What are the solutions to the equation x^2 + 6x - 4 = 0? | SAT Nonlinear equations in one variable and systems of equations in two variables Question, Answer Explained

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Math · Advanced MathMedium

What are the solutions to the equation $x^2 + 6x - 4 = 0$ ?

$x = -3 \pm \sqrt{13}$

$x = -3 \pm \sqrt{5}$

$x = 3 \pm \sqrt{13}$

$x = -6 \pm \sqrt{13}$

Show answer and explanation

Answer: A

Complete the square. Add $4$ to both sides: $x^2 + 6x = 4$ . Add $9$ to both sides: $x^2 + 6x + 9 = 13$ , so $(x + 3)^2 = 13$ . Then $x + 3 = \pm\sqrt{13}$ , so $x = -3 \pm \sqrt{13}$ .

Choice B solves $x^2 + 6x + 4 = 0$ , which has the wrong sign on the constant term.

Why the other choices are wrong

B. This solves x² + 6x + 4 = 0. With -4, completing the square gives (x + 3)² = 13.
C. This has the wrong sign. Since (x + 3)² = 13, x = -3 ± √13.
D. Completing the square uses (x + 3)², so the solutions are centered at -3, not -6.

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