-\sqrt{x-5}\). Square: \(x = 25-10\sqrt{x-5}+x-5\). \(10\sqrt{x-5} = 20\), \(\sqrt{x-5}=2\), \(x-5=4\), \(x=9\). Check: \(3+2=5\). Correct.

4^(1/2) means the square root of 4. Since 2^2 = 4, the answer is 2. Choice D (16) is a common error from multiplying 4 by 4 instead of taking the square root.

The 4th root of 81 is the number that when raised to the 4th power equals 81. Since 3^4 = 81, the answer is 3. Choice B (9) is the square root of 81, not the 4th root.

25^(1/2) means the square root of 25. Since 5^2 = 25, the answer is 5. Choice B (12.5) comes from treating the exponent as division (25 divided by 2), which is an incorrect interpretation of rational exponents.

Odd-indexed roots of negative numbers are real. Since (-2)^5 = -32, the 5th root of -32 is -2. Choice D is incorrect because only even-indexed roots of negative numbers are undefined in the real numbers; odd-indexed roots are always defined for all real inputs.

The \(n\)th root of \(x\) is written as \(x^{1/n}\). So the 6th root of \(x\) equals \(x^{1/6}\). Choice B, \(x^6\), is \(x\) raised to the 6th power, which is the opposite operation of taking a root.

\(\sqrt{144} = 12\) because \(12 \times 12 = 144\). Choice B (\(72\)) is simply \(144\) divided by \(2\), which is not the same as taking the square root.

The cube root is an odd-indexed radical and is defined for all real numbers, including negative inputs. Unlike square roots, cube roots do not require non-negative inputs. Choice B describes the domain of \(f(x) = \sqrt{x}\), not the cube root function.

16^(3/4) = (16^(1/4))^3. The 4th root of 16 is 2 since 2^4 = 16, and then 2^3 = 8. Choice D (64) comes from computing 16^3 first and then dividing by 4, which is not how rational exponents are evaluated.

Factor the radicand: \(\sqrt{50x^4} = \sqrt{25 \cdot 2 \cdot x^4} = \sqrt{25} \cdot \sqrt{x^4} \cdot \sqrt{2} = 5 \cdot x^2 \cdot \sqrt{2}\). Choice B, \(5x\sqrt{2}\), incorrectly extracts only \(x\) instead of \(x^2\), since \(\sqrt{x^4} = x^2\), not \(x\).

Square both sides to get \(3x - 2 = 16\). Then \(3x = 18\) and \(x = 6\). Check: \(\sqrt{3(6) - 2} = \sqrt{16} = 4\), which is confirmed. Choice C (\(x = 7\)) comes from forgetting to subtract \(2\) from \(16\) before dividing by \(3\).

When multiplying powers with the same base, add the exponents: 1/2 + 1/3 = 3/6 + 2/6 = 5/6. So x^(1/2) * x^(1/3) = x^(5/6). Choice B (x^(1/6)) comes from subtracting 1/3 from 1/2 instead of adding.

Apply the exponent \(1/3\) to each factor inside: \((x^3)^{1/3} = x^{3/3} = x\), and \((y^6)^{1/3} = y^{6/3} = y^2\). Choice C makes the error of applying the \(1/3\) exponent to \(x\) itself rather than to \(x^3\), producing \(x^{1/3}\) instead of \(x\).

The radicand must be non-negative: \(3 - 2x \geq 0\). Solving gives \(-2x \geq -3\). Dividing by \(-2\) and flipping the inequality yields \(x \leq 3/2\). Choice B reverses the inequality direction, which is the most common error when dividing by a negative coefficient.

Square both sides: \(x + 3 = (x - 3)^2 = x^2 - 6x + 9\). Rearranging: \(x^2 - 7x + 6 = 0\), which factors as \((x - 6)(x - 1) = 0\). Checking \(x = 1\): \(\sqrt{4} = 2\), but \(1 - 3 = -2\), so \(x = 1\) is extraneous. Checking \(x = 6\): \(\sqrt{9} = 3 = 6 - 3\), which is confirmed. Choice C is a common error that skips checking for extraneous solutions.

32^(2/5) = (32^(1/5))^2. The 5th root of 32 is 2 since 2^5 = 32, and then 2^2 = 4. Choice B (8) results from mistakenly cubing instead of squaring: 2^3 = 8, which would correspond to an exponent of 3/5, not 2/5.

