y = -x^2 + 4 y = mx + 8 In the system of equations, m is a... | SAT Nonlinear equations in one variable and systems of equations in two variables Question, Answer Explained

Question bank / Nonlinear equations in one variable and systems of equations in two variables

Math · Advanced MathHard

$y = -x^2 + 4$
$y = mx + 8$
In the system of equations, $m$ is a constant. For which of the following values of $m$ does the system have no real solutions?

$-5$

$-4$

$3$

$6$

Show answer and explanation

Answer: C

Set the expressions equal: $-x^2 + 4 = mx + 8$ , so $x^2 + mx + 4 = 0$ . The system has no real solutions when the discriminant is negative: $m^2 - 4(1)(4) < 0$ , so $m^2 < 16$ and $-4 < m < 4$ . Of the choices, only $3$ is in this interval.

Choice B makes the discriminant exactly $0$ , so the line touches the parabola at one point instead of missing it.

Why the other choices are wrong

A. With m = -5, the discriminant is 25 - 16 = 9, which is positive, so there are two solutions.
B. With m = -4, the discriminant is 0, so the line touches the parabola at exactly one point.
D. With m = 6, the discriminant is 36 - 16 = 20, which is positive, so there are two solutions.

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