Quadratic Equations — Free Algebra 1 Review Games.

Take square root of both sides: \(x = \pm 5\).

Factor: \(x(x+3) = 0\), so \(x = 0\) or \(x = -3\).

Standard form of a quadratic is ax^2 + bx + c = 0.

Set each factor to zero: x = 4 or x = -1.

A quadratic equation can have at most 2 real solutions.

Factor: \((x-2)(x-3) = 0\), \(x = 2\) or \(x = 3\).

Discriminant \(= b^2 - 4ac = 4 - 4(1)(-3) = 4 + 12 = 16\).

Discriminant = 0 means one repeated real solution.

\(x = \frac{4 \pm \sqrt{16-12}}{2} = \frac{4 \pm 2}{2}\), so \(x = 3\) or \(x = 1\).

\(x = \frac{-b}{2a} = \frac{6}{2} = 3\). \(y = 9-18+8 = -1\). Vertex: \((3, -1)\).

\(x^2+6x+9 = 16\), \((x+3)^2 = 16\), \(x+3 = \pm 4\), \(x = 1\) or \(-7\).

Factor: (2x+1)(x-2)=0, x=2 or x=-1/2.

Discriminant = 4 - 12k < 0, so k > 1/3.

By Vieta's formulas, sum of roots \(= \frac{-b}{a} = 7\).

\(x^2 = -1\) has no real solutions since squares are non-negative.

When a product equals zero, at least one factor must equal zero. Setting x + 5 = 0 gives x = -5, and setting x - 3 = 0 gives x = 3. A common mistake is to flip the signs: the zero of (x + 5) is x = -5, not x = 5.

The discriminant is b^2 - 4ac, the expression under the radical in the quadratic formula. It determines the nature of the solutions. Choice A (b^2 + 4ac) is incorrect because the term is subtracted, not added.

\(x^2 - 9\) is a difference of two perfect squares: \(x^2 - 3^2 = (x - 3)(x + 3)\). Choice A, \((x - 3)^2\), expands to \(x^2 - 6x + 9\), which includes a middle term and is not equal to \(x^2 - 9\).

Applying the Zero Product Property: x = 0 or x - 6 = 0, so x = 0 or x = 6. Choice D is wrong because x - 6 = 0 gives x = +6, not -6.

Standard form requires all terms on one side equal to zero with terms in descending degree. \(x^2 + 5x + 6 = 0\) satisfies this. Choices A and D have terms on both sides, and Choice B is not yet fully expanded and combined.

In the standard form \(ax^2 + bx + c = 0\), 'a' is the coefficient of the squared term \(x^2\), 'b' is the coefficient of \(x\), and 'c' is the constant. Confusing 'a' with 'b' leads to errors in the formula.

\(x^2 = 36\), so \(x = \pm \sqrt{36} = \pm 6\). There are two solutions because both 6 and -6, when squared, equal 36. Choice D is wrong; dividing 36 by 2 is not the correct operation.

Find two numbers that multiply to 6 and add to 5: those are 2 and 3. So \(x^2 + 5x + 6 = (x + 2)(x + 3) = 0\), giving \(x = -2\) or \(x = -3\). Choice A gives the magnitudes but forgets the signs; the zeros of \((x + 2)\) and \((x + 3)\) are negative.

Find two numbers that multiply to -12 and add to -1: those are -4 and 3. So \((x - 4)(x + 3) = 0\), giving \(x = 4\) or \(x = -3\). Choice B reverses the signs; -4 and 3 add to -1 correctly but the zeros of \((x - 4)\) and \((x + 3)\) are \(+4\) and \(-3\).

With \(a = 1\), \(b = 4\), \(c = -5\): discriminant \(= 16 + 20 = 36\), \(\sqrt{36} = 6\). Then \(x = \frac{-4 + 6}{2} = 1\) or \(x = \frac{-4 - 6}{2} = -5\). Choice B flips the signs of both answers, a common sign error.

With a = 2, b = -4, c = 2: discriminant = (-4)^2 - 4(2)(2) = 16 - 16 = 0. A discriminant of 0 means exactly one repeated real solution. Choice D (16) only computes b^2 without subtracting the 4ac term.

The rule is to add \((b/2)^2\) where \(b\) is the coefficient of \(x\). Here \(b = 8\), so add \((8/2)^2 = 4^2 = 16\). This gives \(x^2 + 8x + 16 = (x + 4)^2\). Choice C (64) incorrectly squares the full coefficient instead of half of it.

Factor by splitting the middle term: 2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1) = 0. Setting each factor to zero gives x = -3/2 or x = -1. Choice C incorrectly applies integer factoring without accounting for the leading coefficient 2.

When the discriminant \(b^2 - 4ac > 0\), the square root is defined and positive, producing two different values: \(\frac{-b + \sqrt{D}}{2a}\) and \(\frac{-b - \sqrt{D}}{2a}\). Choice B describes the case when the discriminant equals exactly 0, giving one repeated root.

By Vieta's formulas, for \(ax^2 + bx + c = 0\), the product of the roots equals \(\frac{c}{a}\). Here \(c = -10\) and \(a = 1\), so the product is \(\frac{-10}{1} = -10\). You can verify by factoring: \((x + 5)(x - 2) = 0\), and \(5 \times (-2) = -10\). Choice C gives the absolute value, ignoring the sign.

Move all terms to one side: \(x^2 - 6x = 0\), then factor: \(x(x - 6) = 0\). By the Zero Product Property, \(x = 0\) or \(x = 6\). A common error is to divide both sides by \(x\), which loses the solution \(x = 0\). Division by a variable is only valid when that variable is confirmed to be nonzero.

With \(a = 2\), \(b = -7\), \(c = 3\): discriminant \(= 49 - 24 = 25\), \(\sqrt{25} = 5\). Then \(x = \frac{7 + 5}{4} = 3\) or \(x = \frac{7 - 5}{4} = 1/2\). Choice C mixes the sign of one root; both solutions are positive here.

Expanding \((x - 3)^2\) gives \(x^2 - 6x + 9\). The constant needed is 9, which equals \((6/2)^2 = 3^2\). Choice C (36) would result from squaring the full coefficient 6 instead of half of it, a frequent completing-the-square error.

Move constant: \(x^2 - 8x = -7\). Add \((8/2)^2 = 16\) to both sides: \(x^2 - 8x + 16 = 9\). Factor: \((x - 4)^2 = 9\). Take square roots: \(x - 4 = \pm 3\), so \(x = 7\) or \(x = 1\). Choice D would result from incorrectly taking the square root of 7 rather than the completed square value of 9.

