PSAT 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 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.

Question 1

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

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 = rac{-3 pm $\sqrt{9 + 16}$}{4} = rac{-3 pm 5}{4}), giving (rac{2}{4}=0.5) and (rac{-8}{4}=-2). Factoring works well with small integers; formula confirms. Common error: incorrect factoring signs. Verify by plugging solutions back in.

Question 2

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

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 3

1. $x=4$
2. $x=6$ (correct answer)
3. $x=8$
4. $x=12$

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 = -rac{b}{2a} = -rac{24}{2(-2)} = rac{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 4

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

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

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 6

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

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 7

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

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 8

1. $t=1$
2. $t=2$ (correct answer)
3. $t=4$
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 9

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

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 10

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

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 11

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

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 12

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

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 = -rac{b}{2a} = -rac{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.

Question 13

1. $y=(x+1)(x+7)$ (correct answer)
2. $y=(x-1)(x-7)$
3. $y=(x+7)^2$
4. $y=(x+4)^2-7$

Explanation: This question asks for the factored form of (y = $x^2$ + 8x + 7). Factor by finding numbers that multiply to 7 and add to 8, which are 1 and 7, so (y = (x + 1)(x + 7)). Verify by expanding: ($x^2$ + 7x + x + 7 = $x^2$ + 8x + 7). The quadratic formula gives roots -1 and -7, confirming factors. Avoid errors like choosing numbers that multiply incorrectly, such as -1 and -7 which add to -8. When factoring quadratics with positive constant and leading coefficient 1, look for positive factors if b>0.

Question 14

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

Explanation: This question asks us to find the solutions to the quadratic equation $2x^2-5x-3=0$. We can solve this by factoring: we need two numbers that multiply to $(2)(-3)=-6$ and add to $-5$. These numbers are $-6$ and $1$, so we rewrite as $2x^2-6x+x-3=0$, then factor by grouping: $2x(x-3)+1(x-3)=0$, giving us $(2x+1)(x-3)=0$. Setting each factor to zero: $2x+1=0$ gives $x=-\frac{1}{2}$, and $x-3=0$ gives $x=3$. A common error is making arithmetic mistakes when factoring or solving for x. When you see a quadratic equation on the PSAT, first check if it factors nicely before using the quadratic formula.

Question 15

1. $t=\frac{3}{2}$ (correct answer)
2. $t=\frac{48}{16}$
3. $t=3$
4. $t=\frac{5}{48}$

Explanation: This question asks for the time when the ball reaches maximum height, given the height function $h(t)=-16t^2+48t+5$. For a quadratic function in the form $at^2+bt+c$, the maximum (when $a<0$) occurs at $t=-\frac{b}{2a}$. Here, $a=-16$ and $b=48$, so the maximum occurs at $t=-\frac{48}{2(-16)}=-\frac{48}{-32}=\frac{48}{32}=\frac{3}{2}$. We can verify this is a maximum since the coefficient of $t^2$ is negative, meaning the parabola opens downward. A common error is forgetting the negative sign in the formula or making arithmetic mistakes with fractions. For projectile motion problems, remember that the vertex formula gives you the time of maximum height directly.

Question 16

1. $t=1$
2. $t=2$ (correct answer)
3. $t=4$
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 = -rac{b}{2a} = -rac{20}{2(-5)} = rac{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 17

1. $t=1$
2. $t=2$ (correct answer)
3. $t=4$
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 18

1. $x=1$ (correct answer)
2. $x=2$
3. $y=1$
4. $y=2$

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 = rac{-1 + 3}{2} = 1). You can also think of it as (x = -rac{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 19

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

Explanation: The vertex form $y=a(x-h)^2+k$ directly shows the vertex at $(h,k)$. In the equation $y=2(x-3)^2-8$, we identify $h=3$ and $k=-8$, so the vertex is at $(3,-8)$. Note that the sign inside the parentheses is opposite to the x-coordinate: $(x-3)$ means the vertex has x-coordinate positive 3. A common error is thinking $(x-3)$ gives vertex at $x=-3$, but remember the form is $(x-h)$, not $(x+h)$. When given vertex form, you can immediately read off the vertex without any calculations.

Question 20

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

Explanation: The axis of symmetry of a parabola is a vertical line that passes through the vertex. For the parabola $y=(x+1)^2-4$ in vertex form, the vertex is at $(-1,-4)$ because we have $(x-h)$ where $h=-1$. The axis of symmetry is the vertical line through this vertex, which has equation $x=-1$. Students often confuse the axis of symmetry (a vertical line $x=$ constant) with horizontal lines ($y=$ constant). Remember that parabolas with vertical axes have vertical lines of symmetry, making the answer format $x=$ value.