SAT Math Quiz: Quadratic Equations

20 questions · exam conditions

0:00

Quadratic EquationsQuestion 1 of 20

Solve $2x^2+3x-2=0$ . What are the solutions?

x=\frac{3\pm\sqrt{17}}{4}

x=\frac{1}{2}\text{ or }-2

x=\frac{-3\pm\sqrt{17}}{4}

x=-\frac{1}{2}\text{ or }2

← Back to quizzes

SAT Math Quiz

SAT Math Quiz: Quadratic Equations

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

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.

All questions

Question 1

Solve $2x^2+3x-2=0$ . What are the solutions?

$x=\frac{3\pm\sqrt{17}}{4}$
$x=\frac{1}{2}\text{ or }-2$ (correct answer)
$x=\frac{-3\pm\sqrt{17}}{4}$
$x=-\frac{1}{2}\text{ or }2$

Explanation: Factor 2x^2 + 3x - 2 = 0 by finding numbers that multiply to 2 times -2, or -4, and add to 3: those are 4 and -1. Rewriting gives 2x^2 + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2) = 0, so x = 1/2 or x = -2, the pair listed together. The quadratic formula agrees: x = (-3 +/- sqrt(9 + 16))/4 = (-3 +/- 5)/4, giving 1/2 and -2. The choice listing -1/2 and 2 flips both signs; testing x = 2 gives 8 + 6 - 2 = 12, not 0. Both choices containing sqrt(17) mis-evaluate the discriminant as 9 + 8 instead of 9 - 4(2)(-2) = 25, so they never reduce to the actual roots.

Question 2

Solve the equation $x^2+6x+9=4$ . What are the solutions for $x$ ?

$x=-3\pm\sqrt{13}$
$x=3\pm2$
$x=-3$
$x=-3\pm2$ (correct answer)

Explanation: The question asks for the solutions of x^2 + 6x + 9 = 4. The left side is already a perfect square, so rewrite the equation as (x + 3)^2 = 4. Taking square roots of both sides gives x + 3 = 2 or x + 3 = -2, so x = -3 ± 2, meaning x = -1 and x = -5. Checking one: (-1)^2 + 6(-1) + 9 = 1 - 6 + 9 = 4. The answer -3 ± sqrt(13) comes from adding the 4 to the 9 instead of moving it across, producing (x + 3)^2 = 13. The single value -3 drops the ± entirely, which solves (x + 3)^2 = 0, not 4. And 3 ± 2 flips the sign of the shift, since the factor (x + 3) centers the solutions at -3; the values 1 and 5 fail the original equation.

Question 3

A ball is thrown upward from a platform. Its height in feet after $t$ seconds is modeled by $h(t)=-16t^2+64t+5$ . At what time does the ball reach its maximum height?

$1$
$2$ (correct answer)
$4$
$\dfrac{5}{2}$

Explanation: We need to find when the ball reaches maximum height using h(t) = -16t² + 64t + 5. For a quadratic function in the form at² + bt + c, the maximum (when a < 0) occurs at t = -b/(2a). Here, a = -16 and b = 64, so t = -64/(2(-16)) = -64/(-32) = 2 seconds. Common mistakes include using the y-intercept (5, when t = 0) or finding the roots instead of the vertex. Since the coefficient of t² is negative, this parabola opens downward, confirming we're finding a maximum. Always use the vertex formula t = -b/(2a) for optimization problems involving quadratics.

Question 4

A rectangle has perimeter $40$ meters. Let its width be $x$ meters and its length be $20-x$ meters. The area is $A(x)=x(20-x)$ . For what value of $x$ is the area maximized?

$5$
$10$ (correct answer)
$15$
$20$

Explanation: We need to maximize the area function A(x) = x(20 - x) = 20x - x². This is a quadratic that opens downward (since the coefficient of x² is negative), so it has a maximum at its vertex. The x-coordinate of the vertex is x = -b/(2a) = -20/(2(-1)) = -20/(-2) = 10. We can verify this makes sense: when x = 10, the length is also 20 - 10 = 10, giving a square with maximum area. Common mistakes include choosing an endpoint (x = 0 or x = 20) instead of the vertex, or confusing perimeter with area. For optimization problems with quadratics, the extremum always occurs at the vertex unless constrained to an interval.