With \(a = 3\), \(b = -5\), \(c = 2\): discriminant \(= 25 - 24 = 1\), \(\sqrt{1} = 1\). Then \(x = \frac{5 + 1}{6} = 1\) or \(x = \frac{5 - 1}{6} = \frac{4}{6} = 2/3\). Choice C arises from factoring errors: \((3x - 1)(x - 2)\) gives roots \(1/3\) and \(2\), but the correct factoring is \((3x - 2)(x - 1)\).

For exactly one real solution, the discriminant must equal zero: \(k^2 - 4(1)(25) = 0\), so \(k^2 = 100\), giving \(k = \pm 10\). Both values work: \((x^2 + 10x + 25) = (x + 5)^2\) and \((x^2 - 10x + 25) = (x - 5)^2\). Choices A and B each give only one of the two valid answers.

Divide by 2: \(x^2 - 4x - 5 = 0\). Move constant: \(x^2 - 4x = 5\). Add \((4/2)^2 = 4\): \(x^2 - 4x + 4 = 9\). Factor: \((x - 2)^2 = 9\). Take square roots: \(x - 2 = \pm 3\), so \(x = 5\) or \(x = -1\). Choice C incorrectly takes \(\sqrt{5}\) instead of \(\sqrt{9}\) after completing the square.

By Vieta's formulas, the equation with roots \(r_1\) and \(r_2\) is \(x^2 - (r_1 + r_2)x + (r_1 r_2) = 0\). Substituting gives \(x^2 - 5x + 6 = 0\). Choice A uses \(+\) instead of \(-\) for the sum term, and Choice B has the wrong sign on the product term.

Recognize the perfect square trinomial: 4x^2 - 12x + 9 = (2x - 3)^2 = 0. This yields only one solution: 2x - 3 = 0, so x = 3/2. The discriminant is 144 - 144 = 0, confirming one repeated root. Choice B incorrectly includes a negative value, and Choice A incorrectly treats this as a difference of squares.

Substitute \(x = 3\): \(9 - 3k - 18 = 0\), so \(-3k = 9\), giving \(k = -3\). By Vieta's formulas, the product of the roots equals \(c/a = -18\), so the other root \(= -18/3 = -6\). Verify the sum: \(3 + (-6) = -3 = k\), which is consistent. Choice C gets \(k\) correct but assigns the wrong sign to the second root.

The standard form of a quadratic equation is ax^2 + bx + c = 0, where a is not equal to 0. Choice A is a linear equation (degree 1). Choice C is a cubic equation (degree 3). Choice D is missing the linear term bx and is not the general standard form, even though it can represent some quadratics.

By the Zero Product Property, if (x - 5)(x + 2) = 0, then either x - 5 = 0 or x + 2 = 0. Solving each factor gives x = 5 or x = -2. A common mistake is reversing the signs: the factor (x - 5) = 0 yields x = +5, not -5.

A perfect square trinomial has the form \(a^2 + 2ab + b^2 = (a + b)^2\). For \(x^2 + 6x + 9\), we have \(a = x\) and \(b = 3\), since \(2(x)(3) = 6x\) and \(3^2 = 9\), so it factors as \((x + 3)^2\). Choice B uses \(36\) instead of the required \(9\). Choice C has the wrong middle term (\(2 \times x \times 3 = 6x\), not \(3x\)). Choice D has a negative constant, which prevents it from being a perfect square.

The correct quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a). Choice B incorrectly adds 4ac inside the radical instead of subtracting it. Choice C omits the negative sign on b. Choice D divides only by a rather than 2a, which would give the wrong result.

When the discriminant b^2 - 4ac = 0, the quadratic formula gives x = -b / (2a), producing a single repeated (double) root — exactly one distinct real solution. A negative discriminant yields no real solutions, and a positive discriminant yields two distinct real solutions. A quadratic equation can never have three solutions.

\(x^2 - 9\) is a difference of squares: \(a^2 - b^2 = (a - b)(a + b)\). With \(a = x\) and \(b = 3\), the factored form is \((x - 3)(x + 3)\). Choice B is wrong because \((x - 9)(x + 1)\) expands to \(x^2 - 8x - 9\), not \(x^2 - 9\). Choices C and D are perfect squares, which expand to include a middle term.

Matching \(3x^2 - 2x + 7 = 0\) to \(ax^2 + bx + c = 0\): \(a = 3\) is the coefficient of \(x^2\), \(b = -2\) is the coefficient of \(x\) (the negative sign is part of \(b\)), and \(c = 7\) is the constant. Choice B incorrectly lists \(b = 2\) instead of \(-2\), which would change the discriminant calculation. Choice D incorrectly assigns \(c = -7\).

When factoring \(x^2 + bx + c\), the goal is to find two numbers that multiply to \(c\) and add to \(b\). For \(x^2 + 7x + 10\), those numbers are \(2\) and \(5\) (since \(2 \times 5 = 10\) and \(2 + 5 = 7\)), giving \((x + 2)(x + 5) = 0\). Choice C describes completing the square, a different method. Choice D is invalid because dividing by \(x\) loses the solution \(x = 0\) if it were a root.

Find two numbers that multiply to \(12\) and add to \(-7\): those are \(-3\) and \(-4\). So \(x^2 - 7x + 12 = (x - 3)(x - 4) = 0\), giving \(x = 3\) or \(x = 4\). Choice B (\(x = -3\) and \(x = -4\)) would correspond to \((x + 3)(x + 4) = x^2 + 7x + 12\), which has a positive middle term, not negative.

By Vieta's formulas, the sum of the roots of ax^2 + bx + c = 0 equals -b/a. Here a = 3 and b = -9, so the sum = -(-9)/3 = 3. A common error is computing c/a = 6/3 = 2, which gives the product of the roots, not the sum.

First simplify the radical: √28 = √(4 times 7) = 2√7. The expression becomes (4 ± 2√7) / 6. Factor 2 from the numerator: 2(2 ± √7) / 6 = (2 ± √7) / 3. Choice B fails to simplify √28. Choice C introduces an extra factor of 2 in the numerator. Choice D incorrectly applies order of operations by not dividing the entire numerator by 6.

The discriminant is \(b^2 - 4ac\). With \(a = 1, b = 5, c = 7\): discriminant \(= 5^2 - 4(1)(7) = 25 - 28 = -3\). Since the discriminant is negative, this equation has no real solutions. Choice A (\(53\)) results from adding \(4ac\) instead of subtracting. Choice D (\(25\)) calculates only \(b^2\) and stops before subtracting \(4ac\).

For roots \(r_1 = 2\) and \(r_2 = -5\): sum \(= 2 + (-5) = -3\) and product \(= (2)(-5) = -10\). Using Vieta's formulas, the equation is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - (-3)x + (-10) = x^2 + 3x - 10 = 0\). Choice A has the sign of the sum reversed. You can also verify directly: \((x - 2)(x + 5) = x^2 + 3x - 10\).