Factor \(54\) as \(27 \cdot 2\): cube root of \((27 \cdot 2 \cdot x^3)\) = cube root of \(27\) times cube root of \(x^3\) times cube root of \(2\) = \(3 \cdot x \cdot\) cube root of \(2\). Choice B uses cube root of \(6\) because \(54 = 9 \cdot 6\), but \(9\) is not a perfect cube, so that factoring does not extract any whole number from under the radical.

f(4) = 4^(3/2) = (4^(1/2))^3 = 2^3 = 8. Choice B (6) comes from treating the exponent as multiplication: 4 times 3/2 = 6. This is incorrect because the base is raised to the exponent, not multiplied by it.

A negative exponent indicates a reciprocal: \(x^{-1/3} = 1 / x^{1/3} = 1\) divided by the cube root of \(x\). Choice B is a common misconception. The negative sign in the exponent does NOT make the expression negative; it places the expression in the denominator as a reciprocal.

Square both sides: \(x + 6 = (\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1\). Simplifying gives \(5 = 2\sqrt{x}\), so \(\sqrt{x} = 5/2\) and \(x = 25/4\). Check: \(\sqrt{25/4 + 6} = \sqrt{49/4} = 7/2\), and \(\sqrt{25/4} + 1 = 5/2 + 1 = 7/2\), which is confirmed. Choice B (\(x = 7\)) results from incorrectly squaring the right side as if \((\sqrt{x} + 1)^2 = x + 1\), omitting the cross term.

Subtract exponents when dividing like bases: x^(2/3 - 1/3) = x^(1/3), and y^(1/2 - 3/2) = y^(-1) = 1/y. The result is x^(1/3) / y. Choice B ignores the negative exponent on y, incorrectly keeping y in the numerator instead of moving it to the denominator.

Isolate one radical: \(\sqrt{x - 3} = 4 - \sqrt{x + 5}\). Square both sides: \(x - 3 = 16 - 8\sqrt{x + 5} + (x + 5)\). Simplify to get \(8\sqrt{x + 5} = 24\), so \(\sqrt{x + 5} = 3\) and \(x = 4\). Check: \(\sqrt{1} + \sqrt{9} = 1 + 3 = 4\), confirmed. Choice B fails because \(\sqrt{4} + \sqrt{12}\) equals \(2 + 2\sqrt{3}\), which is not equal to \(4\).

f(-27) = (-27)^(2/3) = ((-27)^(1/3))^2 = (-3)^2 = 9. The cube root of -27 is -3, which is real, and squaring -3 yields a positive result of 9. Choice B is wrong because (-3)^2 = 9, not -9. Choice D is incorrect because the index 3 is odd, so cube roots of negative numbers are always real.

Isolate the radical: \(\sqrt{2x - 1} = 2\sqrt{x - 1} - 1\). Let \(u = \sqrt{x - 1}\); then \(2x - 1 = 2u^2 + 1\), so \(\sqrt{2u^2 + 1} = 2u - 1\). Squaring both sides: \(2u^2 + 1 = 4u^2 - 4u + 1\), giving \(2u^2 - 4u = 0\), so \(u = 0\) or \(u = 2\). When \(u = 0\), \(x = 1\): checking gives \(2(0) - \sqrt{1} = -1\), which is not \(1\), so \(x = 1\) is extraneous. When \(u = 2\), \(x = 5\): checking gives \(2(2) - \sqrt{9} = 4 - 3 = 1\), confirmed. Choice C ignores the extraneous solution.

Apply the square of a binomial pattern \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = \sqrt{x}\) and \(b = \sqrt{y}\): \((\sqrt{x})^2 + 2\sqrt{x}\sqrt{y} + (\sqrt{y})^2 = x + 2\sqrt{xy} + y\). Choice B is the most common error, omitting the cross term \(2\sqrt{xy}\) entirely. Choice D has the correct structure but divides the cross term by \(2\) instead of multiplying.

Because the cube root function is one-to-one, \(\sqrt[3]{a}\) equals \(\sqrt[3]{b}\) implies \(a = b\). Setting \(x^2 - 4 = 3x\) gives \(x^2 - 3x - 4 = 0\), which factors as \((x - 4)(x + 1) = 0\). Both \(x = 4\) and \(x = -1\) are valid because cube roots accept all real inputs, so no extraneous solutions arise. Choice D replaces \(-1\) with \(1\), but checking \(x = 1\) gives \(\sqrt[3]{-3}\) on the left and \(\sqrt[3]{3}\) on the right, which are not equal.

The cube root of -125 is -5 because (-5)^3 = -125. Odd-index roots of negative numbers are real and negative. Choice B ignores the negative sign. Unlike square roots, cube roots of negative numbers are defined in the real number system.

\(\sqrt{100x^2} = \sqrt{100} \cdot \sqrt{x^2} = 10 \cdot x = 10x\) for \(x \geq 0\). Choice B is wrong because \(\sqrt{100} = 10\), not \(50\). Choice C is wrong because \(\sqrt{x^2} = x\) (not \(x^2\)) when \(x \geq 0\). Choice D mistakenly keeps \(100\) inside instead of taking its square root.

