Quadratic Equations - SAT Math

Solve the equation $2x^2-7x-4=0$ . Which pair of values gives all real solutions for $x$ ?
Answer: $x=4,\,-\frac{1}{2}$ The question asks for the real solutions to the quadratic equation 2x² - 7x - 4 = 0. To solve this, apply the quadratic formula x = [-b ± √(b² - 4ac)] / (2a), where a = 2, b = -7, and c = -4. This gives x = [7 ± √(49 + 32)] / 4 = [7 ± √81] / 4 = [7 ± 9] / 4, resulting in x = 16/4 = 4 and x = -2/4 = -1/2. Factoring could also work: find numbers that multiply to 2*(-4) = -8 and add to -7, which are -8 and 1, so (2x + 1)(x - 4) = 0, yielding the same roots. A common error is misapplying the quadratic formula by using the wrong signs for b or c, leading to incorrect roots like those in choice A. Another mistake is confusing the factoring pairs, resulting in options like C or D. For quadratics that factor nicely, try factoring first before resorting to the formula to save time on tests.
Which of the following equations represents a parabola that opens downward and has a vertex at $(2, -3)$ ?
Answer: $y = -(x - 2)^2 - 3$ This question asks for a parabola that opens downward with vertex at $(2, -3)$ . A parabola opens downward when the coefficient of the squared term is negative, and the vertex form is $y = a(x-h)^2 + k$ where $(h,k)$ is the vertex. For vertex $(2, -3)$ , we need $y = a(x-2)^2 + (-3) = a(x-2)^2 - 3$ , and since it opens downward, $a$ must be negative. Only choice D has the form $y = -(x-2)^2 - 3$ with a negative coefficient and the correct vertex coordinates.
Solve the equation $3x^2+2x-8=0$ . Which pair gives all real solutions?
Answer: $-2,\ \tfrac{4}{3}$ To solve 3x² + 2x - 8 = 0, we can use the quadratic formula: x = (-b ± √(b² - 4ac))/(2a). Here, a = 3, b = 2, and c = -8. The discriminant is b² - 4ac = 4 - 4(3)(-8) = 4 + 96 = 100. So x = (-2 ± √100)/(2·3) = (-2 ± 10)/6. This gives x = (-2 + 10)/6 = 8/6 = 4/3 or x = (-2 - 10)/6 = -12/6 = -2. The solutions are x = -2 and x = 4/3. Common errors include sign mistakes in the formula or arithmetic errors when calculating the discriminant. Always double-check your discriminant calculation and simplify fractions completely.
Which expression is equivalent to $x^2-10x+21$ factored completely?
Answer: $(x-3)(x-7)$ To factor $x^2-10x+21$ , we need two numbers that multiply to 21 and add to -10. These numbers are -3 and -7, since $(-3)×(-7)=21$ and $(-3)+(-7)=-10$ . Therefore, $x^2-10x+21=(x-3)(x-7)$ . Common mistakes include getting the wrong sign combinations or choosing factors of 21 that don't add to -10, like -1 and -21. Always verify your factoring by expanding: $(x-3)(x-7)=x^2-7x-3x+21=x^2-10x+21$ .
The height of a ball (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: $2$ This 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, the vertex occurs at t = -b/(2a), which gives the time of maximum height since a = -5 is negative, indicating a downward-opening parabola. Here, t = -20 / (2 × -5) = -20 / -10 = 2 seconds. You could complete the square: h(t) = -5(t² - 4t) + 1 = -5(t² - 4t + 4 - 4) + 1 = -5((t - 2)² - 4) + 1 = -5(t - 2)² + 20 + 1, showing the vertex at t = 2. A common error is forgetting the negative sign in a, leading to an incorrect t like -20 / (2 × 5) = -2, which is not possible for time. Use the vertex formula for projectile motion problems to quickly find maximum or minimum times.
A parabola has $x$ -intercepts at $x=-1$ and $x=5$ and passes through the point $(0,-10)$ . Which equation represents this parabola in standard form?
Answer: $y=2x^2-8x-10$ This question asks for a parabola's equation given its x-intercepts at $x=-1$ and $x=5$ , passing through $(0,-10)$ . In factored form, the parabola is $y=a(x+1)(x-5)$ for some value $a$ . To find $a$ , substitute the point $(0,-10)$ : $-10=a(0+1)(0-5)=a(1)(-5)=-5a$ , so $a=2$ . The equation is $y=2(x+1)(x-5)$ , which expands to $y=2(x^2-5x+x-5)=2(x^2-4x-5)=2x^2-8x-10$ . Common errors include using $a=-2$ or mixing up the signs in the factors. Always verify your answer by checking that it passes through the given points.
Solve the equation $3x^2+2x-8=0$ . Choose the correct pair of solutions. Possible incorrect paths include factoring incorrectly or making an arithmetic error when applying the quadratic formula.