With \(a = 1, b = -4, c = -5\): discriminant \(= (-4)^2 - 4(1)(-5) = 16 + 20 = 36\). Then \(x = (4 \pm 6) / 2\), giving \(x = 10/2 = 5\) or \(x = -2/2 = -1\). This can be verified by factoring: \((x - 5)(x + 1) = x^2 - 4x - 5\). Choice B reverses the signs of both roots.

By Vieta's formulas, the product of the roots of ax^2 + bx + c = 0 equals c/a. Here a = 5 and c = 2, so the product = 2/5. Choice A (3/5) gives the sum of the roots, which equals -b/a = 3/5. Do not confuse the two Vieta's formulas: sum = -b/a and product = c/a.

Multiply a times c: (3)(-2) = -6. Find two numbers that multiply to -6 and add to 1: those are 3 and -2. Rewrite: 3x^2 + 3x - 2x - 2 = 3x(x + 1) - 2(x + 1) = (3x - 2)(x + 1) = 0. So 3x - 2 = 0 gives x = 2/3, and x + 1 = 0 gives x = -1. Check: 3(4/9) + (2/3) - 2 = 4/3 + 2/3 - 6/3 = 0.

A quadratic has no real solutions when its discriminant \(b^2 - 4ac\) is negative. For \(x^2 + 2x + 5\): discriminant \(= 4 - 20 = -16 < 0\). For choice A: \(16 - 12 = 4 > 0\), giving two real solutions. For choice C: \(36 - 36 = 0\), giving exactly one real solution. For choice D: \(0 - 4(1)(-1) = 4 > 0\), giving two real solutions (\(x = 1\) and \(x = -1\)).

Move the constant: \(x^2 + 6x = 7\). Take half of \(6\) to get \(3\), then add \(3^2 = 9\) to both sides: \(x^2 + 6x + 9 = 16\), so \((x + 3)^2 = 16\). Take the square root: \(x + 3 = \pm 4\). Then \(x = 1\) or \(x = -7\). Choice B reverses the signs of both answers. Choice C only considers the positive square root and gets the negative root wrong.

Two distinct real solutions require a strictly positive discriminant: \(b^2 - 4ac > 0\). Here \(a = 1, b = -6, c = k\): \((-6)^2 - 4(1)(k) > 0 \to 36 - 4k > 0 \to k < 9\). When \(k = 9\), the discriminant \(= 0\), producing exactly one repeated root. When \(k > 9\), the discriminant is negative, giving no real solutions. Choice D includes \(k = 9\), which gives only one root, not two distinct ones.

With a = 2, b = 3, c = -5: discriminant = 3^2 - 4(2)(-5) = 9 + 40 = 49. Then x = (-3 ± 7) / 4. So x = (-3 + 7)/4 = 4/4 = 1 or x = (-3 - 7)/4 = -10/4 = -5/2. Verify: 2(1)^2 + 3(1) - 5 = 0 and 2(25/4) + 3(-5/2) - 5 = 25/2 - 15/2 - 10/2 = 0. Choice B reverses the signs of both solutions.

A double root occurs when the discriminant equals \(0\): \(b^2 - 4ac = 0\). With \(a = 1\) and \(b = -12\): \((-12)^2 - 4(1)(c) = 0 \to 144 = 4c \to c = 36\). This can also be seen by completing the square: \(x^2 - 12x + 36 = (x - 6)^2 = 0\), giving \(x = 6\) as the only solution. Choice D (\(144\)) confuses \(b^2\) alone with the value of \(c\).

Sum of roots \(= (3 + \sqrt{5})/2 + (3 - \sqrt{5})/2 = 6/2 = 3\). Product of roots \(= (3 + \sqrt{5})/2 \times (3 - \sqrt{5})/2 = (9 - 5)/4 = 4/4 = 1\). Using Vieta's formulas, the equation is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - 3x + 1 = 0\). Choice B has the wrong sign on the middle term. Choice C uses product \(= 4\) instead of \(1\). Choice D uses product \(= -1\).

With \(a = 1, b = 4, c = 5\): discriminant \(= 16 - 20 = -4\). Then \(x = (-4 \pm \sqrt{-4}) / 2 = (-4 \pm 2i) / 2 = -2 \pm i\). So \(x = -2 + i\) or \(x = -2 - i\). Choice B has the wrong sign on the real part. Choice C has the imaginary part doubled — \(\sqrt{-4} = 2i\), and after dividing by \(2\) gives \(i\), not \(2i\). Choice D divides only the real part by \(2\) and not the imaginary part.

Sum of roots \(= (3 + 2i) + (3 - 2i) = 6\). Product of roots \(= (3 + 2i)(3 - 2i) = 9 - (2i)^2 = 9 - (-4) = 13\). Using Vieta's formulas: \(x^2 - 6x + 13 = 0\). Choice B uses \(-6x\) instead of \(+6x\) based on the wrong sign for the sum. Choice C uses \(-13\), which would require an imaginary product. Choice D uses product \(= 5\), which would come from real roots \((3 + 2)(3 - 2) = 5\), not complex roots.

Multiply a times c: (6)(-2) = -12. Find two numbers that multiply to -12 and add to -1: those are -4 and 3. Rewrite: 6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (2x + 1)(3x - 2) = 0. So 2x + 1 = 0 gives x = -1/2 and 3x - 2 = 0 gives x = 2/3. Verify: 6(4/9) - (2/3) - 2 = 8/3 - 2/3 - 6/3 = 0. Choice B reverses the signs of both roots.

Standard form is ax^2 + bx + c = 0, where a ≠ 0. Choice A is vertex form, choice C is factored form, and choice D is the quadratic formula — these are all related but distinct forms. Recognizing standard form is essential before applying solution methods.

The expression b^2 - 4ac is called the discriminant. It determines the nature of the roots: positive means two real solutions, zero means exactly one real solution, and negative means no real solutions. The axis of symmetry is x = -b/(2a), which is a different concept entirely.

\(x^2 - 16\) is a difference of squares: \(a^2 - b^2 = (a - b)(a + b)\). Here \(a = x\) and \(b = 4\), so the factored form is \((x - 4)(x + 4)\). Choice A gives \(x^2 - 8x + 16\) and choice B gives \(x^2 + 8x + 16\) — both are perfect square trinomials, not the original expression.

By the Zero Product Property, if (x - 5)(x + 3) = 0, then either x - 5 = 0 or x + 3 = 0. Solving each gives x = 5 or x = -3. A common error is reversing the signs: (x - 5) = 0 gives x = +5, not -5.

When the discriminant b^2 - 4ac = 0, the quadratic formula gives x = -b/(2a) with no ± variation, so there is exactly one real solution (called a repeated or double root). A negative discriminant gives no real solutions, and a positive discriminant gives two distinct real solutions.

