Math 2 Quiz: Solving Quadratics By Factoring

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Solving Quadratics By FactoringQuestion 1 of 15

If $p$ and $q$ are the solutions to $x^2 - 7x + 12 = 0$ , what is the value of $\frac{1}{p} + \frac{1}{q}$ ?

\frac{1}{7}

\frac{12}{7}

\frac{1}{12}

\frac{7}{12}

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Math 2 Quiz

Math 2 Quiz: Solving Quadratics By Factoring

Practice Solving Quadratics By Factoring in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Quadratics By Factoring, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

If $p$ and $q$ are the solutions to $x^2 - 7x + 12 = 0$ , what is the value of $\frac{1}{p} + \frac{1}{q}$ ?

$\frac{1}{7}$
$\frac{12}{7}$
$\frac{1}{12}$
$\frac{7}{12}$ (correct answer)

Explanation: When you encounter a quadratic equation and need to find an expression involving the roots, think about Vieta's formulas—they provide elegant shortcuts without requiring you to solve for the individual roots. For the quadratic

x^2 - 7x + 12 = 0

, Vieta's formulas tell us that if

p

and

q

are the roots, then

p + q = 7

(the coefficient of

x

with opposite sign) and

pq = 12

(the constant term). To find

\frac{1}{p} + \frac{1}{q}

, we need a common denominator:

\frac{1}{p} + \frac{1}{q} = \frac{q + p}{pq}

. Substituting our values:

\frac{7}{12}

. Let's examine why the other answers are incorrect. Choice A gives

\frac{1}{7}

, which you'd get if you mistakenly used

\frac{1}{p + q}

instead of

\frac{p + q}{pq}

. Choice B gives

\frac{12}{7}

, which results from incorrectly flipping the fraction to

\frac{pq}{p + q}

. Choice C gives

\frac{1}{12}

, which you'd get if you used

\frac{1}{pq}

instead of the correct expression. The correct answer is D:

\frac{7}{12}

. Study tip: Memorize that for any quadratic

ax^2 + bx + c = 0

with roots

r

and

s

, we have

r + s = -\frac{b}{a}

and

rs = \frac{c}{a}

. This saves time and reduces computational errors compared to solving for individual roots.

Question 2

A manufacturer determines that the profit $P$ (in thousands of dollars) from producing $x$ thousand units is given by $P = -2x^2 + 16x - 30$ . For what production levels will the company break even (profit equals zero)?

$x = 5$ thousand units and $x = 6$ thousand units
$x = 2$ thousand units and $x = 6$ thousand units
$x = 3$ thousand units and $x = 10$ thousand units
$x = 3$ thousand units and $x = 5$ thousand units (correct answer)

Explanation: This problem tests your ability to find the zeros of a quadratic function, which represents break-even points in a business context. When a company "breaks even," its profit equals zero, so you need to solve

P = 0

. Set the profit equation equal to zero:

-2x^2 + 16x - 30 = 0

. You can solve this by factoring. First, factor out the common factor of -2:

-2(x^2 - 8x + 15) = 0

. This simplifies to

x^2 - 8x + 15 = 0

. To factor

x^2 - 8x + 15

, you need two numbers that multiply to 15 and add to -8. Those numbers are -3 and -5, giving you

(x - 3)(x - 5) = 0

. Therefore,

x = 3

x = 5

. The company breaks even at 3 thousand units and 5 thousand units. Choice A (

x = 5

and

x = 6

) includes one correct value but substitutes 6 for the actual solution of 3. Choice B (

x = 2

and

x = 6

) contains neither correct solution and likely results from factoring errors. Choice C (

x = 3

and

x = 10

) includes one correct value but 10 is far from any reasonable solution to this quadratic. When solving quadratic break-even problems, always verify your solutions by substituting back into the original equation. Both

x = 3

and

x = 5

should yield

P = 0

. This double-check catches algebraic mistakes and confirms you've found the true break-even points.

Question 3

The equation $4x^2 - 4x - 3 = 0$ can be solved by factoring. Which of the following shows the correct factorization?

