PSAT Math Quiz: Quadratic Equations

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Quadratic EquationsQuestion 1 of 20

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=1

t=2

t=4

t=5

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PSAT Math Quiz

PSAT Math Quiz: Quadratic Equations

Practice Quadratic Equations in PSAT 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 PSAT 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

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?

$t=1$
$t=2$ (correct answer)
$t=4$
$t=5$

Explanation: The question asks when the ball reaches maximum height given

h(t)=-5t^2+20t+1

. For a quadratic in the form

at^2+bt+c

, the maximum (when

a<0

) occurs at

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

. Here,

a=-5

and

b=20

, so

t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2

seconds. We can verify this is a maximum since the coefficient of

t^2

is negative (-5), making the parabola open downward. A common mistake is forgetting the negative sign in the formula or confusing when to use maximum vs minimum. For time-based problems, always check that your answer makes physical sense (positive time).

Question 2

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?

$t=1$
$t=2$ (correct answer)
$t=4$
$t=5$

Explanation: This question seeks the time (t) when a ball reaches maximum height, modeled by (h(t) = -5t^2 + 20t + 1). For quadratics, the vertex gives the maximum (since a<0), at (t = -\frac{b}{2a} = -\frac{20}{2(-5)} = \frac{20}{10} = 2). Completing the square: (h(t) = -5(

t^2

- 4t) + 1 = -5(t - 2)^2 + 21), confirming peak at t=2. The quadratic formula isn't needed for vertex time. A common mistake is using positive a or forgetting the negative sign. In projectile problems, remember the vertex formula directly gives the time to max height.

Question 3

A parabola is shown on the coordinate plane. It opens upward and crosses the $x$ -axis at $x=-1$ and $x=3$ . What is the equation of the axis of symmetry of the parabola?

$x=1$ (correct answer)
$x=2$
$y=1$
$y=2$

Explanation: This question requires finding the equation of the axis of symmetry for a parabola that opens upward and crosses the x-axis at (x = -1) and (x = 3). The axis of symmetry for a parabola is the vertical line passing through the midpoint of the roots, calculated as (x =

\frac{-1 + 3}{2}

= 1). You can also think of it as (x = -

\frac{b}{2a}

) in the standard form, but here the roots directly give the symmetry line. Using the quadratic formula isn't necessary since the roots are provided, but it would confirm the same axis. A key error is confusing the axis with the y-intercept or miscalculating the midpoint, such as averaging incorrectly to get 2. For parabolas given by roots, always use the midpoint formula as a quick strategy to find the axis without full equation derivation.

Question 4

A parabola is graphed on the coordinate plane and opens downward. Its vertex is at $(-1,4)$ , and it crosses the $y$ -axis at $(0,3)$ . What are the $x$ -intercepts of the parabola?

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

Explanation: This question requires finding the x-intercepts of a downward-opening parabola with vertex (-1, 4) and y-intercept (0, 3). Use vertex form (y = a(x + 1)^2 + 4); substitute (0, 3): (3 = a(1)^2 + 4), so a = -1, giving (y = -(x + 1)^2 + 4 = -x^2 - 2x + 3). Set y=0: (

x^2

+ 2x - 3 = 0), which factors as (x + 3)(x - 1) = 0, so x = -3, 1. Quadratic formula confirms: (x =

\frac{-2 pm sqrt{4 + 12}}{2}

\frac{-2 pm 4}{2}

), yielding the same. A common mistake is incorrect a or sign errors in expansion. When given vertex and a point, determine a first, then solve for roots using factoring if possible.

Question 5

A rectangle has length $(x+7)\text{ cm}$ and width $(x+3)\text{ cm}$ . If its area is $60\text{ cm}^{2},$ what is the value of $x$ ?

