SAT Math Quiz: Quadratic Equations

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.

Question 1

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

1. $x=-3\pm\sqrt{13}$
2. $x=3\pm2$
3. $x=-3$
4. $x=-3\pm2$ (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

1. $t=2$ (correct answer)
2. $t=4$
3. $t=\frac{20}{5}$
4. $t=1$

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 answers t = 4 and t = 20/5 are the same number, and each comes from dividing 20 by 5 while dropping the factor of 2 in the denominator; that time is actually when the ball has fallen back near its starting height, since h(4) = -80 + 80 + 1 = 1. The answer t = 1 is too early, because h(1) = 16 and the ball is still rising toward its peak.

Question 4

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

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 $x=4$ or $x=5$. 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 5

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

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 $x-2\=0$ so $x\=2$, and $x-3\=0$ so $x\=3$. 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 6

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

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 7

1. (7, -3) (correct answer)
2. (-3, 7)
3. (-7, 3)
4. (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 8

1. -1
2. 1
3. 4
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 $t=3$ or $t=-1$; only $t=3$ is valid. The other positive choices come from algebra slips.

Question 9

1. $k=0$ (correct answer)
2. $k=9$
3. $k=-9$
4. $k=18$

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 10

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

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 11

1. $x=\frac{5\pm\sqrt{49}}{6}$
2. $x=2,\,-\frac{1}{3}$ (correct answer)
3. $x=-2,\,\frac{1}{3}$
4. $x=\frac{5\pm\sqrt{1}}{6}$

Explanation: To solve $3x^2-5x-2=0$, we can use factoring or the quadratic formula. For factoring, we need two numbers that multiply to $(3)(-2)=-6$ and add to $-5$, which are $-6$ and $1$. Rewriting: $3x^2-6x+x-2=3x(x-2)+1(x-2)=(3x+1)(x-2)=0$. This gives us $x=2$ or $x=-\frac{1}{3}$. Students often make errors with the signs when dealing with the fraction, getting $\frac{1}{3}$ instead of $-\frac{1}{3}$. When factoring leads to fractions, double-check by substituting back into the original equation.

Question 12

1. $x=5$
2. $x=15$
3. $x=7.5$ (correct answer)
4. $x=10$

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 13

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

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 14

1. $5$
2. $10$ (correct answer)
3. $15$
4. $20$

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 15

1. $(2,\ -5)$ (correct answer)
2. $(2,\ 3)$
3. $(-2,\ -5)$
4. $(4,\ -5)$

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 16

1. $t=1$
2. $t=2$ (correct answer)
3. $t=4$
4. $t=\frac{20}{5}=4$

Explanation: The question asks for the time t when the ball reaches its maximum height in the model h(t) = -5t² + 20t + 1. For a quadratic function at² + bt + c opening downward (a < 0), the maximum occurs at the vertex t = -b/(2a). Here, a = -5, b = 20, so t = -20 / (2*(-5)) = -20 / -10 = 2 seconds. Completing the square or using calculus isn't needed for this SAT-level problem, but verifies the same result. A common error is using t = b/(2a) without the negative sign, leading to incorrect times like 20/(10) = 2 but ignoring the sign, or miscalculating as in choice D. Confusing maximum with roots might lead to solving h(t) = 0. Remember, for projectile motion problems, the time to maximum height is always at the vertex formula to efficiently find the answer.

Question 17

1. -1
2. 1
3. 4 (correct answer)
4. -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 18

1. 1 sec (correct answer)
2. 2 sec
3. 0.5 sec
4. 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 19

1. $1 - \frac{\sqrt{21}}{4}$
2. $1 + \frac{\sqrt{21}}{4}$ (correct answer)
3. $\frac{4 - \sqrt{21}}{4}$
4. $\frac{4 + \sqrt{21}}{8}$

Explanation: Solve $-16t^2 + 32t + 5 = 0$ to get $t = \frac{4 \pm \sqrt{21}}{4}$. The positive solution is $1 + \frac{\sqrt{21}}{4}$.

Question 20

1. The maximum value of $y$ is 3.
2. The minimum value of $y$ is 3.
3. The maximum value of $y$ is 5.
4. 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.