$(4x + 3)(x - 1) = 0$
$(2x - 3)(2x + 1) = 0$ (correct answer)
$(4x - 3)(x + 1) = 0$
$(2x + 3)(2x - 1) = 0$

Explanation: To factor

4x^2 - 4x - 3

, we need two terms whose product is

4x^2 \cdot (-3) = -12x^2

and whose sum is

-4x

. These are

-6x

and

2x

. So

4x^2 - 4x - 3 = 4x^2 - 6x + 2x - 3 = 2x(2x - 3) + 1(2x - 3) = (2x - 3)(2x + 1)

. We can verify:

(2x - 3)(2x + 1) = 4x^2 + 2x - 6x - 3 = 4x^2 - 4x - 3

✓. Choice A gives

4x^2 - 4x + 3x - 3 = 4x^2 - x - 3

. Choice C gives

4x^2 + 4x - 3x - 3 = 4x^2 + x - 3

. Choice D gives

4x^2 - 2x + 6x - 3 = 4x^2 + 4x - 3

Question 4

For what value of $m$ does the equation $x^2 + mx + 16 = 0$ have exactly one solution?

$m = 4$ only
$m = 8$ only
$m = \pm 8$ (correct answer)
$m = \pm 4$

Explanation: A quadratic has exactly one solution when it's a perfect square trinomial. For

x^2 + mx + 16 = 0

to be a perfect square, we need

16 = (\frac{m}{2})^2

, so

\frac{m^2}{4} = 16

, giving

m^2 = 64

, thus

m = \pm 8

. When

m = 8

x^2 + 8x + 16 = (x + 4)^2 = 0

. When

m = -8

x^2 - 8x + 16 = (x - 4)^2 = 0

. Both give exactly one solution.

Question 5

The expression $4x^2 - 12x + k$ can be factored as $4(x - a)^2$ for some values of $k$ and $a$ . What is the value of $k + a$ ?

$k + a = \frac{15}{2}$
$k + a = \frac{21}{2}$ (correct answer)
$k + a = 12$
$k + a = 15$

Explanation: If 4x² - 12x + k = 4(x - a)², then expanding the right side: 4(x - a)² = 4(x² - 2ax + a²) = 4x² - 8ax + 4a². Comparing coefficients: -8a = -12, so a = 3/2; and 4a² = k, so k = 4(3/2)² = 4(9/4) = 9. Therefore k + a = 9 + 3/2 = 18/2 + 3/2 = 21/2. Choice A would result from miscalculating a or k. Choice C comes from assuming a = 3 (forgetting the factor of 2). Choice D comes from adding incorrectly or using wrong values.

Question 6

If $x^2 - 6x + 8 = 0$ and $x^2 - 6x + k = 5$ , what is the value of $k$ ?

$k = 3$
$k = 8$
$k = 13$ (correct answer)
$k = 18$

Explanation: From the first equation: x² - 6x + 8 = 0, so x² - 6x = -8. Substituting into the second equation: x² - 6x + k = 5 becomes -8 + k = 5, so k = 13. We can verify: the first equation factors as (x-2)(x-4) = 0, giving x = 2 or x = 4. For x = 2: 4 - 12 + k = 5, so k = 13. For x = 4: 16 - 24 + k = 5, so k = 13. Choice A would come from incorrectly computing 5 - (-8) = 5 + 8 = 13, then making an error. Choice B assumes k equals the constant term from the first equation. Choice D comes from computing 5 + 8 + 5 = 18, misunderstanding the relationship.

Question 7

The quadratic equation $3x^2 - kx + 12 = 0$ can be factored as $(3x + p)(x + q) = 0$ where $p$ and $q$ are integers. How many different integer values of $k$ make this possible?

4 different values of $k$
6 different values of $k$
8 different values of $k$ (correct answer)
12 different values of $k$

Explanation: Expanding (3x + p)(x + q) = 3x² + (3q + p)x + pq. Comparing with 3x² - kx + 12: 3q + p = -k and pq = 12. So k = -(3q + p). We need integer factor pairs (p,q) where pq = 12. The pairs are: (1,12), (2,6), (3,4), (4,3), (6,2), (12,1), (-1,-12), (-2,-6), (-3,-4), (-4,-3), (-6,-2), (-12,-1). For each pair, k = -(3q + p): (1,12): k = -(36+1) = -37; (2,6): k = -(18+2) = -20; (3,4): k = -(12+3) = -15; (4,3): k = -(9+4) = -13; (6,2): k = -(6+6) = -12; (12,1): k = -(3+12) = -15; (-1,-12): k = -(-36-1) = 37; (-2,-6): k = -(-18-2) = 20; (-3,-4): k = -(-12-3) = 15; (-4,-3): k = -(-9-4) = 13; (-6,-2): k = -(-6-6) = 12; (-12,-1): k = -(-3-12) = 15. The distinct values are: -37, -20, -15, -13, -12, 37, 20, 15, 13, 12. That's 10 values, but some repeat. Actually: -37, -20, -15, -13, -12, 12, 13, 15, 20, 37. Removing duplicates: 8 distinct values.

