Question

question 15 of 25
which of the following are solutions to the quadratic equation? check all that apply.
$x^2 + 12x + 36 = 7$

a. $sqrt{7} + 6$
b. $-sqrt{7} - 6$
c. $-6$
d. $6$
e. $sqrt{7} - 6$
f. $sqrt{7}$

Explanation:

Step1: Factor the left - hand side

The left - hand side of the equation \(x^{2}+12x + 36=7\) is a perfect square trinomial. Recall that \(a^{2}+2ab + b^{2}=(a + b)^{2}\). For \(x^{2}+12x + 36\), we have \(a=x\), \(2ab = 12x\), so \(b = 6\) (since \(2\times x\times6=12x\)) and \(b^{2}=36\). So \(x^{2}+12x + 36=(x + 6)^{2}\). The equation becomes \((x + 6)^{2}=7\).

Step2: Take the square root of both sides

Taking the square root of both sides of the equation \((x + 6)^{2}=7\), we get \(x+6=\pm\sqrt{7}\) (because if \(y^{2}=a\) (\(a\geq0\)), then \(y = \pm\sqrt{a}\)).

Step3: Solve for x

Case 1: When \(x + 6=\sqrt{7}\), subtract 6 from both sides of the equation. We have \(x=\sqrt{7}-6\) (which can also be written as \(x=- 6+\sqrt{7}\)).
Case 2: When \(x + 6=-\sqrt{7}\), subtract 6 from both sides of the equation. We get \(x=-\sqrt{7}-6\) (which can also be written as \(x=-6 - \sqrt{7}\)).

Answer:

B. \(-\sqrt{7}-6\), E. \(\sqrt{7}-6\)