Question 5

A company's profit (in dollars) from selling $x$ items is $P(x)=-x^2+12x-20$ . For what value of $x$ is the profit maximized?

$4$
$6$ (correct answer)
$12$
$20$

Explanation: This question asks for the value of x that maximizes the profit P(x) = -x² + 12x - 20. Since a = -1 < 0, the parabola opens downward, and the maximum is at the vertex x = -b/(2a) = -12 / (2 × -1) = -12 / -2 = 6. Complete the square: P(x) = -(x² - 12x) - 20 = -( (x - 6)² - 36 ) - 20 = -(x - 6)² + 36 - 20 = -(x - 6)² + 16, confirming maximum at x = 6. Check by evaluating nearby points if needed. A common error is using a positive a, leading to a minimum instead. For optimization problems, apply the vertex formula to find the extremum efficiently.

Question 6

What are the solutions to the equation $x^2-9x+20=0$ ?

$x=4,5$ (correct answer)
$x=-4,-5$
$x=1,20$
$x=9,20$

Explanation: This question asks us to find the solutions to the quadratic equation

x^2-9x+20=0

. To solve this, we can factor the quadratic by finding two numbers that multiply to 20 and add to -9, which are -4 and -5. This gives us

(x-4)(x-5)=0

, so

x=4

x=5

. A common error is to confuse the signs when factoring, which might lead to choosing answer B with negative values. When factoring, always verify your answer by expanding back or substituting the solutions into the original equation.

Question 7

A rectangle has area $48$ square units. Its length is $2$ units more than its width. Which equation can be used to find the width $w$ (in units)?

$w^2+4w-48=0$
$w^2-2w-48=0$
$2w^2+w-48=0$
$w^2+2w-48=0$ (correct answer)

Explanation: This question asks for the equation to find the width

w

of a rectangle with area

48

and length

2

units more than the width. Let length =

w + 2

, so area gives

w(w + 2) = 48

, expanding to

w^2 + 2w - 48 = 0

. This is a quadratic equation solvable by factoring:

(w + 8)(w - 6) = 0

w = 6

(since width

> 0

). The quadratic formula could also be used:

w = [-2 \pm \sqrt{4 + 192}]/2 = [-2 \pm \sqrt{196}]/2 = [-2 \pm 14]/2

, giving

6

-8

. Key errors include incorrect coefficients on the linear term, as in choices A, B, and C. Always define variables clearly and check by solving to ensure positive dimensions.

Question 8

What are the solutions to the equation $x^2-5x+6=0$ ? Choose the pair of values of $x$ that makes the equation true.

$x=2,3$ (correct answer)
$x=-2,-3$
$x=1,6$
$x=2,-3$

Explanation: This question asks us to solve the quadratic equation

x^2-5x+6\=0

by finding the values of

x

that make it true. To solve, we can factor the quadratic: we need two numbers that multiply to 6 and add to -5, which are -2 and -3, giving us

(x-2)(x-3)\=0

. Setting each factor to zero gives

x-2\=0

x\=2

, and

x-3\=0

x\=3

. A common error is to factor as

(x+2)(x+3)\=0

by using the wrong signs, which would give

x\=-2,-3

. When factoring, always check that the middle term matches:

(x-2)(x-3)\=x^2-3x-2x+6\=x^2-5x+6

✓.

Question 9

Which expression is equivalent to $x^2+6x+1$ written in vertex form $a(x-h)^2+k$ ? Watch for distractors that complete the square but change the constant incorrectly.

$(x+3)^2+10$
$(x+3)^2-8$ (correct answer)
$(x-3)^2-8$
$(x-3)^2+10$

Explanation: To write x² + 6x + 1 in vertex form, we need to complete the square. First, take half of the coefficient of x and square it: (6/2)² = 9. Add and subtract 9: x² + 6x + 9 - 9 + 1 = (x + 3)² - 8. The vertex form is (x + 3)² - 8, which can also be written as (x - (-3))² + (-8). Common errors include forgetting to subtract the added term (getting (x + 3)² + 10) or making sign errors with the h-value. When completing the square, always remember to balance what you add by subtracting the same amount.