\(x^2 + 4x + 4 = (x + 2)^2\) because the constant term equals \((\text{half the middle coefficient})^2\): \((4/2)^2 = 4\). Choice A factors as \((x+2)(x+4)\), not a perfect square. For choice D, \((5/2)^2 = 6.25\), not \(25\), so it is not a perfect square trinomial.

Taking the square root of both sides of \(x^2 = 25\) gives \(x = \pm 5\), so both \(x = 5\) and \(x = -5\) are solutions. A common mistake is forgetting the negative root. Choice D is incorrect because \(\sqrt{25} = 5\), not \(\sqrt{5}\).

To complete the square for \(x^2 + bx\), add \((b/2)^2\). Here \(b = 10\), so add \((10/2)^2 = 5^2 = 25\). This creates the perfect square trinomial \(x^2 + 10x + 25 = (x + 5)^2\). Choice A is just \(b/2\), which is a frequent arithmetic error.

Find two numbers that multiply to \(10\) and add to \(7\): those are \(2\) and \(5\). So \(x^2 + 7x + 10 = (x + 2)(x + 5) = 0\). Setting each factor to zero gives \(x = -2\) and \(x = -5\). Choice A reverses the signs — \((x - 2)(x - 5)\) would give \(x^2 - 7x + 10\), a different equation.

By Vieta's formulas, the sum of the roots of ax^2 + bx + c = 0 equals -b/a. Here a = 3 and b = -9, so the sum = -(-9)/3 = 9/3 = 3. Choice D (9) forgets to divide by a, and choice B (4/3) is the product of the roots, which equals c/a = 4/3.

Add \(36\) to both sides: \(4x^2 = 36\). Divide by \(4\): \(x^2 = 9\). Take the square root: \(x = \pm 3\). Choice D incorrectly takes the square root of \(36\) without first dividing by \(4\). This can also be solved by factoring: \(4(x^2 - 9) = 4(x-3)(x+3) = 0\).

With \(a = 1, b = 2, c = -8\): discriminant \(= 4 - 4(1)(-8) = 4 + 32 = 36\). So \(x = (-2 \pm 6)/2\). This gives \(x = 4/2 = 2\) and \(x = -8/2 = -4\). Choice B reverses the signs. This can be confirmed by factoring: \((x - 2)(x + 4) = 0\).

Move the constant: \(x^2 - 6x = -7\). Add \((-6/2)^2 = 9\) to both sides: \(x^2 - 6x + 9 = 2\). Factor: \((x - 3)^2 = 2\). Choice B incorrectly adds 9 to only the left side. Choice D has the wrong sign on the right side.

The discriminant is b^2 - 4ac with a = 3, b = -6, c = 3. So (-6)^2 - 4(3)(3) = 36 - 36 = 0. This means the equation has exactly one real (repeated) root. Choice C lists only b^2 without subtracting 4ac.

Find two numbers that multiply to -12 and add to -1: those are -4 and 3. So \(x^2 - x - 12 = (x - 4)(x + 3) = 0\), giving \(x = 4\) or \(x = -3\). Choice A reverses the roots — \((x+3)(x-4)\) still gives the same factored form, so the roots are \(x = 4\) and \(x = -3\), not 3 and -4.

Set h = 0: -16t^2 + 32t = 0. Factor: -16t(t - 2) = 0. So t = 0 (launch) or t = 2 (landing). The ball returns to the ground at t = 2 seconds. Choice A (t = 1) is when the ball reaches maximum height, not when it lands.

If the two roots are r1 and r2, their product is r1 * r2 = c/a. Here r1 = 3 and r1 * r2 = -6, so 3 * r2 = -6, giving r2 = -2. Choice C (-18) incorrectly multiplies 3 by -6 instead of dividing.

\(x^2 + 8x + 16 = (x + 4)^2 = 0\) gives the single repeated root \(x = -4\). Choice B, \(x^2 - 8x + 16 = (x - 4)^2 = 0\), gives \(x = 4\) (not -4). Choice D factors as \((x-4)(x+4) = 0\), giving two roots: \(x = 4\) and \(x = -4\).

Divide through by 2: \(x^2 - 4x - 5 = 0\). Move the constant: \(x^2 - 4x = 5\). Complete the square: add \((-4/2)^2 = 4\) to both sides: \((x - 2)^2 = 9\). Take square roots: \(x - 2 = \pm 3\), so \(x = 5\) or \(x = -1\). Choice C is wrong because forgetting to divide by 2 first leads to incorrect constants.

For exactly one real solution, the discriminant must equal zero: \(b^2 - 4ac = 0\). Here \(k^2 - 4(1)(16) = 0\), so \(k^2 = 64\), giving \(k = \pm 8\). Since the problem asks for the positive value, \(k = 8\). Choice A (\(k = 4\)) gives a discriminant of \(16 - 64 = -48\), which means no real solutions.

With a = 3, b = -5, c = -2: discriminant = 25 - 4(3)(-2) = 25 + 24 = 49. So x = (5 ± 7)/6. This gives x = 12/6 = 2 and x = -2/6 = -1/3. Choice B reverses the signs of both roots. You can verify by factoring: (3x + 1)(x - 2) = 3x^2 - 6x + x - 2 = 3x^2 - 5x - 2.

Move the constant: \(x^2 + 4x = -7\). Add \((4/2)^2 = 4\): \((x + 2)^2 = -7 + 4 = -3\). So \(k = -3\). Since a perfect square cannot equal a negative number in the reals, this equation has no real solutions. Choice A incorrectly changes the sign of \(k\). Choice C adds 4 to -7 incorrectly.

When a quadratic equation has rational coefficients, irrational roots must come in conjugate pairs. If one root is 3 - √7, the other must be its conjugate: 3 + √7. This ensures that the coefficients formed by (x - (3-√7))(x - (3+√7)) remain rational. Choice C would change the rational part from 3 to -3, which is not conjugate.

Use Vieta's formulas. Sum of roots = \((1+\sqrt{5})/2 + (1-\sqrt{5})/2 = 1\). Product of roots = \((1+\sqrt{5})/2 \cdot (1-\sqrt{5})/2 = (1-5)/4 = -1\). The equation is \(x^2 - (\text{sum})x + (\text{product}) = 0\), which gives \(x^2 - x - 1 = 0\). Choice A has the wrong sign on the middle term, which would give a root sum of -1.

Move the constant: \(x^2 - 8x = -10\). Add \((-8/2)^2 = 16\) to both sides: \(x^2 - 8x + 16 = 6\). Factor: \((x - 4)^2 = 6\). Take square roots: \(x - 4 = \pm\sqrt{6}\), so \(x = 4 \pm \sqrt{6}\). Choice B incorrectly uses the original constant (10) inside the radical. Choice C forgets to divide the coefficient by 2 when building the binomial.

