Quadratic Equations - PSAT Math

1. A parabola is defined by $y=(x-4)^2-9$. Which statement gives the vertex of the parabola? Do not confuse the vertex coordinates with the axis of symmetry or with an $x$-intercept.
   Answer: $(4,-9)$ The question requires identifying the vertex of the parabola given by y = (x - 4)² - 9. The equation is in vertex form y = a(x - h)² + k, where the vertex is (h, k), so here h = 4 and k = -9. To verify, note that the vertex occurs at the minimum or maximum point, and since a = 1 > 0, it opens upward with the minimum at (4, -9). Completing the square is unnecessary as it's already in vertex form, but if starting from standard form, group the x terms and add/subtract (b/2a)². A key error is mistaking the signs, such as thinking it's ( -4, 9) by misreading the form. For test-taking, recognize vertex form quickly to avoid unnecessary calculations.
2. A parabola has x-intercepts at $x=-2$ and $x=4$ and passes through the point $(0,-8)$. Which equation could represent the parabola?
   Answer: $y=(x+2)(x-4)$ This question asks for the equation of a parabola with x-intercepts at $x=-2$ and $x=4$ that passes through $(0,-8)$. If the x-intercepts are at $-2$ and $4$, the parabola can be written as $y=a(x+2)(x-4)$ for some value $a$. To find $a$, we use the fact that the parabola passes through $(0,-8)$: $-8=a(0+2)(0-4)=a(2)(-4)=-8a$. Solving for $a$: $a=1$. Therefore, the equation is $y=(x+2)(x-4)$. A common error is forgetting to find the value of $a$ or assuming $a=1$ without checking. When given x-intercepts and another point, always use the factored form and substitute to find the leading coefficient.
3. Which factorization is equivalent to $6x^{2}-7x-3$ ?
   Answer: $(3x+1)(2x-3)$ When you encounter a quadratic expression that needs to be factored, you're looking for two binomials that multiply to give the original expression. The key is understanding how the coefficients and signs work together. To verify which factorization is correct, let's expand the given options using FOIL (First, Outer, Inner, Last). For option D, $(3x+1)(2x-3)$: - First: $3x \cdot 2x = 6x^2$ - Outer: $3x \cdot (-3) = -9x$ - Inner: $1 \cdot 2x = 2x$ - Last: $1 \cdot (-3) = -3$ Combining: $6x^2 - 9x + 2x - 3 = 6x^2 - 7x - 3$ ✓ Now let's see why the other options fail: Option A: $(2x-1)(3x+3)$ expands to $6x^2 + 6x - 3x - 3 = 6x^2 + 3x - 3$. The middle term is $+3x$, not $-7x$. Option B: $(3x-1)(2x+3)$ expands to $6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3$. The middle term has the wrong sign (positive instead of negative). Option C: $(2x+1)(3x-3)$ expands to $6x^2 - 6x + 3x - 3 = 6x^2 - 3x - 3$. The middle term is $-3x$, not $-7x$. Study tip: When checking factorizations, always expand your answer using FOIL to verify. Pay special attention to the middle term—it's where most errors occur. The middle term comes from adding the outer and inner products, so track those signs carefully.
4. Which equation is equivalent to $y=x^2+8x+7$ written in factored form?
   Answer: $y=(x+1)(x+7)$ 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.
5. 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.
   Answer: $x=3,\,-\tfrac{1}{2}$ 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.
6. What is the axis of symmetry of the parabola $y=-3x^2+12x-7$?
   Answer: $x=2$ 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.
7. 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.
   Answer: $t=2$ 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.
8. The parabola defined by $y=x^{2}+4x+c$ intersects the $x$-axis at exactly one point. What is the value of $c$ ?
   Answer: $4$ When a parabola intersects the x-axis at exactly one point, this means the quadratic equation has exactly one solution, which occurs when the discriminant equals zero. For any quadratic $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$. In the equation $y = x^2 + 4x + c$, we have $a = 1$, $b = 4$, and the coefficient $c$ is what we need to find. Setting the discriminant equal to zero: $4^2 - 4(1)(c) = 0$, which gives us $16 - 4c = 0$. Solving for $c$: $16 = 4c$, so $c = 4$. We can verify this by completing the square: $y = x^2 + 4x + 4 = (x + 2)^2$. This parabola has its vertex at $(-2, 0)$, touching the x-axis at exactly one point. Choice A ($c = -4$) would give discriminant $16 - 4(-4) = 32 > 0$, meaning two x-intercepts. Choice B ($c = 0$) yields discriminant $16 - 0 = 16 > 0$, also giving two intercepts. Choice C ($c = 8$) produces discriminant $16 - 32 = -16 < 0$, meaning no real x-intercepts at all. The answer is D. Study tip: Remember that for a parabola to touch the x-axis at exactly one point, set the discriminant $b^2 - 4ac = 0$. This concept appears frequently on standardized tests, so practice identifying when a quadratic has one, two, or zero real solutions.