Question 10

What is the vertex of the parabola formed by $f(x)\=(x-7)^2-3$ ??

(7, -3) (correct answer)
(-3, 7)
(-7, 3)
(3, -7)

Explanation: The equation of a parabola is

f(x)\=(x-a)^2+b

where

a

is the location of the x-coordinate of the vertex and

b

is the location of the y-coordinate of the vertex. Using this information and the equation of our quadratic, we can see the vertex of this quadratic is (7, -3)..

Question 11

A ball is thrown upward from a platform; its height in feet is given by $h(t)=-16t^2+32t+48$ , where $t$ is time in seconds. When does it hit the ground?

-1
1
4
3 (correct answer)

Explanation: Set

h(t)=0

-16t^2+32t+48=0\Rightarrow t^2-2t-3=0\Rightarrow (t-3)(t+1)=0

, so

t=3

t=-1

; only

t=3

is valid. The other positive choices come from algebra slips.

Question 12

What is the value of $k$ such that the parabola $y=x^2+kx+9$ has its vertex on the $y$ -axis? A common incorrect path is to set the vertex's $y$ -coordinate to 0 instead of making the vertex's $x$ -coordinate equal to 0.

$k=0$ (correct answer)
$k=9$
$k=-9$
$k=18$

Explanation: For the parabola

y=x^2+kx+9

to have its vertex on the y-axis, the x-coordinate of the vertex must be 0. The x-coordinate of the vertex is

x=-\frac{b}{2a}=-\frac{k}{2(1)}=-\frac{k}{2}

. Setting this equal to 0:

-\frac{k}{2}=0

, which means

k=0

. With

k=0

, the equation becomes

y=x^2+9

, which has vertex at

(0,9)

. A common error is trying to make the y-coordinate of the vertex equal to 0, but the question asks for the vertex to be ON the y-axis (where

x=0

), not at the origin.

Question 13

On a coordinate plane, the parabola $y=x^2-2x-3$ is graphed. What is the minimum value of $y$ for this parabola?

$-4$ (correct answer)
$1$
$-3$
$2$

Explanation: We need the minimum value of

y

for the parabola

y=x^2-2x-3

. Since the parabola opens upward (positive leading coefficient), the minimum occurs at the vertex. The x-coordinate of the vertex is

x=-\frac{b}{2a}=-\frac{-2}{2(1)}=1

. Substituting

x=1

into the equation:

y=(1)^2-2(1)-3=1-2-3=-4

. The minimum value of

y

is -4. A common error is to report the x-coordinate (1) instead of the y-coordinate (-4) as the minimum value. When asked for minimum or maximum values, always give the y-value unless specifically asked for the x-value.

Question 14

A rectangle has perimeter 30. If its length is $x$ and its width is $15-x$ , its area is $A(x)=x(15-x)$ . For what value of $x$ is the area maximized?

$x=5$
$x=15$
$x=7.5$ (correct answer)
$x=10$

Explanation: The area function

A(x)=x(15-x)=15x-x^2

represents a downward-opening parabola. To find the maximum, we use the vertex formula

x=-\frac{b}{2a}

where the function is written as

A(x)=-x^2+15x

. Here

a=-1

and

b=15

, so

x=-\frac{15}{2(-1)}=\frac{15}{2}=7.5

. This makes sense physically: for a rectangle with fixed perimeter, the maximum area occurs when it's a square (length equals width). Common errors include using the wrong formula or not recognizing that this is an optimization problem requiring the vertex.

Question 15

A parabola is shown on the coordinate plane with vertex at $(0,3)$ and it passes through $(2,-1)$ . Which equation represents the parabola? Use vertex form $y=a(x-0)^2+3$ and solve for $a$ using the given point.

$y=x^2+3$
$y=-x^2+3$ (correct answer)
$y=-x^2-3$
$y=\tfrac{1}{2}x^2+3$

Explanation: This question asks for the equation of a parabola with vertex at (0, 3) that passes through (2, -1). Using vertex form y = a(x - 0)² + 3 = ax² + 3, we substitute the point (2, -1) to find a: -1 = a(2)² + 3 = 4a + 3, so 4a = -4 and a = -1. Therefore, the equation is y = -x² + 3. A common mistake is to use the wrong sign for a or to forget that a negative value of a means the parabola opens downward. When finding a parabola's equation from its vertex and another point, always substitute carefully and solve for the leading coefficient.