Rewrite as (16^(1/4))^3. Since 2^4 = 16, the fourth root of 16 is 2. Then 2^3 = 8. Choice B = 4 results from computing 16^(1/2) instead of 16^(3/4). Choice D = 64 would equal 16^(3/2), which uses the wrong denominator in the exponent.

For a square root to be defined, the radicand must be non-negative: \(x + 9 \geq 0\), which gives \(x \geq -9\). Choice B incorrectly makes the bound positive. Choice C would only be correct if the function were \(\sqrt{x}\). Choice D applies to cube root (odd-index) functions, not square roots.

The nth root of \(x^m\) equals \(x^{m/n}\). Here \(n = 3\) and \(m = 2\), so the result is \(x^{2/3}\). Choice B = \(x^{3/2}\) swaps numerator and denominator, which is the square root of \(x^3\), not the cube root of \(x^2\). Choice C = \(x^{1/3}\) is the cube root of \(x\), missing the exponent on \(x\).

The fourth root of 81 equals 81^(1/4). Since 3^4 = 81, the answer is 3. Choice B = 9 is the square root of 81, not the fourth root. Choice C = 27 is 3^3, not 3^4. Choice D = 4 confuses the index of the radical with its value.

\(\sqrt{x} \cdot \sqrt{x} = (x^{1/2})^2 = x^1 = x\). Alternatively, \(\sqrt{x \cdot x} = \sqrt{x^2} = x\) for \(x \geq 0\). Choice B = \(x^2\) would require \((\sqrt{x})^4\). Choice C = \(2x\) incorrectly treats multiplication of two radicals as adding coefficients. Choice D = \(2\sqrt{x}\) confuses multiplication with doubling.

By the power of a power rule, multiply the exponents: \((x^4)^{1/2} = x^{4 \cdot 1/2} = x^2\). Choice B = \(x^{1/2}\) results from dividing exponents (\(4 / (1/2)\)) is not what the rule says — the rule is multiply, not divide. Choice C removes the outer exponent entirely. Choice D = \(x^8\) results from multiplying \(4\) by the denominator alone (\(4 \cdot 2 = 8\)) instead of by the fraction \(1/2\).

Factor out the largest perfect square: \(72 = 36 \cdot 2\), so \(\sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}\). Choice D = \(36\sqrt{2}\) results from pulling out the factor \(36\) without taking its square root. Choice C = \(4\sqrt{18}\) simplifies further to \(4 \cdot 3\sqrt{2} = 12\sqrt{2}\), which is not fully simplified and is not equal to \(6\sqrt{2}\).

Square both sides: \(3x + 4 = 49\). Subtract 4: \(3x = 45\). Divide by 3: \(x = 15\). Verify: \(\sqrt{3(15) + 4} = \sqrt{49} = 7\). Choice B = \(1\) results from not squaring the right side and solving \(3x + 4 = 7\) directly. Choice D = \(45\) is the intermediate value \(3x = 45\) before the final division by 3.

Apply the rational exponent to each factor. For 27: \(27^{2/3} = (27^{1/3})^2 = 3^2 = 9\). For \(x^6\): \((x^6)^{2/3} = x^{6 \cdot 2/3} = x^4\). Result: \(9x^4\). Choice B = \(3x^4\) stops after the cube root step without squaring (\(27^{1/3} = 3\) only). Choice C = \(9x^9\) uses the reciprocal exponent on x: \(x^{6 \cdot 3/2} = x^9\).

The domain is \(x \geq 2\), but the range describes the output values. A square root always produces non-negative outputs, so \(f(x) \geq 0\) for all valid x, giving range \(y \geq 0\). Choice B confuses the domain restriction \(x \geq 2\) with the range. Choice D applies to cube root functions, which can output all real numbers, but not to square root functions.

When dividing powers with the same base, subtract exponents: 3/4 - 1/4 = 2/4 = 1/2. Result: x^(1/2). Choice C = x results from adding exponents (3/4 + 1/4 = 1) instead of subtracting — a common confusion between multiplication and division rules. Choice D = x^(3/16) results from multiplying the exponents (3/4 * 1/4 = 3/16), which applies to a power of a power, not a quotient.

Write \(x^3 = x^2 \cdot x\), so \(\sqrt{x^3} = \sqrt{x^2} \cdot \sqrt{x} = x\sqrt{x}\). In rational exponent notation: \(x^{3/2} = x^1 \cdot x^{1/2} = x\sqrt{x}\). Choice B = \(x^2\sqrt{x} = x^{5/2}\), which would equal \(\sqrt{x^5}\). Choice C = \(x^2\) discards the remaining radical entirely. Choice D = \(x^{2/3}\) is the cube root of \(x^2\), not the square root of \(x^3\).