Standard form is ax^2 + bx + c = 0, where a ≠ 0. Choice A is vertex form, Choice C is intercept (factored) form, and Choice D is the quadratic formula — a tool for solving, not a form of the equation itself.

To factor \(x^2 + 5x + 6\), find two numbers that multiply to 6 and add to 5. Those numbers are 2 and 3, giving \((x + 2)(x + 3)\). Check: \((x + 2)(x + 3) = x^2 + 5x + 6\). Choice A gives \(x^2 + 7x + 6\), which is incorrect.

By the Zero Product Property, either x - 5 = 0 or x + 3 = 0. Solving each gives x = 5 or x = -3. Choice A is wrong because x + 3 = 0 yields x = -3, not x = 3. Always set each factor equal to zero and solve.

The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a. Choice A adds 4ac inside the radical instead of subtracting it. Choice B omits the negative sign on b. Choice D divides by a instead of 2a, which gives an incorrect result.

\(x^2 - 16\) is a difference of two squares: \(a^2 - b^2 = (a - b)(a + b)\), so \(x^2 - 4^2 = (x - 4)(x + 4)\). Choice A gives \((x - 4)^2 = x^2 - 8x + 16\), which is not equal to \(x^2 - 16\). Choice D gives \(x^2 + 6x - 16\), also incorrect.

Factor as \((x - 7)(x + 7) = 0\) to get \(x = 7\) or \(x = -7\). Equivalently, \(x^2 = 49\) means \(x = \pm\sqrt{49} = \pm 7\). Choice A omits the negative solution. Choice D results from incorrectly taking \(x = \pm 49\) rather than \(\pm\sqrt{49}\).

In standard form \(ax^2 + bx + c = 0\), \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant. Here \(a = 2\), \(b = -3\) (the minus sign belongs to \(b\)), and \(c = 5\). Choice A incorrectly drops the negative sign, giving \(b = 3\).

Factor out \(x\): \(x(x - 8) = 0\). By the Zero Product Property, \(x = 0\) or \(x - 8 = 0\), giving \(x = 0\) or \(x = 8\). Choice A is incomplete since \(x = 8\) is also valid. Choice D incorrectly gives \(x = -8\) by misreading the sign of the factored term.

Find two numbers that multiply to -15 and add to 2: those are 5 and -3. So \(x^2 + 2x - 15 = (x + 5)(x - 3) = 0\), giving \(x = -5\) or \(x = 3\). Choice B reverses the signs — \((x - 5)(x + 3) = x^2 - 2x - 15\), which has a middle term of \(-2x\), not \(+2x\).

With a = 2, b = -7, c = 3, the discriminant is (-7)^2 - 4(2)(3) = 49 - 24 = 25. So x = (7 ± 5) / 4, giving x = 12/4 = 3 and x = 2/4 = 1/2. Choice C uses the correct formula but incorrectly takes one root as negative. Choice D uses b/2a and c/2a directly rather than applying the full formula.

The discriminant b^2 - 4ac determines the nature of solutions. When it is negative, the square root is not a real number, so there are no real solutions. Choice A describes the case when the discriminant equals zero (one repeated real root). Choice B describes the case when the discriminant is positive (two distinct real roots).

Find two numbers that multiply to 20 and add to -9: those are -4 and -5. So \((x - 4)(x - 5) = 0\), giving \(x = 4\) or \(x = 5\). Choice B has incorrect signs: \((x + 4)(x + 5) = x^2 + 9x + 20\), which has a positive middle term, not \(-9x\).

Multiply a times c: 2(-3) = -6. Find two numbers that multiply to -6 and add to 5: those are 6 and -1. Rewrite: 2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) = 0. So 2x - 1 = 0 gives x = 1/2, and x + 3 = 0 gives x = -3. Choice B reverses the signs of both solutions.

Move the constant: \(x^2 + 10x = -3\). Add \((10/2)^2 = 25\) to both sides: \(x^2 + 10x + 25 = 22\). This gives \((x + 5)^2 = 22\). Choice B adds 25 only to the left side, computing \(-3 + 31 = 28\) incorrectly. Choice D has a sign error, placing -22 on the right instead of +22.

Factor out the GCF: 3x(x + 4) = 0. By the Zero Product Property, 3x = 0 gives x = 0 and x + 4 = 0 gives x = -4. Choice A incorrectly gives x = 4 instead of x = -4. Choice C omits x = 0, which is always a solution when there is no constant term.

To complete the square for \(x^2 + bx\), add \((b/2)^2\). Here \(b = 6\), so add \((6/2)^2 = 9\). Then \(x^2 + 6x + 9 = (x + 3)^2\). Choice A uses \(b/2 = 3\) instead of \((b/2)^2 = 9\). Choice D uses \(b^2 = 36\) instead of \((b/2)^2\).

With \(a = 1, b = 4, c = -5\): discriminant \(= 16 - 4(1)(-5) = 36\). So \(x = (-4 \pm 6) / 2\), giving \(x = 2/2 = 1\) and \(x = -10/2 = -5\). This can also be verified by factoring: \((x - 1)(x + 5) = 0\). Choice B reverses the signs of both solutions.

First divide by 3: \(x^2 - 4x + 2 = 0\). Move the constant: \(x^2 - 4x = -2\). Add \((4/2)^2 = 4\) to both sides: \(x^2 - 4x + 4 = 2\), giving \((x - 2)^2 = 2\). Choice C correctly forms the left side but obtains a negative right side by subtracting 4 instead of adding it to both sides.

Sum of roots = -(-8)/2 = 4. Product of roots = 6/2 = 3. Using the identity r1^2 + r2^2 = (r1 + r2)^2 - 2(r1)(r2): sum of squares = 4^2 - 2(3) = 16 - 6 = 10. Choice A gives only the sum of the roots, not the sum of their squares. Choice D gives (sum)^2 = 16 without subtracting twice the product.

Divide by 2: \(x^2 + 6x - 7 = 0\). Move the constant: \(x^2 + 6x = 7\). Add \((6/2)^2 = 9\): \(x^2 + 6x + 9 = 16\), giving \((x + 3)^2 = 16\). Take square roots: \(x + 3 = \pm 4\), so \(x = 1\) or \(x = -7\). Choice B reverses the signs — that would result from \((x - 3)^2 = 16\) instead.

Multiply the factors: \((x - (2 + \sqrt{5}))(x - (2 - \sqrt{5})) = ((x - 2) - \sqrt{5})((x - 2) + \sqrt{5}) = (x - 2)^2 - (\sqrt{5})^2 = x^2 - 4x + 4 - 5 = x^2 - 4x - 1 = 0\). Choice A incorrectly computes \(4 - 5\) as \(+9\). Choice D has \(+1\) as the constant, which would require a product of roots equal to 1, but here the product is \((2 + \sqrt{5})(2 - \sqrt{5}) = 4 - 5 = -1\).

