Math 2 Quiz: Quadratic Inequalities

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Quadratic InequalitiesQuestion 1 of 13

Consider the quadratic inequality $2x^2 + kx - 3 > 0$ . If this inequality has no solution, what is the range of possible values for $k$ ?

k ∈ ℝ

(all real numbers)

-2\sqrt{6} < k < 2\sqrt{6}

k < -2\sqrt{6}

k > 2\sqrt{6}

No such values of

k

exist

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Math 2 Quiz

Math 2 Quiz: Quadratic Inequalities

Practice Quadratic Inequalities 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 Quadratic Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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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

Consider the quadratic inequality $2x^2 + kx - 3 > 0$ . If this inequality has no solution, what is the range of possible values for $k$ ?

$k ∈ ℝ$ (all real numbers)
$-2\sqrt{6} < k < 2\sqrt{6}$
$k < -2\sqrt{6}$ or $k > 2\sqrt{6}$
No such values of $k$ exist (correct answer)

Explanation: For the quadratic

2x^2 + kx - 3

with

a = 2 > 0

, the parabola opens upward. For the inequality

2x^2 + kx - 3 > 0

to have no solution, the parabola would need to be entirely below the x-axis, meaning

2x^2 + kx - 3 ≤ 0

for all

x

. However, since

a = 2 > 0

, the parabola opens upward and goes to

+∞

x → ±∞

. Therefore, there will always be some values of

x

for which

2x^2 + kx - 3 > 0

, regardless of the value of

k

. No such values of

k

exist. Choice B gives the range where the quadratic has no real roots. Choice C gives the complement of that range. Choice A incorrectly suggests all values work.

Question 2

The inequality $x^2 - kx + k ≥ 0$ is satisfied by all real numbers. Which of the following describes the possible values of $k$ ?

$k ≤ 0$ or $k ≥ 4$
$0 ≤ k ≤ 4$ (correct answer)
$k < 0$ or $k > 4$
$k = 0$ or $k = 4$

Explanation: For the quadratic

x^2 - kx + k

to be non-negative for all real numbers, its discriminant must be non-positive (so it has at most one real root and opens upward). The discriminant is

(-k)^2 - 4(1)(k) = k^2 - 4k = k(k-4)

. For

k(k-4) ≤ 0

, we need

k

and

(k-4)

to have opposite signs or one to be zero. This occurs when

0 ≤ k ≤ 4

. Choice A gives the complement condition. Choice C excludes the boundary cases. Choice D only includes the boundary points where the discriminant equals zero.

Question 3

Consider the system of inequalities: $x^2 - 5x + 6 < 0$ and $x^2 - x - 12 \geq 0$ . What is the solution set for this system?

$(-\infty, -3] \cup [4, \infty)$
$(2, 3)$ only
$\emptyset$ (no solution) (correct answer)
$(-\infty, -3] \cup (2, 3)$

Explanation: For the first inequality:

x^2 - 5x + 6 = (x - 2)(x - 3) < 0

. This quadratic is negative between its roots, so the solution is

2 < x < 3

. For the second inequality:

x^2 - x - 12 = (x - 4)(x + 3) \geq 0

. This quadratic is non-negative outside its roots, so the solution is

x \leq -3

x \geq 4

. The system requires both conditions to hold simultaneously:

x \in (2, 3)

AND

x \in (-\infty, -3] \cup [4, \infty)

. Since the interval

(2, 3)

does not intersect with

(-\infty, -3] \cup [4, \infty)

, there is no solution.

Question 4

Given that $f(x) = x^2 + px + q$ where $p$ and $q$ are constants, and the solution to $f(x) \leq 0$ is $[-3, 2]$ , what is the solution to $f(x + 1) > 0$ ?

$(-\infty, -4) \cup (1, \infty)$ (correct answer)
$(-\infty, -2) \cup (3, \infty)$
$(-4, 1)$ with endpoints excluded
$(-2, 3)$ with endpoints excluded

Explanation: Since

f(x) \leq 0

has solution

[-3, 2]

, the quadratic

f(x)

has zeros at

x = -3

and

x = 2

, and opens upward. So

f(x) = a(x + 3)(x - 2)

where

a > 0

. Thus

f(x) > 0

when

x < -3

x > 2

. For

f(x + 1) > 0

, we substitute

u = x + 1

, giving

f(u) > 0

when

u < -3

u > 2

. Converting back to

x

x + 1 < -3

gives

x < -4

, and

x + 1 > 2

gives

x > 1

. Therefore,

f(x + 1) > 0

when

x \in (-\infty, -4) \cup (1, \infty)

. The horizontal shift left by 1 unit moves the solution intervals accordingly.