Cube both sides to undo the cube root: x = (-3)^3 = (-3)*(-3)*(-3) = -27. Check: the cube root of -27 = -3. Choice B = 27 results from ignoring the negative sign during cubing. Choice C = -9 confuses cubing with multiplying: (-3) * 3 = -9, but cubing means raising to the third power, not multiplying by 3. Choice D = -3 applies no operation at all.

Apply the fifth root to each factor using rational exponents: \((x^{10})^{1/5} \cdot (y^5)^{1/5} = x^{10/5} \cdot y^{5/5} = x^2 y\). Choice B uses \(x^{10/2} = x^5\), applying a square root to \(x\) instead of a fifth root. Choice D leaves \(y^5\) unchanged, forgetting to apply the fifth root to the \(y\) factor.

Square both sides: \(x^2 + 7 = 16\), so \(x^2 = 9\). Taking the square root of both sides gives \(x = \pm 3\). Both values are valid since \(x^2 + 7 \geq 0\) for all real x. Choice B misses the negative solution. Choice C = \(\pm 9\) results from finding \(x^2 = 9\) but then writing \(x = \pm 9\) instead of taking the square root of 9 to get \(\pm 3\).

Raise both sides to the \(3/2\) power: \(x = \pm 9^{3/2} = \pm(\sqrt{9})^3 = \pm 3^3 = \pm 27\). Both are valid: \((27)^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9\), and \((-27)^{2/3} = (\sqrt[3]{-27})^2 = (-3)^2 = 9\). Choice B misses the negative solution. Choice C = \(\pm 729\) comes from raising 9 to the 3rd power instead of the \(3/2\). Choice D = \(\pm 3\) stops at \(9^{1/2} = 3\) without completing the cube.

Use \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = x^{1/2}\) and \(b = x^{-1/2}\). Then \(a^2 = x\), the middle term \(2ab = 2 \cdot x^{1/2} \cdot x^{-1/2} = 2 \cdot x^{1/2 + (-1/2)} = 2 \cdot x^0 = 2\), and \(b^2 = x^{-1}\). Result: \(x - 2 + x^{-1}\). Choice B omits the middle term entirely. Choice C has a sign error on \(x^{-1}\), making it negative instead of positive. Choice D incorrectly squares the outer exponents (\(1/2\) becomes \(2\) and \(-1/2\) becomes \(-2\)) rather than squaring the terms.

Need \(x^2 - 9 \geq 0\), so \(x^2 \geq 9\), meaning \(|x| \geq 3\). This holds when \(x \leq -3\) or \(x \geq 3\). Choice B correctly identifies one branch but misses \(x \leq -3\) entirely. Choice C is exactly backward — those values make \(x^2 - 9\) negative, which is undefined under a square root. Choice D results from solving \(x^2 \geq 9\) as \(x \geq 9\) without correctly taking the square root of both sides.

Isolate one radical: \(\sqrt{x + 3} = 1 + \sqrt{x - 2}\). Square both sides: \(x + 3 = 1 + 2\sqrt{x - 2} + (x - 2)\), which simplifies to \(x + 3 = x - 1 + 2\sqrt{x - 2}\). Then \(4 = 2\sqrt{x - 2}\), so \(\sqrt{x - 2} = 2\), giving \(x - 2 = 4\) and \(x = 6\). Check: \(\sqrt{9} - \sqrt{4} = 3 - 2 = 1\). Choice C: \(\sqrt{10} - \sqrt{5}\) is approximately \(0.92\), not \(1\). Choice D: \(\sqrt{5} - \sqrt{0} = \sqrt{5}\) is not equal to \(1\).

Need \(x^2 - 5x + 6 \geq 0\). Factor: \((x - 2)(x - 3) \geq 0\). This product is non-negative when both factors are non-negative (giving \(x \geq 3\)) or both are non-positive (giving \(x \leq 2\)). The domain is \(x \leq 2\) or \(x \geq 3\). Choice B is the interval between the roots where the quadratic is negative, making the radicand negative. Choices C and D each capture only one of the two parts of the domain.

Subtract exponents for each base. For x: \(3/4 - (-1/4) = 3/4 + 1/4 = 1\). For y: \(-1/2 - 1/2 = -1\). Result: \(x^1 \cdot y^{-1} = x/y\). Choice B computes \(x^{3/4 - 1/4} = x^{1/2}\), forgetting to add the double negative for the x denominator exponent. Choice C ignores that the combined y exponent is \(-1\), incorrectly writing \(y^1\) in the numerator.