For two distinct real solutions, the discriminant must be strictly positive: \(b^2 - 4ac > 0\). With \(a = 1, b = -6, c = k\): \(36 - 4k > 0\), so \(k < 9\). Choice C (\(k = 9\)) gives discriminant = 0, meaning exactly one real solution. Choice D includes \(k = 9\), but that boundary value yields only one solution, not two distinct ones.

With a = 2, b = 3, c = -2: discriminant = 9 - 4(2)(-2) = 9 + 16 = 25. So x = (-3 ± 5) / 4, giving x = 2/4 = 1/2 and x = -8/4 = -2. This can be confirmed by factoring: (2x - 1)(x + 2) = 0. Choice B reverses the signs of both solutions.

Let w = width; then length = 2w + 3. Setting up the area equation: w(2w + 3) = 35, so 2w^2 + 3w - 35 = 0. Discriminant = 9 + 280 = 289 = 17^2. So w = (-3 + 17)/4 = 7/2. The other root w = -5 is rejected since width must be positive. Checking Choice B: if w = 5, length = 13 and area = 65, not 35.

For exactly one real solution, the discriminant must equal zero. For \(x^2 - 6x + 9 = 0\): discriminant \(= 36 - 36 = 0\), and \((x - 3)^2 = 0\) gives \(x = 3\). Choice B has discriminant \(16 - 20 = -4\), meaning no real solutions. Choice C has discriminant \(25 - 24 = 1 > 0\), giving two distinct solutions \(x = 2\) and \(x = 3\).

The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a). Choice B incorrectly adds 4ac inside the radical instead of subtracting it. Choice C omits the negative sign before b. Choice D divides only by a rather than 2a, which is a common denominator error.

A perfect square trinomial has the form \((x + k)^2 = x^2 + 2kx + k^2\). For the middle term \(6x\), we need \(2k = 6\) so \(k = 3\), and \(k^2 = 9\). Thus \(x^2 + 6x + 9 = (x + 3)^2\). Choice D is the most common error — students use \(6^2 = 36\) instead of \((6/2)^2 = 9\).

The Zero Product Property says at least one of the factors must equal zero — it does not require both to be zero. This is the foundation for solving factored quadratics: once (x - r)(x - s) = 0, set each factor to zero separately. Choice B is overly restrictive; only one factor needs to be zero.

Standard form of a quadratic is \(ax^2 + bx + c = 0\), requiring \(a \ne 0\) (if \(a = 0\), the \(x^2\) term vanishes and the equation becomes linear). Choice B is slope-intercept form of a line. Choice C is standard form of a linear equation. Choice D is an incomplete quadratic that is missing the \(bx\) term.

Apply the Zero Product Property: x - 3 = 0 gives x = 3, and x + 5 = 0 gives x = -5. Choice B is the most common error — students read the number directly from the factor without solving, treating -3 and +5 as the answers rather than solving x - 3 = 0 and x + 5 = 0.

The expression b^2 - 4ac is the discriminant. It determines the nature of the solutions: if positive, two distinct real roots; if zero, exactly one real root; if negative, no real roots. It is not related to the leading coefficient (which is a), the vertex of a parabola, or a conjugate pair.

The Zero Product Property requires a product equal to zero. If the equation is not set equal to zero first (e.g., \(x^2 - x = 6\) factored as \(x(x-1) = 6\)), you cannot simply set each factor to 6 — that reasoning is invalid. Rearranging to \(x^2 - x - 6 = 0\) and factoring \((x-3)(x+2) = 0\) is the correct approach.

Find two numbers that multiply to 12 and add to -7: those are -3 and -4. So \(x^2 - 7x + 12 = (x - 3)(x - 4) = 0\), yielding \(x = 3\) or \(x = 4\). Choice B results from writing the factors as \((x + 3)(x + 4)\) and then reading the solutions as -3 and -4 — sign errors in both the factoring and the solving steps.

\(x^2 - 16\) is a difference of squares: \((x - 4)(x + 4) = 0\), giving \(x = 4\) or \(x = -4\). Choice B omits the negative root. Choice C incorrectly divides 16 by 2 rather than taking its square root. Choice D sets \(x^2 = 16\) but then uses 16 as the answer instead of taking \(\sqrt{16} = 4\).

To complete the square, take half of the x-coefficient and square it: \((12/2)^2 = 6^2 = 36\). Adding 36 gives \(x^2 + 12x + 36 = (x + 6)^2\). Choice B is only the intermediate step (\(12/2 = 6\)) before squaring. Choice C uses \(12^2 = 144\), skipping the halving step entirely.

With a = 2, b = -4, c = 3: discriminant = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8. Because the discriminant is negative, this equation has no real solutions. Choice B reverses the subtraction: 24 - 16 = 8. Choice C incorrectly uses b^2 + 4ac.

Find two numbers that multiply to -12 and add to 1: those are 4 and -3. So \((x + 4)(x - 3) = 0\), giving \(x = -4\) or \(x = 3\). Choice B reverses both signs — a frequent error where students read the numbers directly from the factors (4 and -3) instead of solving \(x + 4 = 0\) and \(x - 3 = 0\).

With \(a = 1\), \(b = -4\), \(c = -5\): discriminant \(= 16 + 20 = 36\). Then \(x = (4 \pm 6) / 2\), giving \(x = 10/2 = 5\) or \(x = -2/2 = -1\). Choice B negates both correct answers — an error from using \(+b\) instead of \(-b\) in the numerator, turning 4 into -4 before applying \(\pm 6\).

Find two numbers that multiply to 6 and add to 5: those are 2 and 3. So \((x + 2)(x + 3) = 0\) is correct. Choice B gives \(x^2 - 5x + 6\), which has the wrong sign on the middle term. Choice C gives \(x^2 + 7x + 6\), where \(6 + 1 = 7\), not 5. Choice D gives \(x^2 - x - 6\), a different trinomial entirely.

The discriminant is \(b^2 - 4ac\) with \(a = 1\), \(b = -2\), \(c = -8\): \((-2)^2 - 4(1)(-8) = 4 + 32 = 36\). Note that subtracting a negative \(4ac\) term adds to the discriminant. Choice B only computes \(b^2 = 4\), ignoring the \(-4ac\) portion. Since \(\sqrt{36} = 6\), the full formula gives \(x = (2 \pm 6)/2 = 4\) or \(-2\).

Take half of the coefficient of x and square it: \((-8/2)^2 = (-4)^2 = 16\). This gives \(x^2 - 8x + 16 = (x - 4)^2\). Choice B is only the intermediate value (\(-8/2 = -4\)), not the squared result. Choice C computes \(8^2 = 64\), skipping the required halving step.