9. On the coordinate plane, the parabola $y=(x+1)^2-4$ is shown. What is the axis of symmetry of the parabola?
   Answer: $x=-1$ 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.
10. Solve the equation $x^2+4x+1=0$. Which pair of solutions is correct?
    Answer: $x=-2\pm\sqrt{3}$ This question asks us to solve $x^2+4x+1=0$ using the quadratic formula. With $a=1$, $b=4$, and $c=1$, we get $x=\frac{-4\pm\sqrt{16-4}}{2}=\frac{-4\pm\sqrt{12}}{2}=\frac{-4\pm 2\sqrt{3}}{2}=-2\pm\sqrt{3}$. We can verify by completing the square: $x^2+4x+1=0$ becomes $(x+2)^2-4+1=0$, or $(x+2)^2=3$, giving $x+2=\pm\sqrt{3}$, so $x=-2\pm\sqrt{3}$. A common error is making arithmetic mistakes under the square root or forgetting to simplify the fraction. When using the quadratic formula, always simplify your answer completely by factoring out common terms from the numerator.
11. Which quadratic equation has exactly $4$ and $-\dfrac{2}{3}$ as its roots?
    Answer: $3x^{2}-10x-8=0$ When you see a question asking which quadratic equation has specific roots, you need to work backwards from the roots to build the equation. If a quadratic has roots $r_1$ and $r_2$, then the equation can be written as $(x - r_1)(x - r_2) = 0$. With roots $4$ and $-\frac{2}{3}$, start by writing: $(x - 4)(x - (-\frac{2}{3})) = 0$, which simplifies to $(x - 4)(x + \frac{2}{3}) = 0$. Expanding this expression: $x^2 + \frac{2}{3}x - 4x - 4 \cdot \frac{2}{3} = x^2 + \frac{2}{3}x - 4x - \frac{8}{3} = x^2 - \frac{10}{3}x - \frac{8}{3} = 0$ To eliminate fractions, multiply everything by 3: $3x^2 - 10x - 8 = 0$. This matches choice C. Choice A has the wrong sign on the middle term ($-\frac{14}{3}x$ instead of $-\frac{10}{3}x$) and uses $\frac{14}{3}$ instead of the correct $\frac{10}{3}$. Choice B has the correct middle term magnitude but wrong sign ($+10x$ instead of $-10x$). Choice D has both the wrong magnitude ($\frac{14}{3}$) and wrong sign (positive instead of negative) for the middle term. Remember: you can always verify your answer by substituting the given roots back into your equation. Both $x = 4$ and $x = -\frac{2}{3}$ should make the equation equal zero. This double-check method is especially useful when working with fractions.
12. For which values of $m$ does the equation $x^{2}+(m-3)x+m=0$ have two distinct real solutions?
    Answer: $m<1 \text{ or } m>9$ When you encounter a quadratic equation and need to determine when it has two distinct real solutions, you're working with the discriminant. For any quadratic $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$. Two distinct real solutions exist when the discriminant is positive. In the equation $x^2 + (m-3)x + m = 0$, we have $a = 1$, $b = (m-3)$, and $c = m$. The discriminant becomes: $(m-3)^2 - 4(1)(m) > 0$ Expanding: $m^2 - 6m + 9 - 4m > 0$, which simplifies to $m^2 - 10m + 9 > 0$. Factoring: $(m-1)(m-9) > 0$. This inequality is satisfied when both factors have the same sign. Since the parabola $y = (m-1)(m-9)$ opens upward, it's positive when $m < 1$ or $m > 9$. Choice A ($m \leq 1$ and $m \geq 9$) incorrectly uses "and" instead of "or," and includes the boundary points where the discriminant equals zero (giving one repeated solution, not two distinct solutions). Choice B ($1 < m < 9$) gives the interval where the discriminant is negative, resulting in no real solutions. Choice C ($m = 1$ or $m = 9$) represents exactly where the discriminant equals zero, giving one repeated real solution rather than two distinct ones. Choice D correctly identifies $m < 1$ or $m > 9$ as the solution. Remember: For quadratics, discriminant > 0 means two distinct real roots, discriminant = 0 means one repeated root, and discriminant < 0 means no real roots.
13. 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.
    Answer: $y=x^2-2x-3$ 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.
14. Rewrite $y=x^2+6x+1$ in vertex form $y=a(x-h)^2+k$. Which expression is equivalent?
    Answer: $y=(x+3)^2-8$ 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.
15. The parabola $y=2(x-3)^2-8$ is written in vertex form. What is the vertex of the parabola?
    Answer: $(3,-8)$ 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.
16. What are the solutions to the equation $2x^2-5x-3=0$?
    Answer: $x=-\frac{1}{2},\ 3$ 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.
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?
    Answer: $t=2$ 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).
18. 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?
    Answer: $t=2$ 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.
19. 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?
    Answer: $x=1$ 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.
20. 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?
    Answer: Two; $x=0$ and $x=1$ 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 $x = 0$ or $x = 1$. 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.

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