$13$
$5$
$6$
$3$ (correct answer)

Explanation: When you see a geometry problem involving algebraic expressions for dimensions, you're dealing with a system where you need to set up an equation using the given measurements and solve for the unknown variable. Since area equals length times width, you can write:

(x+7)(x+3) = 60

. Expanding this quadratic equation:

x^2 + 3x + 7x + 21 = 60

, which simplifies to

x^2 + 10x + 21 = 60

. Subtracting 60 from both sides gives you

x^2 + 10x - 39 = 0

. To factor this quadratic, you need two numbers that multiply to

-39

and add to

10

. Those numbers are

13

and

-3

, so the equation factors as

(x + 13)(x - 3) = 0

. This gives you

x = -13

x = 3

. Since length and width must be positive,

x = 3

is the only valid solution. You can verify: length =

3 + 7 = 10

cm, width =

3 + 3 = 6

cm, and area =

10 \times 6 = 60

cm². Choice A (

13

) is the other solution to the quadratic equation, but it would make the dimensions unreasonably large (

20 \times 16 = 320

cm²). Choice B (

5

) would give an area of

12 \times 8 = 96

cm², which is too large. Choice C (

6

) would result in

13 \times 9 = 117

cm², also too large. Remember to always check that your solution makes sense in the real-world context—negative dimensions are impossible, and your final answer should satisfy the original area condition.

Question 6

The system of equations $y = x^{2}+2x+1$ and $y = 3x+1$ has how many intersection points, and what are the $x$ -coordinates of those points?

One; $x=1$ only
Two; $x=-1$ and $x=3$
Two; $x=0$ and $x=1$ (correct answer)
None; the curves never intersect

Explanation: When you encounter a system involving a quadratic and linear equation, you're looking for intersection points where both equations have the same

x

and

y

values. This means setting the equations equal to each other. Since both equations equal

y

, set them equal:

x^2 + 2x + 1 = 3x + 1

. Rearranging gives you

x^2 + 2x + 1 - 3x - 1 = 0

, which simplifies to

x^2 - x = 0

. Factor out

x

x(x - 1) = 0

. This gives you

x = 0

x = 1

. Since you found two distinct

x

-values, there are two intersection points. Choice A is incorrect because while

x = 1

is one solution, it's not the only one. This answer misses the

x = 0

solution. Choice B is wrong because neither

x = -1

nor

x = 3

satisfies the equation

x^2 - x = 0

. You can verify:

(-1)^2 - (-1) = 2 \neq 0

and

3^2 - 3 = 6 \neq 0

. Choice D is incorrect because we clearly found solutions. A quadratic and linear equation typically intersect at two points unless the line is tangent to the parabola or they don't intersect at all. Choice C correctly identifies both solutions:

x = 0

and

x = 1

. Strategy tip: When solving systems with quadratic and linear equations, always set them equal and solve the resulting quadratic equation completely. Don't stop after finding one solution—quadratics typically have two solutions.

Question 7

Which completed-square form is equivalent to $2x^{2}+8x+5$ ?

$2(x+2)^{2}+5$
$2(x+4)^{2}+5$
$(x+2)^{2}+1$
$2(x+2)^{2}-3$ (correct answer)

Explanation: When you see a quadratic expression that needs to be converted to completed-square form, you're working with the process of completing the square. This technique rewrites

ax^2 + bx + c

a(x + h)^2 + k

, which reveals the vertex of the parabola and makes certain algebraic operations easier. To complete the square for

2x^2 + 8x + 5

, first factor out the coefficient of

x^2

from the first two terms:

2(x^2 + 4x) + 5

. Next, complete the square inside the parentheses by taking half of the coefficient of

x

(which is 4), squaring it to get 4, then adding and subtracting this value:

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

. Distributing the 2 gives:

2(x + 2)^2 - 8 + 5 = 2(x + 2)^2 - 3

. Choice A incorrectly keeps the original constant term 5 instead of adjusting it after completing the square. Choice B uses the wrong value inside the squared term—it should be half of 4 (which is 2), not 4 itself. Choice C fails to factor out the leading coefficient of 2, making the squared term coefficient incorrect. Choice D correctly shows

2(x + 2)^2 - 3

, which matches our work. Remember: when completing the square with a leading coefficient other than 1, always factor it out first, complete the square inside the parentheses, then carefully distribute and combine constants at the end.

Question 8

Which set lists all real solutions to the quadratic equation $x^{2}-5x+6=0$ ?

