Question

question 3 of 10
which values of x are solutions to the equation below?
check all that apply.
$3x^2 - 42 = 33$

a. $x = -sqrt{15}$
b. $x = 5$
c. $x = -5$
d. $x = -4$
e. $x = 4$
f. $x = sqrt{15}$

Explanation:

Step1: Isolate the quadratic term

Add 42 to both sides of the equation $3x^{2}-42 = 33$.
$3x^{2}-42 + 42=33 + 42$
$3x^{2}=75$

Step2: Solve for \(x^{2}\)

Divide both sides by 3.
$\frac{3x^{2}}{3}=\frac{75}{3}$
$x^{2} = 25$ (Wait, no, 75 divided by 3 is 25? Wait, original equation: $3x^{2}-42 = 33$. So $3x^{2}=33 + 42=75$. Then $x^{2}=\frac{75}{3} = 25$? Wait, no, 75 divided by 3 is 25? Wait, 325=75. Then $x^{2}=25$, so $x=\pm5$. Wait, but let's check again. Wait, maybe I made a mistake. Wait, 3x² -42 =33. So 3x²=33 +42=75. Then x²=25. So x=5 or x=-5. Wait, but the options have A and F with sqrt(15). Wait, maybe I miscalculated. Wait, 3x² -42 =33. Let's do it again. 3x²=33 +42=75. Then x²=25. So x=±5. So B and C are correct. Wait, but let's check the options. Option B: x=5, option C: x=-5. Let's verify by plugging back. For x=5: 3(5)² -42=325 -42=75 -42=33. Correct. For x=-5: 3(-5)² -42=3*25 -42=75 -42=33. Correct. So B and C are solutions. Wait, but why are A and F there? Wait, maybe I made a mistake in the equation. Wait, the equation is 3x² -42 =33? Or is it 3x² -42 = 33? Wait, 3x²=75, x²=25, x=±5. So B and C are correct.

Wait, maybe the original equation was 3x² - 42 = 33? Let's check again. 3x² = 33 +42=75. x²=25. x=5 or x=-5. So B (x=5) and C (x=-5) are correct.

Answer:

B. \( x = 5 \), C. \( x = -5 \)