Quadratic Equations
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SAT Math › Quadratic Equations
A ball's height (in feet) is modeled by $h(t)=-16t^2+64t+80$. At what time $t$ (in seconds) is the ball at a height of 120 feet for the later of the two times?
(4 + $\sqrt{5}$)/2
(4 + $\sqrt{6}$)/2
(4 + 2$\sqrt{6}$)/4
(4 - $\sqrt{6}$)/2
Explanation
Solve $-16t^2+64t+80=120$ to get $2t^2-8t+5=0$, so $t=\frac{4\pm\sqrt{6}}{2}$ and the later time is $\frac{4+\sqrt{6}}{2}$. The other choices use a wrong discriminant, incorrect simplification, or select the earlier time.
A ball is thrown upward from a platform; its height in feet is given by $h(t)=-16t^2+32t+48$, where $t$ is time in seconds. When does it hit the ground?
-1
1
3
4
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.
Solve $\sqrt{x+5}=x-1$. Which value of $x$ satisfies the equation?
-4
-1
1
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.
For the quadratic $2x^2 + kx + 8 = 0$ to have exactly one real solution, what must $k$ equal?
k = -8 or k = 8
k = 0
k = 8 only
k = 4 or k = -4
Explanation
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.
For the quadratic function $y = x^2 - 6x + 5$, what is the $x$-coordinate of the vertex?
-6
-3
3
6
Explanation
The vertex $x$-coordinate is $-\frac{b}{2a}=\frac{6}{2}=3$. The other options come from sign or factor-of-2 errors.
The graph of $y=2x^2+px+8$ crosses the x-axis at $x=-2$. What is the value of $p$?
-8
4
8
16
Explanation
Set $y=0$ at $x=-2$: $2(-2)^2+p(-2)+8=0\Rightarrow 16-2p=0$, so $p=8$. The other choices reflect sign or arithmetic mistakes (e.g., taking $p=-8$ or miscomputing $2(-2)^2$).
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?
0.5
1
2
32
Explanation
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 $b$.
The quadratic $f(x)=x^2+kx+9$ has exactly one real solution. What is the positive value of $k$?
-6
0
6
36
Explanation
Exactly one real root means the discriminant $k^2-36=0$, so $|k|=6$ and the positive value is 6. -6 ignores 'positive,' while 0 and 36 come from misusing the discriminant.
For $y = x^2 - 6x + 11$, what is the minimum value of $y$?
-2
2
3
11
Explanation
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.
A ball's height is modeled by $h(t) = -16t^2 + 32t + 5$, where $t$ is in seconds. When does it hit the ground? (Give the positive solution.)
$1 + \frac{\sqrt{21}}{4}$
$\frac{4 - \sqrt{21}}{4}$
$\frac{4 + \sqrt{21}}{8}$
$1 - \frac{\sqrt{21}}{4}$
Explanation
Solve $-16t^2 + 32t + 5 = 0$ to get $t = \frac{4 \pm \sqrt{21}}{4}$. The positive solution is $1 + \frac{\sqrt{21}}{4}$.