Question 5

If $p(x) = x^2 + mx + n$ and the solution set to $p(x) < 0$ is $(-4, 1)$ , what is the solution set to $p(x - 3) \geq 0$ ?

$(-\infty, -1] \cup [4, \infty)$
$(-\infty, 1] \cup [7, \infty)$ (correct answer)
$(-7, -2)$ with endpoints included
$(-1, 4)$ with endpoints excluded

Explanation: Since

p(x) < 0

(-4, 1)

, the quadratic

p(x)

has zeros at

x = -4

and

x = 1

, and opens upward (positive leading coefficient). Thus

p(x) \geq 0

when

x \leq -4

x \geq 1

. For

p(x - 3) \geq 0

, we substitute

u = x - 3

, giving

p(u) \geq 0

when

u \leq -4

u \geq 1

. Converting back:

x - 3 \leq -4

gives

x \leq -1

, and

x - 3 \geq 1

gives

x \geq 4

. The horizontal shift moves the solution set 3 units right.

Question 6

The quadratic inequality $ax^2 + bx + c \leq 0$ has solution set $[2, 5]$ . If the same quadratic is written as $a(x - h)^2 + k \leq 0$ , what are the values of $h$ and $k$ ?

$h = 3.5, k = -2.25a$ where $a > 0$
$h = 3.5, k = 2.25a$ where $a < 0$
$h = 7, k = -6a$ where $a > 0$
$h = 3.5, k = -2.25a$ where $a < 0$ (correct answer)

Explanation: Since the solution set is the closed interval

[2, 5]

, the quadratic must be negative or zero on this finite interval, which means the parabola opens downward, so

a < 0

. The vertex (where the maximum occurs) is at

h = \frac{2 + 5}{2} = 3.5

. Since the quadratic has roots at

x = 2

and

x = 5

, we can write it as

a(x - 2)(x - 5)

. At the vertex

x = 3.5

k = a(3.5 - 2)(3.5 - 5) = a(1.5)(-1.5) = -2.25a

. Therefore

h = 3.5

and

k = -2.25a

where

a < 0

Question 7

A student creates a sign chart for $\frac{(x + 2)^2(x - 1)}{x - 3} \leq 0$ and concludes the solution is $x \in (-\infty, 1] \cup (3, \infty)$ . What error did the student most likely make?

Failed to exclude $x = 3$ from the domain completely
Incorrectly treated $(x + 2)^2$ as changing sign at $x = -2$
Forgot to include $x = -2$ in the final solution set (correct answer)
Misidentified the intervals where the expression is negative

Explanation: The correct analysis:

(x + 2)^2 \geq 0

always and equals 0 only at

x = -2

. The factor

(x - 1)

changes from negative to positive at

x = 1

. The denominator

(x - 3)

changes from negative to positive at

x = 3

(excluded from domain). Sign analysis shows the expression is negative on

(1, 3)

and zero at

x = -2

and

x = 1

. The correct solution is

\{-2\} \cup [1, 3)

. The student's answer

(-\infty, 1] \cup (3, \infty)

suggests they correctly found most intervals but forgot that

x = -2

makes the entire expression equal to 0, satisfying the inequality.

Question 8

The solution to the inequality $x^2 - 4x + k > 0$ is all real numbers except for a single point. What is the value of $k$ , and what is that exceptional point?

$k = 4$ and the point is $x = 2$ (correct answer)
$k = -4$ and the point is $x = 2$
$k = 4$ and the point is $x = 0$
$k = 2$ and the point is $x = 4$

Explanation: For a quadratic

ax^2 + bx + c > 0

to be positive for all real numbers except one point, the discriminant must equal zero (perfect square) and

a > 0

. Here,

\Delta = 16 - 4k = 0

, so

k = 4

. The quadratic becomes

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

. Since

(x - 2)^2 \geq 0

for all real

x

and equals zero only when

x = 2

, the inequality

(x - 2)^2 > 0

holds for all real numbers except

x = 2

Question 9

A quadratic inequality $ax^2 + bx + c ≤ 0$ has solution set $[-4, 2]$ . If $a > 0$ , what is the value of $\frac{b}{a}$ ?