Solve \(a^{4/3} = 16\) by raising both sides to the \(3/4\) power: \(a = \pm 16^{3/4}\). Compute \(16^{3/4} = (16^{1/4})^3 = 2^3 = 8\). Both values work: \(8^{4/3} = (8^{1/3})^4 = 2^4 = 16\), and \((-8)^{4/3} = ((-8)^{1/3})^4 = (-2)^4 = 16\). Choice B misses the negative solution. Choice D = \(\pm 64\) comes from mistakenly computing \(16^{3/2} = (\sqrt{16})^3 = 4^3 = 64\), using exponent \(3/2\) instead of \(3/4\).

Convert to rational exponents and apply the product rule: \(x^{4/6} \cdot x^{2/6}\). Add the exponents: \(4/6 + 2/6 = 6/6 = 1\), giving \(x^1 = x\). Choice B \(= x^{1/3}\) results from subtracting exponents instead of adding: \(4/6 - 2/6 = 2/6 = 1/3\). Choice C \(= x^2\) comes from incorrectly halving the index (using 3 instead of 6), which gives \(x^{4/3} \cdot x^{2/3} = x^2\). Choice D \(= x^6\) comes from multiplying the radicands without applying the sixth root: \(x^4 \cdot x^2 = x^6\).

The principal square root of 49 is 7, because 7² = 49. The √ symbol always refers to the principal (non-negative) root, so √49 = 7, not ±7. The value ±7 is the solution set for x² = 49, which is a different question from evaluating √49.

By the definition of rational exponents, \(a^{1/n}\) equals the nth root of a. So \(x^{1/4}\) equals \(\sqrt[4]{x}\). The choices \(x/4\) and \(4x\) confuse the exponent \(1/4\) with division or multiplication, and \(x^4\) is a completely different expression.

The cube root of 27 asks: what number cubed equals 27? Since 3³ = 3 × 3 × 3 = 27, the answer is 3. The value 9 is incorrect because 9³ = 729, not 27. The choices √3 and 3√3 are irrational and cannot equal the exact cube root of 27.

The exponent 1/2 means square root, so 64^(1/2) = √64 = 8, because 8² = 64. A common mistake is dividing 64 by 2 to get 32, but raising to the 1/2 power means taking the square root, not dividing by 2.

Using the product property of radicals: √(36x²) = √36 · √(x²) = 6 · x = 6x. This simplification is valid because x ≥ 0 ensures √(x²) = x rather than |x|. The result is not 6x² because √(x²) = x, not x².

Odd-index roots, such as cube roots and fifth roots, are defined for all real numbers including negative values. For example, ⁵√(-32) = -2, since (-2)⁵ = -32. Only even-index roots require non-negative radicands. The restriction x ≥ 0 applies to square roots and fourth roots, not fifth roots.

When a power is raised to another power, multiply the exponents: (x^(1/2))^4 = x^(1/2 · 4) = x^(4/2) = x². Treating 4 as a coefficient to get 4x^(1/2) is wrong — the 4 is an exponent applied to the whole expression, not a multiplier. Adding exponents to get x^(9/2) is the rule for multiplying like bases, not raising a power to a power.

Factor 50 as 25 · 2, where 25 is a perfect square: √50 = √(25 · 2) = √25 · √2 = 5√2. The choice 25√2 is incorrect because (25√2)² = 1250 ≠ 50. The choice 10√5 is also wrong: (10√5)² = 500 ≠ 50.

Square both sides to eliminate the radical: x + 5 = 9. Subtract 5: x = 4. Check: √(4 + 5) = √9 = 3 ✓. The choice x = 14 results from adding 5 to 9 instead of subtracting it. The choice x = -2 comes from incorrectly subtracting 3 from 5 instead of solving the squared equation properly.

For a square root to produce a real number, the radicand must be ≥ 0. Set 2x - 6 ≥ 0 and solve: 2x ≥ 6, so x ≥ 3. The value x = 3 is included because √0 = 0 is defined. The choice x > 3 incorrectly excludes x = 3. The choice x ≥ 6 comes from only dividing one side of the inequality by 2.

Interpret the rational exponent as a root then a power: 16^(3/4) = (16^(1/4))³ = (⁴√16)³. First find the fourth root: ⁴√16 = 2, since 2⁴ = 16. Then cube: 2³ = 8. The choice 4 is just the fourth root without cubing. The choice 12 comes from multiplying 3 by 4 instead of applying the exponent correctly.

When raising a power to a power, multiply the exponents: (x^(2/3))^(3/2) = x^(2/3 · 3/2) = x^(6/6) = x¹ = x. The exponents 2/3 and 3/2 are reciprocals, so their product is 1. The choice x^(4/9) incorrectly multiplies only the numerators together and denominators together without recognizing that 2/3 · 3/2 simplifies to 1.