Answer: $x=\frac{4}{3},-2$ We need to solve $3x^2+2x-8=0$ . Using the quadratic formula with $a=3$ , $b=2$ , $c=-8$ : $x=\frac{-2\pm\sqrt{4-4(3)(-8)}}{2(3)}=\frac{-2\pm\sqrt{4+96}}{6}=\frac{-2\pm\sqrt{100}}{6}=\frac{-2\pm 10}{6}$ . This gives $x=\frac{-2+10}{6}=\frac{8}{6}=\frac{4}{3}$ and $x=\frac{-2-10}{6}=\frac{-12}{6}=-2$ . We can verify by factoring: $3x^2+2x-8=(3x-4)(x+2)=0$ . Common errors include sign mistakes in the discriminant or arithmetic errors when simplifying fractions. Always check your solutions by substituting back into the original equation.
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?
Answer: $2$ 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.
For the quadratic $2x^2 + kx + 8 = 0$ to have exactly one real solution, what must $k$ equal?
Answer: k = -8 or k = 8 Exactly one real solution requires discriminant $k^2-4(2)(8)=0\Rightarrow k^2=64$ , so $k=\pm 8$ . Other options miss one value or use the wrong discriminant.
A projectile's height is given by $h(t) = -16t^2 + 32t + 48$ , where $t$ is time in seconds. At what time does the projectile reach its maximum height?
Answer: 1 The maximum occurs at the vertex with $t = -\frac{b}{2a} = -\frac{32}{2(-16)} = 1$ . Other choices result from sign errors or using an incorrect formula.
For the quadratic function $f(x)=-2x^2+8x+1$ , what is the axis of symmetry?
Answer: $x=2$ We need to find the axis of symmetry for $f(x)=-2x^2+8x+1$ . The axis of symmetry for a quadratic $ax^2+bx+c$ is the vertical line $x=-\frac{b}{2a}$ . Here, $a=-2$ and $b=8$ , so the axis of symmetry is $x=-\frac{8}{2(-2)}=-\frac{8}{-4}=2$ . A common mistake is forgetting the negative sign in the formula or confusing it with a horizontal line (y = constant). Remember that the axis of symmetry is always a vertical line passing through the vertex's x-coordinate.
A projectile's height (in feet) is modeled by $h(t)=-16t^2+32t+5$ , where $t$ is time in seconds. At what time does it reach its maximum height?
Answer: 1 The maximum occurs at the vertex $t=-\frac{b}{2a}=-\frac{32}{2(-16)}=1$ . 0.5 and 2 result from arithmetic slips with $-b/(2a)$ , and 32 is just $b$ .
What is the vertex of the parabola $y=2x^2-8x+3$ ? Give your answer as an ordered pair $(h,k)$ .
Answer: $(2,\ -5)$ 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.
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 asks when the ball reaches maximum height given $h(t)=-5t^2+20t+1$ . Since the parabola opens downward (negative leading coefficient), the maximum occurs at the vertex. Using $t=-\frac{b}{2a}$ where $a=-5$ and $b=20$ , we get $t=-\frac{20}{2(-5)}=-\frac{20}{-10}=2$ seconds. We can verify: at $t=2$ , $h(2)=-5(4)+20(2)+1=-20+40+1=21$ meters. A common error is to use the discriminant formula or to find when $h(t)=0$ instead of finding the vertex. For optimization problems with quadratics, always identify whether you need a maximum (vertex of downward parabola) or minimum (vertex of upward parabola).
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?
Answer: $6$ 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.
What are the solutions to the equation $x^2-5x+6=0$ ? Choose the pair of values of $x$ that makes the equation true.
Answer: $x=2,3$ 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: $(x-2)(x-3)=x^2-3x-2x+6=x^2-5x+6$ ✓.
In the coordinate plane, consider the parabola $y=-(x+1)^2+9$ . What are the coordinates of its vertex?
Answer: $(-1,9)$ For the parabola $y=-(x+1)^2+9$ in vertex form $y=a(x-h)^2+k$ , we identify $a=-1$ , $h=-1$ , and $k=9$ . The vertex is at $(h,k)=(-1,9)$ . Note that when we have $(x+1)^2$ , this means $(x-(-1))^2$ , so $h=-1$ , not $h=1$ . A common mistake is to misread the signs and report $(-1,-9)$ or $(1,9)$ . Remember: in vertex form $y=a(x-h)^2+k$ , the vertex is always at $(h,k)$ , and $(x+1)^2$ means $h=-1$ .
A rectangle has area $48$ square units. Its length is $x$ units and its width is $(x-2)$ units. Which equation can be used to find possible values of $x$ ?
Answer: $x^2-2x-48=0$ This question asks for the equation to find x where a rectangle has area 48, length x, width x-2. Set up x(x-2)=48, so $x^2$ -2x -48=0. This is a quadratic from the area formula. Solving via factoring: (x-8)(x+6)=0, x=8 or -6 (discard negative), but the question wants the equation. Completing the square or formula isn't needed here. A key error is sign mistakes, like B's +2x giving wrong width relation. Choice D assumes a square, ignoring the x-2. Model real-world problems into quadratics and check signs carefully.
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 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.
For $y = x^2 - 6x + 11$ , what is the minimum value of $y$ ?
Answer: 2 The vertex is at $x=\frac{-b}{2a}=\frac{6}{2}=3$ , and $y(3)=9-18+11=2$ , the minimum. 3 is the x-coordinate, -2 is a sign error, and 11 is the constant term.

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