First factor out the GCF of 2: \(2(x^2 - 5x + 4) = 0\). Then factor the trinomial: \(2(x - 1)(x - 4) = 0\). Setting each variable factor to zero gives \(x = 1\) or \(x = 4\). Choice B results from incorrectly dividing only \(-10x\) and 8 by 2 and treating the resulting numbers as roots without properly factoring.

Since 25 > 0, the quadratic formula produces two distinct real values: x = (-b + √25)/(2a) and x = (-b - √25)/(2a). Exactly one solution occurs only when the discriminant equals zero (the ± yields the same value). No real solutions occur when the discriminant is negative.

Move the constant: \(x^2 - 6x = -7\). Add \((6/2)^2 = 9\) to both sides: \(x^2 - 6x + 9 = -7 + 9 = 2\). This gives \((x - 3)^2 = 2\), so \(k = 2\). Choice C is the number added during completing the square (9), not the resulting right-hand side. The solutions are \(x = 3 \pm \sqrt{2}\).

Sum of roots \(= (-3 + \sqrt{7}) + (-3 - \sqrt{7}) = -6\). Product of roots \(= (-3)^2 - (\sqrt{7})^2 = 9 - 7 = 2\). By Vieta's formulas, the equation is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - (-6)x + 2 = x^2 + 6x + 2 = 0\). Choice B has \(-6x\), which corresponds to a root sum of \(+6\), not \(-6\).

Move the constant: \(x^2 - 10x = -21\). Add \((10/2)^2 = 25\) to both sides: \(x^2 - 10x + 25 = 4\), giving \((x - 5)^2 = 4\). Taking the square root: \(x - 5 = \pm 2\), so \(x = 7\) or \(x = 3\). Choice C results from incorrectly writing \(x - 5 = \pm 4\) (treating 4 as the square root of itself rather than computing \(\sqrt{4} = 2\)). Choice D uses \(\sqrt{21}\) as if the constant 21 were never moved.

Exactly one real solution requires a discriminant of zero: \(b^2 - 4(1)(9) = 0\), so \(b^2 = 36\), giving \(b = \pm 6\). When \(b = 6\), the equation factors as \((x + 3)^2 = 0\); when \(b = -6\), it factors as \((x - 3)^2 = 0\). Choice B incorrectly uses \(\pm\sqrt{9} = \pm 3\) instead of solving \(b^2 = 36\).

Let the smaller integer be \(x\); then the next consecutive integer is \(x + 1\). Their product is \(x(x + 1) = 56\), which expands to \(x^2 + x - 56 = 0\). Factoring: \((x + 8)(x - 7) = 0\) gives \(x = 7\) as the positive solution (so the integers are 7 and 8, and \(7 \times 8 = 56\)). Choice B models \(x(x - 1) = 56\), which describes two consecutive integers where the second is one less.

With a = 3, b = 7, c = -6: discriminant = 49 + 72 = 121 = 11^2. Then x = (-7 ± 11) / 6. This gives x = 4/6 = 2/3 or x = -18/6 = -3. Choice B negates both roots — an error from using +b = +7 in the numerator instead of -b = -7.

Vieta's formulas state that for \(ax^2 + bx + c = 0\), the product of the roots \(= c/a\). Here \(c = -12\) and \(a = 1\), so the product \(= -12\). Verification: the roots are \(x = 6\) and \(x = -2\), and \(6 \times (-2) = -12\). Choice B describes the sum of the roots (\(-b/a = 4\)), not the product. Choice D is a partial truth — it only works when \(a = 1\); for \(2x^2 - x - 6 = 0\) the constant is \(-6\) but the product of roots is \(c/a = -3\).

The standard form of a quadratic equation is ax^2 + bx + c = 0, where a ≠ 0. Choice B is a linear equation. Choice C is vertex form, which is equivalent but not the standard form. Choice D is the quadratic formula, a tool for solving quadratics, not the equation form itself.

Matching \(3x^2 - 5x + 2 = 0\) to \(ax^2 + bx + c = 0\): \(a = 3\) (coefficient of \(x^2\)), \(b = -5\) (coefficient of \(x\), including the negative sign), and \(c = 2\) (constant term). A common mistake is dropping the negative sign on the middle term and writing \(b = 5\), which is incorrect.

\(x^2 - 16\) is a difference of squares: \(a^2 - b^2 = (a - b)(a + b)\), where \(a = x\) and \(b = 4\), giving \((x - 4)(x + 4)\). Choice B expands to \(x^2 - 6x - 16\), which includes a middle term. Choice C expands to \(x^2 - 8x + 16\), also including a middle term. Choice D expands to \(x^2 + 6x - 16\).

Factor out \(x\): \(x(x - 5) = 0\). By the zero product property, either \(x = 0\) or \(x - 5 = 0\), giving \(x = 0\) and \(x = 5\). A common error is dividing both sides by \(x\), which loses the solution \(x = 0\). Never divide both sides of an equation by a variable, as it eliminates solutions.

A quadratic equation has the form \(ax^2 + bx + c = 0\) where \(a \neq 0\), meaning the highest power of \(x\) is exactly 2. Choice B is linear (degree 1). Choice C is cubic (degree 3). Choice D is a rational equation, not a polynomial. Only Choice A has \(x^2\) as the highest degree term.

The expression b^2 - 4ac is called the discriminant. It determines the nature of the solutions: positive means two distinct real solutions, zero means exactly one real solution, and negative means no real solutions. The vertex and axis of symmetry are features of the parabola's graph, not components of the quadratic formula.

Taking the square root of both sides gives \(x = \pm\sqrt{25} = \pm 5\), so \(x = 5\) and \(x = -5\). Both positive and negative roots must be included. Choice B misses the negative root. Choice C incorrectly sets \(x = \pm 25\) instead of \(\pm\sqrt{25}\). Choice D takes the square root of 5 rather than 25.

To complete the square, take half the coefficient of \(x\) and square it: \((12/2)^2 = 6^2 = 36\). So \(c = 36\), making \(x^2 + 12x + 36 = (x + 6)^2\). Choice B (6) is half of 12 without squaring. Choice C (144) squares 12 without halving first. Choice D (24) doubles the coefficient rather than halving and squaring.

Find two numbers that multiply to 20 and add to -9: those are -4 and -5. So \(x^2 - 9x + 20 = (x - 4)(x - 5) = 0\). Setting each factor to zero: \(x = 4\) and \(x = 5\). Choice B uses the wrong signs; \((x + 4)(x + 5) = x^2 + 9x + 20\), which has a positive middle term. Always check that the factors produce the correct middle term.

Multiply a and c: 2 × (-3) = -6. Find two numbers that multiply to -6 and add to 5: 6 and -1. Rewrite the middle term: 2x^2 + 6x - x - 3. Factor by grouping: 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) = 0. So x = 1/2 and x = -3. Choice B uses opposite signs for both solutions.

