Question #3d12b1e0 - SAT Suite Question Bank

ID: 3d12b1e0

Modded SAT Question Bank by Abdullah Mallik

$- 16 x^{2} - 8 x + c = 0$

In the given equation, $c$ is a constant. The equation has exactly one solution. What is the value of $c$ ?

Toggle Answer

Tip: Press CTRL/Command to toggle answer

Correct Answer: -1

Rationale

The correct answer is $negative 1$ . A quadratic equation in the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants, has exactly one solution when its discriminant, $b squared minus 4 a c$ , is equal to $0$ . In the given equation, $- 16 x^{2} - 8 x + c = 0$ , $a = - 16$ and $b = - 8$ . Substituting $negative 16$ for $a$ and $negative 8$ for $b$ in $b squared minus 4 a c$ yields ${(- 8)}^{2} - 4 (- 16) (c)$ , or $64 plus 64 c$ . Since the given equation has exactly one solution, $64 + 64 c = 0$ . Subtracting $64$ from both sides of this equation yields $64 c = - 64$ . Dividing both sides of this equation by $64$ yields $c = - 1$ . Therefore, the value of $c$ is $negative 1$ .

Question Difficulty: Hard