Simplify each radical separately: √12 = √(4 · 3) = 2√3, and √27 = √(9 · 3) = 3√3. Since both terms share the same radicand, add the coefficients: 2√3 + 3√3 = 5√3. The choice √39 incorrectly adds the radicands (12 + 27 = 39). The choice 6√3 incorrectly multiplies the coefficients 2 and 3 instead of adding them.

Cube both sides to eliminate the cube root: 2x - 1 = 3³ = 27. Add 1: 2x = 28. Divide by 2: x = 14. Check: the cube root of (28 - 1) = the cube root of 27 = 3 ✓. The choice x = 1 results from using 3 instead of 27 (forgetting to cube). The choice x = 28 stops one step early, before dividing by 2.

The expression √(x + 4) has a minimum value of 0 (when x = -4), so √(x + 4) ≥ 0 for all x in the domain. Adding 1 shifts every output up by 1: √(x + 4) + 1 ≥ 1. Therefore the range is y ≥ 1. The choice y ≥ -4 confuses the domain bound with the range. The choice y ≥ 5 incorrectly computes the range minimum as 4 + 1 = 5.

Use the product property of radicals: √(2x) · √(8x) = √(2x · 8x) = √(16x²). Then simplify: √16 · √(x²) = 4 · x = 4x (since x ≥ 0). The choice 16x is incorrect because √(16x²) ≠ 16x — only the square root of 16 is taken. The choice 4x² incorrectly keeps x² instead of taking its square root.

By the rule \(a^{m/n} = \sqrt[n]{a^m}\), the expression \(x^{2/5}\) equals \(\sqrt[5]{x^2}\). Equivalently, \(x^{2/5} = (x^{1/5})^2\), meaning the fifth root of x, then squared. The choice \(\sqrt{x^5}\) equals \(x^{5/2}\), not \(x^{2/5}\). The choice \(2x^{1/5}\) treats the numerator 2 as a coefficient instead of an exponent.

Interpret the rational exponent step by step: (-8)^(2/3) = ((-8)^(1/3))² = (cube root of -8)². The cube root of -8 is -2, since (-2)³ = -8. Then (-2)² = 4. The result is positive because squaring any real number yields a non-negative value. The choice 'not a real number' is incorrect — cube roots of negative numbers are real, unlike square roots of negative numbers.

Isolate the radical: √(3x + 4) = x - 2. For the square root to equal x - 2, we need x - 2 ≥ 0, so x ≥ 2. Square both sides: 3x + 4 = x² - 4x + 4, giving x² - 7x = 0, so x(x - 7) = 0. The candidates are x = 0 and x = 7. Since x = 0 fails x ≥ 2 (check: √4 + 2 = 4 ≠ 0), it is extraneous. Only x = 7 is valid: √25 + 2 = 7 ✓.

To find f(f(x)), substitute f(x) into f: f(f(x)) = f(x^(2/3)) = (x^(2/3))^(2/3) = x^(2/3 · 2/3) = x^(4/9). A common mistake is to add the exponents and get x^(4/3), but that would be f(x) · f(x) = [f(x)]², not f composed with f. Composition means applying the exponent rule for power of a power, which multiplies exponents.

Raise both sides to the power \(2/3\), the reciprocal of \(3/2\): \(x = 27^{2/3} = (27^{1/3})^2 = 3^2 = 9\). Verify: \(9^{3/2} = (9^{1/2})^3 = 3^3 = 27\) ✓. The choice \(x = 3\) is the cube root of \(27\), which solves \(x^3 = 27\), not \(x^{3/2} = 27\). The choice \(x = 81\) comes from incorrectly squaring \(27\) (\(27^2 = 729 \neq 27^{2/3}\)).

Set the radicand ≥ 0: x² - 4x + 3 ≥ 0. Factor: (x - 1)(x - 3) ≥ 0. This product is non-negative when both factors are non-negative (x ≥ 3) or both are non-positive (x ≤ 1). So the domain is x ≤ 1 or x ≥ 3. Choice A (1 ≤ x ≤ 3) is actually where the radicand is negative — for example, at x = 2: 4 - 8 + 3 = -1 < 0.

Multiply numerator and denominator by the conjugate (√5 + √2): [3(√5 + √2)] / [(√5 - √2)(√5 + √2)] = [3(√5 + √2)] / (5 - 2) = [3(√5 + √2)] / 3 = √5 + √2. Choice B uses 7 in the denominator, which results from incorrectly computing (√5)² + (√2)² = 7 instead of the difference of squares (√5)² - (√2)² = 3.