$2$ (correct answer)
$-2$
$6$
$8$

Explanation: Since

a > 0

and the solution set is

[-4, 2]

, the parabola opens upward and is non-positive between its roots at

x = -4

and

x = 2

. The quadratic can be written as

a(x + 4)(x - 2) ≤ 0

. Expanding:

a(x^2 + 2x - 8) = ax^2 + 2ax - 8a

. Comparing with

ax^2 + bx + c

, we have

b = 2a

, so

\frac{b}{a} = 2

. Choice B gives the negative of the correct answer. Choice C represents

b

when

a = 3

. Choice D represents

c

when

a = 1

Question 10

The inequality $|x^2 - 4x| ≤ 5$ is equivalent to which of the following compound inequalities?

$-1 ≤ x ≤ 5$ (correct answer)
$-1 ≤ x ≤ 0$ or $4 ≤ x ≤ 5$
$x ≤ -1$ or $x ≥ 5$
$0 ≤ x ≤ 4$

Explanation: The inequality

|x^2 - 4x| ≤ 5

is equivalent to

-5 ≤ x^2 - 4x ≤ 5

. This gives us two inequalities: (1)

x^2 - 4x ≥ -5

, which becomes

x^2 - 4x + 5 ≥ 0

. The discriminant is

16 - 20 = -4 < 0

, so this is always true. (2)

x^2 - 4x ≤ 5

, which becomes

x^2 - 4x - 5 ≤ 0

. Factoring:

(x - 5)(x + 1) ≤ 0

. This gives

-1 ≤ x ≤ 5

. Since the first condition is always satisfied, the solution is

-1 ≤ x ≤ 5

. Choice B incorrectly splits the interval. Choice C gives the complement of the correct answer. Choice D represents the interval where

x^2 - 4x ≤ 0

Question 11

The solution set of $(x-1)^2(x+2) ≥ 0$ includes which of the following intervals?

$(-∞, -2)$
$(-2, 1)$
$[1, ∞)$ (correct answer)
$(-2, ∞)$

Explanation: The expression

(x-1)^2(x+2)

has zeros at

x = 1

(with multiplicity 2) and

x = -2

(with multiplicity 1). Since

(x-1)^2 ≥ 0

for all

x

, the sign of the expression depends only on the sign of

(x+2)

. For

x < -2

(x+2) < 0

, so the expression is negative. For

x > -2

(x+2) > 0

, so the expression is non-negative. At

x = -2

: the expression equals 0. At

x = 1

: the expression equals 0. Therefore,

(x-1)^2(x+2) ≥ 0

when

x ∈ [-2, ∞)

. This includes the interval

[1, ∞)

. Choice A represents where the expression is negative. Choice B represents part of where the expression is positive but excludes endpoints. Choice D is too broad as it should start at

-2

, not exclude it.

Question 12

The quadratic function $f(x) = -2x^2 + 8x - 6$ is graphed on a coordinate plane. For which values of $x$ is $f(x) > 0$ ?

$x < 1$ or $x > 3$
$1 < x < 3$ (correct answer)
$x ≤ 1$ or $x ≥ 3$
$x < -1$ or $x > 3$

Explanation: To solve

f(x) > 0

, first find where

f(x) = 0

. Setting

-2x^2 + 8x - 6 = 0

, divide by -2:

x^2 - 4x + 3 = 0

. Factoring:

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

, so

x = 1

x = 3

. Since the coefficient of

x^2

is negative (-2), the parabola opens downward. Therefore,

f(x) > 0

between the roots:

1 < x < 3

. Choice A gives the region where

f(x) < 0

. Choice C includes the boundary points where

f(x) = 0

. Choice D uses incorrect roots.

Question 13

A quadratic function $f(x) = ax^2 + bx + c$ satisfies $f(x) < 0$ for $x ∈ (-2, 5)$ and $f(x) > 0$ for $x ∈ (-∞, -2) ∪ (5, ∞)$ . If $f(0) = -10$ , what is the value of $a$ ?

$a = 1$ (correct answer)
$a = 2$
$a = -1$
$a = \frac{1}{2}$

Explanation: Since

f(x) > 0

for

x ∈ (-∞, -2) ∪ (5, ∞)

and

f(x) < 0

for

x ∈ (-2, 5)

, the parabola opens upward (

a > 0

) with roots at

x = -2

and

x = 5

. Therefore,

f(x) = a(x + 2)(x - 5) = a(x^2 - 3x - 10)

. Given

f(0) = -10

f(0) = a(0 + 2)(0 - 5) = a(2)(-5) = -10a = -10

. Solving:

a = 1

. Choice B would give

f(0) = -20

. Choice C would make the parabola open downward, contradicting the given information. Choice D would give

f(0) = -5