Question 8

A rectangular garden has dimensions where the length is 3 feet more than the width. If the area of the garden is 88 square feet, what are the possible dimensions of the garden?

Width = 8 feet, Length = 11 feet (correct answer)
Width = 11 feet, Length = 8 feet
Width = 8 feet, Length = 11 feet or Width = -11 feet, Length = -8 feet
Width = 11 feet, Length = 14 feet or Width = -8 feet, Length = -5 feet

Explanation: Let w = width, then length = w + 3. Area equation: w(w + 3) = 88, which gives w² + 3w - 88 = 0. Factoring: (w + 11)(w - 8) = 0, so w = -11 or w = 8. Since width must be positive, w = 8 feet and length = 11 feet. Choice B switches the dimensions incorrectly. Choice C includes negative dimensions which are not physically meaningful. Choice D uses wrong factorization.

Question 9

If $(2x - 3)(x + k) = 2x^2 + 5x - 12$ , what is the value of $k$ ?

$k = -4$
$k = 4$ (correct answer)
$k = \frac{8}{3}$
$k = -\frac{8}{3}$

Explanation: Expanding the left side:

(2x - 3)(x + k) = 2x^2 + 2kx - 3x - 3k = 2x^2 + (2k - 3)x - 3k

. Comparing with

2x^2 + 5x - 12

: the coefficient of

x

gives

2k - 3 = 5

, so

k = 4

. We can verify with the constant term:

-3k = -3(4) = -12

✓. Choice A gives coefficient

2(-4) - 3 = -11

. Choice C gives

2(\frac{8}{3}) - 3 = \frac{7}{3}

. Choice D gives

2(-\frac{8}{3}) - 3 = -\frac{25}{3}

Question 10

Consider the equation $(x-1)^2 + (x-1) - 6 = 0$ . If we let $u = x-1$ , then after solving for $u$ and substituting back, what are the solutions for $x$ ?

$x = 1$ and $x = 5$
$x = -1$ and $x = 4$
$x = 0$ and $x = 2$
$x = -2$ and $x = 3$ (correct answer)

Explanation: When you encounter an equation with repeated expressions like

(x-1)^2 + (x-1) - 6 = 0

, substitution is your best friend. This technique transforms a complex equation into a simpler quadratic form. Let's substitute

u = x-1

, which converts our equation to

u^2 + u - 6 = 0

. This is now a standard quadratic that we can factor. We need two numbers that multiply to

-6

and add to

1

. Those numbers are

3

and

-2

, so we get

(u+3)(u-2) = 0

. This gives us

u = -3

u = 2

. Now we substitute back: since

u = x-1

, we have

x-1 = -3

x-1 = 2

. Solving these gives us

x = -2

x = 3

. Let's check why the other answers are wrong. Choice A (

x = 1, 5

) would come from incorrectly solving

u = 2

x = 1+2 = 3

and making an error with the other root. Choice B (

x = -1, 4

) suggests errors in the back-substitution step, possibly confusing signs. Choice C (

x = 0, 2

) might result from factoring the quadratic incorrectly or making arithmetic mistakes in the substitution process. Strategy tip: Always verify your substitution method by checking one solution in the original equation. For example, when

x = -2

(-2-1)^2 + (-2-1) - 6 = 9 - 3 - 6 = 0

✓. This confirms you've handled the substitution correctly throughout the entire process.

Question 11

The solutions to $x^2 - 6x + 5 = 0$ are $x = a$ and $x = b$ where $a < b$ . What is the value of $a^2 + b^2$ ?

$26$ (correct answer)
$36$
$30$
$11$

Explanation: Factoring:

x^2 - 6x + 5 = (x - 1)(x - 5) = 0

, so

x = 1

x = 5

. Since

a < b

, we have

a = 1

and

b = 5

. Therefore

a^2 + b^2 = 1^2 + 5^2 = 1 + 25 = 26

. Choice B is

(a + b)^2 = 6^2 = 36

. Choice C is

5 \times 6 = 30

. Choice D is

a + b + ab = 1 + 5 + 5 = 11

Question 12

If $x^2 - 8x + k = 0$ has solutions that differ by 6, what is the value of $k$ ?

$k = -9$
$k = 7$ (correct answer)
$k = 16$
$k = -7$

Explanation: If the solutions are

r

and

s

with

r - s = 6

, then by Vieta's formulas:

r + s = 8

and

rs = k

. Solving the system:

r = 7, s = 1

. Therefore

k = rs = 7(1) = 7

. We can verify:

x^2 - 8x + 7 = (x-7)(x-1) = 0

, giving solutions

x = 7

and

x = 1

, which differ by 6.