$x=2 \text{ and } x=-3$
$x=-2 \text{ and } x=-3$
$x=-2 \text{ and } x=3$
$x=2 \text{ and } x=3$ (correct answer)

Explanation: When you encounter a quadratic equation like

x^{2}-5x+6=0

, you need to find the values of

x

that make the equation true. The most efficient approach here is factoring, since the coefficients suggest clean factors. To factor

x^{2}-5x+6

, you need two numbers that multiply to give 6 (the constant term) and add to give -5 (the coefficient of

x

). Those numbers are -2 and -3, since

(-2) \times (-3) = 6

and

(-2) + (-3) = -5

. This gives you the factored form:

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

. Using the zero product property, if the product of two factors equals zero, then at least one factor must equal zero. So either

x-2=0

x-3=0

, which means

x=2

x=3

. Choice A gives

x=2

and

x=-3

. The second value is incorrect—it confuses the factor

-3

with the solution. Choice B gives

x=-2

and

x=-3

. Both values are wrong, representing the factors themselves rather than solving

x-2=0

and

x-3=0

. Choice C gives

x=-2

and

x=3

. This mixes up signs—getting one solution right but making the same factor-versus-solution error with the other. Choice D correctly identifies both solutions:

x=2

and

x=3

. Study tip: When factoring quadratics, remember that if your factored form is

(x-a)(x-b)=0

, the solutions are

x=a

and

x=b

—not

x=-a

and

x=-b

. Always solve each factor equation separately.

Question 9

Solve the equation $2x^2+3x-2=0$ and select the correct pair of solutions.

$x=\tfrac{1}{2},\ -2$ (correct answer)
$x=-\tfrac{1}{2},\ 2$
$x=\tfrac{2}{3},\ -1$
$x=-\tfrac{2}{3},\ 1$

Explanation: This question asks to solve (

2x^2

+ 3x - 2 = 0) and select the solutions. Factor: find numbers for (2x + a)(x + b) where a*b=-2, and cross terms give 3x; try (2x - 1)(x + 2) = 2x^2 + 4x - x - 2 = 2x^2 + 3x - 2, yes, so x=1/2 or x=-2. Quadratic formula: (x =

\frac{-3 pm sqrt{9 + 16}}{4}

\frac{-3 pm 5}{4}

), giving (

\frac{2}{4}

=0.5) and (

\frac{-8}{4}

=-2). Factoring works well with small integers; formula confirms. Common error: incorrect factoring signs. Verify by plugging solutions back in.

Question 10

The parabola $y=-2x^2+8x-3$ is graphed on the coordinate plane. What is the axis of symmetry of this parabola? Do not confuse the axis of symmetry with the $y$ -intercept or with the vertex's $y$ -coordinate.

$x=-2$
$x=2$ (correct answer)
$y=2$
$y=-3$

Explanation: The question asks for the axis of symmetry of y = -2x² + 8x - 3. The axis is the vertical line x = -b/(2a), with a = -2, b = 8, so x = -8/(2*(-2)) = -8/-4 = 2. This is the line through the vertex, confirmed by vertex form after completing the square: y = -2(x² - 4x) - 3 = -2(x² - 4x + 4 - 4) - 3 = -2(x - 2)² + 8 - 3 = -2(x - 2)² + 5. Mistaking it for y-values like the y-intercept -3 or another number is common. Confusing with positive b sign can lead to x = -2. Always use the formula x = -b/(2a) for quick identification in graphing questions.

Question 11

A rectangular garden is to be built against a wall, so only three sides need fencing. If 24 meters of fencing are available, the area is $A(x)=x(24-2x)$ where $x$ is the length of each side perpendicular to the wall. What value of $x$ maximizes the area? Choose the correct $x$ -value.

$x=4$
$x=6$ (correct answer)
$x=8$
$x=12$

Explanation: This question asks for the value of x that maximizes the area of a garden, given A(x) = x(24 - 2x) = 24x - 2x^2. This is a quadratic that opens downward (a = -2), so the maximum occurs at the vertex, x = -\frac{b}{2a} = -\frac{24}{2(-2)} = \frac{24}{4} = 6. Completing the square: A(x) = -2(

x^2

- 12x) = -2(x - 6)^2 + 72, confirming x=6. A key error is treating it as minimizing instead of maximizing or misexpanding the area formula. Another mistake is solving for roots rather than the vertex. In optimization problems, convert to quadratic form and use the vertex formula to find the extremum efficiently.