Square both sides: x + 1 = (x - 1)² = x² - 2x + 1. Rearranging: 0 = x² - 3x = x(x - 3), giving x = 0 or x = 3. Check x = 0: √1 = 0 - 1 → 1 = -1, which is false — x = 0 is extraneous. Check x = 3: √4 = 3 - 1 → 2 = 2 ✓. Only x = 3 is valid. Extraneous solutions arise because squaring both sides can introduce values that satisfy the squared equation but not the original.

Expand the numerator by distributing the outer exponent: (a^(1/2))^4 · (b^(3/4))^4 = a^(1/2 · 4) · b^(3/4 · 4) = a² · b³. Now divide by (a · b²) using the quotient rule for exponents: a² / a = a^(2-1) = a, and b³ / b² = b^(3-2) = b. The result is a · b = ab. The choice a²b omits dividing a², and ab³ omits dividing b³.

The principal square root of 144 is 12 because 12^2 = 144. The principal square root is always the non-negative root by definition, so -12 is excluded. 72 is simply 144/2, and 14.4 is a rough decimal estimate — neither equals the exact principal root.

The cube root of -125 is -5 because (-5)^3 = -125. Unlike square roots, cube roots (and all odd-index roots) are defined for negative numbers. The result is negative when the radicand is negative. -25 is incorrect because (-25)^3 = -15625, not -125.

The nth root of \(x^m\) equals \(x^{m/n}\), where the power becomes the numerator and the root index becomes the denominator. So \(\sqrt[5]{x^3}\) equals \(x^{3/5}\). Choosing \(x^{5/3}\) reverses the numerator and denominator, a common error.

A rational exponent of 1/2 means square root: 25^(1/2) = the square root of 25 = 5. The value 12.5 comes from dividing 25 by 2, which confuses division with taking a root. 625 = 25^2 and 50 = 25 times 2 — both use multiplication or squaring instead.

Factor out the largest perfect square: 72 = 36 times 2, so the square root of 72 = the square root of 36 times the square root of 2 = 6 times the square root of 2. Choosing 8 times the square root of 2 is wrong because 8^2 times 2 = 128, not 72. The correct perfect square factor is 36, not 4 or 9.

The index of a radical is the small number placed in the notch of the radical symbol indicating which root to take. For the fourth root of x, the index is 4. When no index is written on a radical, the index is assumed to be 2 (a square root). The index is not the same as the exponent 1/4, although they are related.

27^(1/3) means the cube root of 27. Since 3^3 = 27, the answer is 3. The value 9 is the square root of 81, not the cube root of 27. Raising to the 1/3 power is equivalent to finding the cube root, not dividing by 3.

The radicand of a square root must be non-negative: x minus 3 is greater than or equal to 0, which gives x is greater than or equal to 3. The smallest value in the domain is x = 3, at which point f(3) = the square root of 0 = 0. Choosing x = -3 would make the radicand -6, which is negative and not allowed.

Square both sides to eliminate the radical: 2x + 3 = 25. Subtract 3: 2x = 22. Divide by 2: x = 11. Verify: the square root of (22 + 3) = the square root of 25 = 5. The value 5.5 likely comes from dividing 11 by 2 prematurely. Always verify solutions after squaring, as extraneous solutions can appear.

Apply the exponent \(2/3\) to each factor separately. \(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\). For \(x^6\): \((x^6)^{2/3} = x^{6 \times 2/3} = x^4\). So the result is \(4x^4\). Choosing \(8x^4\) keeps \(8\) unchanged, forgetting to apply the rational exponent to the coefficient.

First simplify each radical: the square root of 18 = 3 times the square root of 2, and the square root of 8 = 2 times the square root of 2. Subtract like terms: 3 times the square root of 2 minus 2 times the square root of 2 = 1 times the square root of 2. A common error is subtracting the radicands directly to get the square root of 10, but radicals can only be combined when their radicands match.

Set the radicand greater than or equal to zero: 7 minus 2x is greater than or equal to 0. Subtract 7: negative 2x is greater than or equal to negative 7. Divide by negative 2 and flip the inequality: x is less than or equal to 7/2. Forgetting to reverse the inequality sign when dividing by a negative number leads to the incorrect answer x greater than or equal to 7/2.

Cube both sides: x minus 2 = (-3)^3 = -27. Add 2 to both sides: x = -25. Verify: the cube root of (-25 minus 2) = the cube root of (-27) = -3. Since cube roots are defined for negative numbers, this equation has a valid real solution. x = -1 comes from ignoring the cube and just solving x minus 2 = -3.

When multiplying powers with the same base, add the exponents: 1/3 plus 1/2. Find a common denominator: 2/6 plus 3/6 = 5/6. So x^(1/3) times x^(1/2) = x^(5/6). The choice x^(1/6) comes from multiplying the exponents (1/3 times 1/2) instead of adding them, which would apply only when raising a power to a power.