Question 16

What is the vertex of the parabola $y=2x^2-8x+3$ ? Give your answer as an ordered pair $(h,k)$ . Some incorrect paths come from finding only the axis of symmetry ( $x=h$ ) or making a sign error when completing the square.

$(2,\ -5)$ (correct answer)
$(2,\ 3)$
$(-2,\ -5)$
$(4,\ -5)$

Explanation: To find the vertex of y = 2x² - 8x + 3, we can use the vertex formula or complete the square. Using the formula, the x-coordinate of the vertex is h = -b/(2a) = -(-8)/(2(2)) = 8/4 = 2. To find the y-coordinate, substitute x = 2 back into the equation: y = 2(2)² - 8(2) + 3 = 8 - 16 + 3 = -5. Therefore, the vertex is (2, -5). A common error is stopping after finding only the x-coordinate or making sign errors when completing the square. When using the vertex formula, always substitute back to find the y-coordinate of the vertex.

Question 17

A ball's height in meters after $t$ seconds is modeled by $h(t)=-5t^2+20t+1$ . At what time does the ball reach its maximum height?

$t=2$ (correct answer)
$t=4$
$t=20$
$t=1$

Explanation: The model h(t) = -5t^2 + 20t + 1 is a downward-opening parabola because a = -5 is negative, so its vertex is the maximum height. The vertex time is t = -b/(2a) = -20/(2 * -5) = 20/10 = 2, so t = 2 is correct. Completing the square confirms it: h(t) = -5(t-2)^2 + 21, a peak of 21 meters at t = 2. The answer t = 4 comes from dividing 20 by 5 while dropping the factor of 2 in the denominator. The answer t = 20 comes from mistakenly using the coefficient b = 20 directly instead of applying the full vertex formula t = -b/(2a). The answer t = 1 is too early, because h(1) = 16 and the ball is still rising toward its peak.

Question 18

Solve $\sqrt{x+5}=x-1$ . Which value of $x$ satisfies the equation?

-1
1
4 (correct answer)
-4

Explanation: Squaring gives

x+5=x^2-2x+1\Rightarrow x^2-3x-4=0\Rightarrow (x-4)(x+1)=0

. Checking in the original shows

x=4

works but

x=-1

is extraneous.

Question 19

At what time does the object reach maximum height if its height is modeled by $h(t) = -16t^2 + 32t + 5$ ?

1 sec (correct answer)
2 sec
0.5 sec
3 sec

Explanation: This question asks when an object reaches maximum height, which occurs at the vertex of the parabola. For a quadratic in the form

h(t) = at^2 + bt + c

, the time at maximum height is

t = -\frac{b}{2a}

. With

h(t) = -16t^2 + 32t + 5

, we have

a = -16

and

b = 32

, so

t = -\frac{32}{2(-16)} = -\frac{32}{-32} = 1

second. A common error is forgetting the negative sign in the formula or mixing up the values of

a

and

b

. For projectile motion problems, always use the vertex formula to find the time of maximum height.

Question 20

The equation of a parabola in the $xy$ -plane is $y \= 2(x - 3)^2 + 5$ . Which of the following statements is true about the parabola?

The maximum value of $y$ is 3.
The minimum value of $y$ is 3.
The maximum value of $y$ is 5.
The minimum value of $y$ is 5. (correct answer)

Explanation: The equation is in vertex form

y \= a(x-h)^2 + k

, where

(h,k)

is the vertex. Here, the vertex is

(3, 5)

. Since the leading coefficient (

a\=2

) is positive, the parabola opens upward, meaning the vertex represents the **minimum** value. The minimum y-value is 5. SAT Strategy: Use The Calculator Graph it in Desmos. You will visually see the bottom of the curve is at

y\=5

. Importantly - many students see a question like this and feel lost because they've forgotten or never really knew the equation of a parabola. But so many questions can be solved with the calculator, so "when in doubt, graph it out" to at least give yourself a chance.