When b^2 - 4ac < 0, the square root of a negative number is not real, so the equation has no real solutions. Exactly one real solution (a repeated root) occurs when the discriminant equals zero. Two distinct real solutions occur when the discriminant is positive. Choices B and C both describe the case when the discriminant equals zero.

To complete the square, take half the coefficient of \(x\) and square it: \((10/2)^2 = 5^2 = 25\). Adding 25 gives \(x^2 + 10x + 25 = (x + 5)^2\). Choice B (10) is the original coefficient unchanged. Choice C (5) is half of 10 but not squared. Choice D (100) is \(10^2\), squaring the full coefficient without halving first.

With \(a = 1\), \(b = 4\), \(c = -5\): discriminant \(= 4^2 - 4(1)(-5) = 16 + 20 = 36\). Then \(x = (-4 \pm \sqrt{36}) / 2 = (-4 \pm 6) / 2\). This gives \(x = (-4 + 6)/2 = 1\) and \(x = (-4 - 6)/2 = -5\). Choice B reverses both signs. This can be verified by factoring: \((x + 5)(x - 1) = 0\) confirms \(x = 1\) and \(x = -5\).

Isolate \(x^2\): \(2x^2 = 8\), so \(x^2 = 4\). Taking the square root of both sides: \(x = \pm 2\). Choice B results from skipping the division step: treating \(x^2 = 8\) gives \(x = \pm 2\sqrt{2}\), not \(\pm 4\). Choice C misses the negative root. Choice D results from not dividing by 2 before taking the square root: \(\sqrt{8} = 2\sqrt{2}\).

With \(a = 1\), \(b = -6\), \(c = 9\): discriminant \(= b^2 - 4ac = (-6)^2 - 4(1)(9) = 36 - 36 = 0\). A discriminant of \(0\) means the equation has exactly one repeated real solution (\(x = 3\)). Choice B (\(27\)) results from computing \(b^2 - ac = 36 - 9\) instead of \(b^2 - 4ac\). Choice C (\(72\)) results from adding \(36 + 36\) instead of subtracting.

Move the constant: \(x^2 + 6x = -5\). Add \((6/2)^2 = 9\) to both sides: \(x^2 + 6x + 9 = -5 + 9 = 4\). Factor the left side: \((x + 3)^2 = 4\). Choice B adds \(9\) only to the left side, computing the right side as \(-5 + 9 = 14\) incorrectly. Choice C uses \(6\) instead of \(3\) in the factored form. Choice D forgets to add \(9\) to the right side.

4x^2 - 25 is a difference of squares: (2x)^2 - 5^2 = (2x - 5)(2x + 5). Verify: (2x - 5)(2x + 5) = 4x^2 + 10x - 10x - 25 = 4x^2 - 25. Choice B expands to 4x^2 + 20x - 5x - 25 = 4x^2 + 15x - 25, which has a nonzero middle term. Choice C is a perfect square trinomial, not a difference of squares.

Move the constant: \(x^2 + 8x = 3\). Add \((8/2)^2 = 16\) to both sides: \(x^2 + 8x + 16 = 3 + 16 = 19\). Factor: \((x + 4)^2 = 19\). Take square roots: \(x + 4 = \pm\sqrt{19}\), so \(x = -4 \pm \sqrt{19}\). Choice B has the wrong sign inside the parentheses. Choice C results from subtracting \(3\) from \(16\) (\(16 - 3 = 13\)) instead of adding. Choice D adds \(16\) only to the left side: \((x + 4)^2 = 16 \to x = 0\) or \(x = -8\).

Let width \(= x\), then length \(= x + 3\). Area equation: \(x(x + 3) = 40\), so \(x^2 + 3x - 40 = 0\). Factor: \((x + 8)(x - 5) = 0\), giving \(x = 5\) or \(x = -8\). Since width must be positive, \(x = 5\) feet. Choice B (8 feet) comes from the other factor but gives a negative solution for width. Always reject negative values in geometric context.

For exactly one real solution, the discriminant must equal zero: \(b^2 - 4ac = 0\). With \(a = 1\), \(b = k\), \(c = 16\): \(k^2 - 4(1)(16) = 0\), so \(k^2 = 64\) and \(k = \pm 8\). Both values work: \(k = 8\) gives \((x + 4)^2 = 0\), and \(k = -8\) gives \((x - 4)^2 = 0\). Choice B gives discriminant \(= 16 - 64 = -48 \neq 0\). Choice D misses the negative solution \(k = -8\).

With a = 2, b = -4, c = -1: discriminant = (-4)^2 - 4(2)(-1) = 16 + 8 = 24. Then x = (4 ± √24) / (2 · 2) = (4 ± 2√6) / 4 = (2 ± √6) / 2. Choice B uses 2a = 2 instead of 4 in the denominator. Choice C uses -b = -4 instead of +4 in the numerator. Choice D computes the discriminant as 16 - 8 = 8 by treating c as positive 1.

Find two numbers that multiply to \(-10\) and add to \(-3\): those are \(-5\) and \(2\). So the correct factoring is \((x - 5)(x + 2) = 0\), giving \(x = 5\) and \(x = -2\). The student correctly found \(x = 5\) but wrote \(x = -3\) instead of \(x = -2\). Verify: \((-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 \neq 0\), confirming \(x = -3\) is not a solution.

Let \(x\) be the number. "Sum of a number and its square equals 42" means \(x + x^2 = 42\), which rearranges to \(x^2 + x - 42 = 0\). Factor: \((x + 7)(x - 6) = 0\), giving \(x = 6\) or \(x = -7\). The positive answer is \(x = 6\). Choice B uses the correct equation but the wrong solution — \((x - 7)(x + 6)\) expands to \(x^2 - x - 42\), not \(x^2 + x - 42\). Choice C sets up the wrong equation entirely.

Check the discriminant: \(b^2 - 4ac = 6^2 - 4(1)(10) = 36 - 40 = -4\). Since the discriminant is negative, the equation has no real solutions whatsoever. Choice C is incorrect because completing the square on a negative-discriminant equation also yields no real solutions. Choice D is wrong because substituting \(x = -3\) gives \(9 - 18 + 10 = 1 \neq 0\).

First divide every term by \(2\): \(x^2 + 4x - 5 = 0\). Move the constant: \(x^2 + 4x = 5\). Add \((4/2)^2 = 4\) to both sides: \(x^2 + 4x + 4 = 9\). Factor: \((x + 2)^2 = 9\). Take square roots: \(x + 2 = \pm 3\), giving \(x = 1\) or \(x = -5\). Choice B reverses both signs. Choice C results from adding \(2\) to \(3\) instead of subtracting: \(x + 2 = 3 \to x = 3 - 2\), but a sign error gives \(x = 3 - 1 = 2\).