In the rational exponent \(x^{m/n}\), the denominator \(n\) is the root index and the numerator \(m\) is the power. So \(x^{5/3} = (x^{1/3})^5 = (\sqrt[3]{x})^5\). The choice \((\sqrt{x})^5\) equals \(x^{5/2}\), not \(x^{5/3}\). The fifth root of \(x^3\) equals \(x^{3/5}\), which swaps the numerator and denominator.

Multiply the coefficients and the radicands separately: (2 times 3) times the square root of (5 times 15) = 6 times the square root of 75. Then simplify: the square root of 75 = the square root of (25 times 3) = 5 times the square root of 3. So the final answer is 6 times 5 times the square root of 3 = 30 times the square root of 3. The choice 6 times the square root of 75 is partially correct but not fully simplified.

Rewrite using roots and powers: 32^(3/5) = (32^(1/5))^3 = (the fifth root of 32)^3. Since 2^5 = 32, the fifth root of 32 is 2. Then 2^3 = 8. The choice 4 corresponds to 32^(2/5) = 2^2, using the wrong numerator. The choice 16 corresponds to 2^4, adding one too many to the exponent.

Square both sides: x + 3 = 2x minus 5. Subtract x from both sides and add 5: 8 = x. Verify: the square root of (8 + 3) = the square root of 11, and the square root of (2 times 8 minus 5) = the square root of 11. Both sides match, so x = 8 is valid. x = -8 makes the radicands negative, so it is not in the domain. Always verify solutions after squaring.

Raise both sides to the 3/2 power: x = plus or minus 4^(3/2) = plus or minus (the square root of 4)^3 = plus or minus 2^3 = plus or minus 8. Both values are valid because the exponent 2/3 includes an even power: 8^(2/3) = (cube root of 8)^2 = 4 and (-8)^(2/3) = (cube root of -8)^2 = (-2)^2 = 4. Selecting only x = 8 misses the negative solution.

The parent function square root of x has range y greater than or equal to 0. The negative sign reflects it vertically, giving negative square root of x with range y less than or equal to 0. Shifting up by 4 gives range y less than or equal to 4. The maximum value is 4, achieved when x = -1 (making the square root equal to 0). As x increases, the output decreases without bound. Choosing y greater than or equal to 4 ignores the reflection.

Apply the outer exponent \(6\) to each factor using the power rule (multiply exponents): \((x^{1/2})^6 = x^{6/2} = x^3\), and \((y^{2/3})^6 = y^{12/3} = y^4\). The result is \(x^3 \times y^4\). Choosing \(x^3 \times y^3\) applies the exponent \(6\) incorrectly to \(y\), treating \(2/3\) as if the \(6\) cancels only the \(3\) but ignores the numerator \(2\).

Isolate one radical: the square root of (x + 5) = 5 minus the square root of x. Square both sides: x + 5 = 25 minus 10 times the square root of x plus x. Simplify: 5 = 25 minus 10 times the square root of x, so 10 times the square root of x = 20, giving the square root of x = 2, and x = 4. Verify: the square root of 9 plus the square root of 4 = 3 plus 2 = 5. The choice x = 0 gives the square root of 5 plus 0, which does not equal 5.

Raise both sides to the 3/4 power: x = plus or minus 16^(3/4) = plus or minus (the fourth root of 16)^3 = plus or minus 2^3 = plus or minus 8. Both are valid: 8^(4/3) = (cube root of 8)^4 = 2^4 = 16, and (-8)^(4/3) = (cube root of -8)^4 = (-2)^4 = 16. The even numerator in the final power makes both signs valid. x = 64 comes from incorrectly squaring 16 then taking a root.

Isolate one radical: the square root of (3x + 1) = 2 plus the square root of (x minus 1). Square both sides: 3x + 1 = 4 plus 4 times the square root of (x minus 1) plus x minus 1. Simplify: 2x minus 2 = 4 times the square root of (x minus 1), or x minus 1 = 2 times the square root of (x minus 1). Let u = the square root of (x minus 1): u^2 = 2u, so u(u minus 2) = 0, giving u = 0 or u = 2, meaning x = 1 or x = 5. Both solutions check out when substituted back.

To find the inverse, replace \(g(x)\) with \(y\) and swap \(x\) and \(y\): \(x = y^{3/2}\). Raise both sides to the \(2/3\) power (the reciprocal of \(3/2\)): \(x^{2/3} = y\). The inverse is \(g^{-1}(x) = x^{2/3}\). The exponent of the inverse is the reciprocal of the original exponent: \(3/2\) and \(2/3\) are reciprocals. Choosing \(x^{-3/2}\) gives the reciprocal of the function, not the inverse function.