SAT Nonlinear equations in one variable and systems of equations in two variables Practice Question 16 (Hard)

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

Math · Advanced MathHard

$y = x^2 + 4$
$y = mx$
In the system of equations above, $m$ is a positive constant. If the system has exactly one solution, what is the value of $m$ ?

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Answer: 4

Substitute: $x^2 + 4 = mx$ , so $x^2 - mx + 4 = 0$ . The system has exactly one solution when this quadratic has exactly one solution, so the discriminant is zero: $m^2 - 16 = 0$ . Since $m > 0$ , $m = 4$ .

Answering 16 reports $m^2$ instead of $m$ .

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