Question 13

The quadratic equation $x^2 + kx + k = 0$ has two distinct real solutions. If both solutions are negative, what is the range of possible values for $k$ ?

$k > 0$ and $k \neq 4$
$0 < k < 4$ only
$k > 4$ or $k < 0$
$k > 4$ only (correct answer)

Explanation: When analyzing quadratic equations with constraints on their solutions, you need to consider three key conditions: the discriminant (for real solutions), the sum of roots, and the product of roots. For the equation

x^2 + kx + k = 0

to have two distinct real solutions, the discriminant must be positive:

k^2 - 4(1)(k) > 0

, which simplifies to

k^2 - 4k > 0

k(k-4) > 0

. This means

k < 0

k > 4

. For both solutions to be negative, you need two additional conditions. Using Vieta's formulas: the sum of roots is

-k

and the product of roots is

k

. If both roots are negative, their sum must be negative (so

-k < 0

, meaning

k > 0

) and their product must be positive (so

k > 0

). Combining all three constraints:

k > 4

(from the discriminant condition) AND

k > 0

(from both root conditions). The intersection is simply

k > 4

. Choice A (

k > 0

and

k \neq 4

) includes values like

k = 1

, which gives complex solutions, not two distinct real ones. Choice B (

0 < k < 4

) fails the discriminant test entirely. Choice C (

k > 4

k < 0

) includes negative values of

k

, which would make at least one root positive since the sum of roots would be positive. Remember: when quadratics have constraints on their roots, always check the discriminant first, then apply Vieta's formulas to test the sign conditions systematically.

Question 14

The equation $3x^2 - 12x - 15 = 0$ is equivalent to which of the following factored forms?

$3(x - 5)(x + 1) = 0$ (correct answer)
$3(x - 3)(x + 1) = 0$
$(3x - 15)(x + 1) = 0$
$3(x + 5)(x - 1) = 0$

Explanation: First factor out the GCF:

3x^2 - 12x - 15 = 3(x^2 - 4x - 5)

. Then factor the quadratic:

x^2 - 4x - 5 = (x - 5)(x + 1)

. So the complete factorization is

3(x - 5)(x + 1) = 0

. Choice B gives

3(x^2 - 2x - 3) = 3x^2 - 6x - 9

. Choice C gives

3x^2 - 12x - 15

but isn't fully factored. Choice D gives

3(x^2 + 4x - 5) = 3x^2 + 12x - 15

Question 15

Which equation has solutions $x = -\frac{2}{3}$ and $x = \frac{1}{4}$ ?

$12x^2 - 5x - 2 = 0$
$12x^2 - 5x + 2 = 0$
$12x^2 + 5x - 2 = 0$ (correct answer)
$12x^2 + 5x + 2 = 0$

Explanation: When you're given the roots of a quadratic equation and need to find the equation itself, you can work backwards using the relationship between roots and coefficients. If a quadratic has roots

r_1

and

r_2

, it can be written as

a(x - r_1)(x - r_2) = 0

. Let's substitute our given roots:

x = -\frac{2}{3}

and

x = \frac{1}{4}

. Starting with

a(x - (-\frac{2}{3}))(x - \frac{1}{4}) = 0

, we get

a(x + \frac{2}{3})(x - \frac{1}{4}) = 0

. Expanding:

(x + \frac{2}{3})(x - \frac{1}{4}) = x^2 - \frac{1}{4}x + \frac{2}{3}x - \frac{2}{12} = x^2 + \frac{5}{12}x - \frac{1}{6}

To match the answer choices (which have integer coefficients), multiply by 12:

12x^2 + 5x - 2 = 0

. This is choice C. Let's verify by checking why the other options fail: Choice A (

12x^2 - 5x - 2 = 0

) has the wrong sign on the middle term. Using the quadratic formula would give different roots. Choice B (

12x^2 - 5x + 2 = 0

) has both wrong signs for the middle and constant terms, leading to entirely different solutions. Choice D (

12x^2 + 5x + 2 = 0

) has the wrong sign on the constant term. The discriminant

25 - 96 = -71

is negative, meaning no real solutions exist. Study tip: When working backwards from roots to equations, remember that the sum of roots equals

-\frac{b}{a}

and the product equals

\frac{c}{a}

. This can help you quickly check your work or eliminate wrong answers.