Question 12

The function $f(x)=x^2-10x+21$ is rewritten in vertex form as $f(x)=(x-h)^2+k$ . What are the values of $h$ and $k$ ? A common error is completing the square but forgetting to keep the expression equivalent.

$h=5,\ k=-4$ (correct answer)
$h=-5,\ k=-4$
$h=5,\ k=4$
$h=-5,\ k=4$

Explanation: The question requires rewriting f(x) = x² - 10x + 21 in vertex form (x - h)² + k to find h and k. Complete the square: x² - 10x = (x - 5)² - 25, so f(x) = (x - 5)² - 25 + 21 = (x - 5)² - 4, thus h = 5, k = -4. The quadratic formula isn't needed, but vertex x = -b/(2a) = 10/2 = 5 confirms h. A common error is adding/subtracting the completion term incorrectly, like forgetting to balance -25 + 21. Mismanaging signs can lead to h = -5. Use completing the square for vertex form conversions, especially when factoring doesn't yield integers.

Question 13

Rewrite $y=x^2+6x+1$ in vertex form $y=a(x-h)^2+k$ . Which expression is equivalent?

$y=(x+3)^2-8$ (correct answer)
$y=(x-3)^2+10$
$y=(x+6)^2-35$
$y=(x+3)^2+8$

Explanation: To convert

y=x^2+6x+1

to vertex form, we complete the square. First, focus on

x^2+6x

: we take half of 6 (which is 3) and square it (getting 9). So

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

. Therefore,

y = x^2+6x+1 = (x+3)^2 - 9 + 1 = (x+3)^2 - 8

. The vertex form is

y=(x+3)^2-8

, showing the vertex at

(-3,-8)

. A common error is forgetting to subtract the 9 when completing the square, or making sign errors. Always expand your final answer to verify it matches the original equation.

Question 14

On the coordinate plane, a parabola opens downward and has vertex at $(2,5)$ . It passes through the point $(0,1)$ . What is the equation of the parabola in vertex form?

$y=-(x-2)^2+5$ (correct answer)
$y=-(x+2)^2+5$
$y=-(x-2)^2+1$
$y=-(x-2)^2+9$

Explanation: This question asks for the equation of a downward-opening parabola with vertex at

(2,5)

that passes through

(0,1)

. Since it opens downward and has vertex at

(2,5)

, we can write

y=a(x-2)^2+5

where

a<0

. To find

a

, substitute the point

(0,1)

1=a(0-2)^2+5=a(4)+5=4a+5

. Solving:

4a=-4

, so

a=-1

. Therefore, the equation is

y=-(x-2)^2+5

. A common mistake is using the wrong sign for

a

or making arithmetic errors when substituting the given point. For vertex form problems, always verify your answer by checking that both the vertex and the given point satisfy the equation.

Question 15

What are the solutions to the equation $2x^2-5x-3=0$ ? Choose the pair of $x$ -values that makes the equation true. Be careful with signs when factoring or using the quadratic formula, since switching a sign can produce a plausible but incorrect pair of solutions.

$x=3,\,-\tfrac{1}{2}$ (correct answer)
$x=-3,\,\tfrac{1}{2}$
$x=\tfrac{3}{2},\,-1$
$x=-\tfrac{3}{2},\,1$

Explanation: The question asks for the solutions to the quadratic equation 2x² - 5x - 3 = 0. To solve, use the quadratic formula x = [-b ± √(b² - 4ac)] / (2a), where a = 2, b = -5, c = -3, giving x = [5 ± √(25 + 24)] / 4 = [5 ± √49] / 4 = [5 ± 7] / 4. This yields x = (5 + 7)/4 = 3 and x = (5 - 7)/4 = -1/2. Factoring could also work by finding numbers that multiply to 2*(-3) = -6 and add to -5, such as -6 and 1, leading to (2x + 1)(x - 3) = 0 after grouping. A common error is switching signs during factoring or in the quadratic formula, resulting in incorrect pairs like those in choices B or D. When dealing with quadratics that factor nicely, try factoring first before resorting to the formula for efficiency in tests.

Question 16

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? Use the vertex idea for quadratics and avoid mixing up the time of maximum with the maximum height value.

