Question

question 3 of 10
which of the following are solutions to the equation below?
check all that apply.
$x^2 + 6x + 9 = 20$

a. $x = 2\sqrt{5} - 3$
b. $x = -\sqrt{3} + 20$
c. $x = \sqrt{3} + 20$
d. $x = -2\sqrt{5} + 3$
e. $x = 2\sqrt{5} + 3$
f. $x = -2\sqrt{5} - 3$

Explanation:

Step1: Factor the left - hand side

The left - hand side of the equation \(x^{2}+6x + 9=20\) is a perfect square trinomial. We know that \(a^{2}+2ab + b^{2}=(a + b)^{2}\), for \(x^{2}+6x + 9\), where \(a=x\) and \(b = 3\) (since \(2ab=2\times x\times3 = 6x\)), so \(x^{2}+6x + 9=(x + 3)^{2}\). The equation becomes \((x + 3)^{2}=20\).

Step2: Take the square root of both sides

Taking the square root of both sides of the equation \((x + 3)^{2}=20\), we get \(x+3=\pm\sqrt{20}\).

Step3: Simplify \(\sqrt{20}\)

We know that \(\sqrt{20}=\sqrt{4\times5}=\sqrt{4}\times\sqrt{5}=2\sqrt{5}\). So \(x + 3=\pm2\sqrt{5}\).

Step4: Solve for \(x\)

Case 1: When \(x + 3=2\sqrt{5}\), then \(x=2\sqrt{5}-3\).

Case 2: When \(x + 3=- 2\sqrt{5}\), then \(x=-2\sqrt{5}-3\).

Answer:

A. \(x = 2\sqrt{5}-3\), F. \(x=-2\sqrt{5}-3\)