Question #77c0cced - SAT Suite Question Bank

Modded SAT Question Bank by Abdullah Mallik

$y = 2 x^{2} - 21 x + 64$

$y = 3 x + a$

In the given system of equations, $a$ is a constant. The graphs of the equations in the given system intersect at exactly one point, $(x, y)$ , in the xy-plane. What is the value of $x$ ?

$negative 8$
$negative 6$
$6$
$8$

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Correct Answer: C

Rationale

Choice C is correct. It's given that the graphs of the equations in the given system intersect at exactly one point, $(x, y)$ , in the xy-plane. Therefore, $(x, y)$ is the only solution to the given system of equations. The given system of equations can be solved by subtracting the second equation, $y = 3 x + a$ , from the first equation, $y = 2 x^{2} - 21 x + 64$ . This yields $y - y = (2 x^{2} - 21 x + 64) - (3 x + a)$ , or $0 = 2 x^{2} - 24 x + 64 - a$ . Since the given system has only one solution, this equation has only one solution. A quadratic equation in the form $r x^{2} + s x + t = 0$ , where $r$ , $s$ , and $t$ are constants, has one solution if and only if the discriminant, $s squared minus 4 r t$ , is equal to zero. Substituting $2$ for $r$ , $negative 24$ for $s$ , and $- a + 64$ for $t$ in the expression $s squared minus 4 r t$ yields ${(- 24)}^{2} - (4) (2) (64 - a)$ . Setting this expression equal to zero yields ${(- 24)}^{2} - (4) (2) (64 - a) = 0$ , or $8 a + 64 = 0$ . Subtracting $64$ from both sides of this equation yields $8 a = - 64$ . Dividing both sides of this equation by $8$ yields $a = - 8$ . Substituting $negative 8$ for $a$ in the equation $0 = 2 x^{2} - 24 x + 64 - a$ yields $0 = 2 x^{2} - 24 x + 64 + 8$ , or $0 = 2 x^{2} - 24 x + 72$ . Factoring $2$ from the right-hand side of this equation yields $0 = 2 (x^{2} - 12 x + 36)$ . Dividing both sides of this equation by $2$ yields $0 = x^{2} - 12 x + 36$ , which is equivalent to $0 = (x - 6) (x - 6)$ , or $0 = {(x - 6)}^{2}$ . Taking the square root of both sides of this equation yields $0 = x - 6$ . Adding $6$ to both sides of this equation yields $x equals 6$ .

Choice A is incorrect. This is the value of $a$ , not $x$ .

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

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