$t=1$
$t=2$ (correct answer)
$t=4$
$t=5$

Explanation: The question asks for the time t when the ball reaches its maximum height in h(t) = -5t² + 20t + 1. For a quadratic at² + bt + c, the vertex gives the maximum or minimum at t = -b/(2a), here a = -5, b = 20, so t = -20/(2*(-5)) = -20/-10 = 2. Since a < 0, the parabola opens downward, confirming a maximum at t = 2. Completing the square or derivatives aren't needed, but vertex form h(t) = -5(t - 2)² + 41 verifies the time. A common mistake is confusing the time with the height value or miscalculating the sign in -b/(2a). In projectile problems, always use the vertex formula for peak time to save steps.

Question 17

The graph of a parabola is shown on the coordinate plane. The parabola opens upward and has $x$ -intercepts at $x=-1$ and $x=3$ , and it passes through the point $(0,-3)$ . Which equation represents this parabola? Choose the correct equation in standard form.

$y=x^2-2x-3$ (correct answer)
$y=x^2+2x-3$
$y=x^2-2x+3$
$y=-x^2+2x+3$

Explanation: This question asks for the equation in standard form of a parabola that opens upward with x-intercepts at x=-1 and x=3, passing through (0, -3). Since the x-intercepts are the roots, the factored form is y = a(x + 1)(x - 3); expanding gives y = a(

x^2

- 2x - 3). Substitute (0, -3) to find a: -3 = a(-3), so a=1, yielding y = x^2 - 2x - 3. Errors often occur by assuming a=1 without verifying the point or flipping signs in the factored form. Another mistake is selecting an equation that opens downward or mismatches the y-value at x=0. When given intercepts and a point, use factored form first, then verify with the point to determine the leading coefficient.

Question 18

Solve the equation $3x^2+4x-2=0$ using the quadratic formula. Which set gives the solutions?

$\frac{-2\pm\sqrt{10}}{3}$ (correct answer)
$\frac{-4\pm\sqrt{40}}{6}$
$\frac{-4\pm\sqrt{16-24}}{6}$
$\frac{2\pm\sqrt{10}}{3}$

Explanation: To solve

3x^2+4x-2=0

using the quadratic formula

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

, we identify

a=3

b=4

, and

c=-2

. Substituting:

x = \frac{-4 \pm \sqrt{16-4(3)(-2)}}{2(3)} = \frac{-4 \pm \sqrt{16+24}}{6} = \frac{-4 \pm \sqrt{40}}{6}

. We can simplify

\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}

, giving

x = \frac{-4 \pm 2\sqrt{10}}{6} = \frac{-2 \pm \sqrt{10}}{3}

. Common errors include sign mistakes in the discriminant calculation or forgetting to simplify the final fraction. Always double-check your arithmetic when using the quadratic formula.

Question 19

What is the axis of symmetry of the parabola $y=-3x^2+12x-7$ ?

$x=-2$
$x=2$ (correct answer)
$y=2$
$x=\frac{12}{3}$

Explanation: This question asks for the axis of symmetry of the parabola

y=-3x^2+12x-7

. For a quadratic in the form

y=ax^2+bx+c

, the axis of symmetry is the vertical line

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

. Here,

a=-3

and

b=12

, so the axis of symmetry is

x=-\frac{12}{2(-3)}=-\frac{12}{-6}=2

. This means the parabola is symmetric about the vertical line

x=2

. A common error is forgetting the negative sign in the formula or confusing the axis of symmetry (a vertical line) with a horizontal line. For the PSAT, remember that the axis of symmetry always passes through the vertex of the parabola.

Question 20

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

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

Explanation: The question asks for the value of x that maximizes the area of a rectangle with perimeter 30 and one side x, given (A(x) = x(15 - x) = -x^2 + 15x). This is a quadratic opening downward, so the maximum is at the vertex (x = -

\frac{b}{2a}

= -

\frac{15}{2 imes -1}

= 7.5). Completing the square: (A(x) = -(

x^2

- 15x) = -(

x^2

- 15x + 56.25 - 56.25) = -(x - 7.5)^2 + 56.25), confirming x=7.5. Factoring isn't needed here, as it's about optimization. A key error is choosing endpoints like 5 or 15, which give zero area. For real-world optimization with quadratics, use the vertex formula to find maxima or minima quickly.