Question

solve for x.
\sqrt{(36x^2 - 19)} + 20 = 36
a no solution
b \frac{5\sqrt{11}}{6} only
c \pm \frac{5\sqrt{11}}{6}
d -\frac{5\sqrt{11}}{6} only

Explanation:

Step1: Isolate the square root term

Subtract 20 from both sides of the equation $\sqrt{36x^{2}-19}+20 = 36$.
We get $\sqrt{36x^{2}-19}=36 - 20=16$.

Step2: Eliminate the square root

Square both sides of the equation $\sqrt{36x^{2}-19}=16$ to get rid of the square root.
So, $(\sqrt{36x^{2}-19})^{2}=16^{2}$, which simplifies to $36x^{2}-19 = 256$.

Step3: Solve for \(x^{2}\)

Add 19 to both sides of the equation $36x^{2}-19 = 256$.
We have $36x^{2}=256 + 19=275$.
Then divide both sides by 36: $x^{2}=\frac{275}{36}$.

Step4: Solve for \(x\)

Take the square root of both sides. Remember that when we take the square root of a number, we get both positive and negative solutions.
So, $x=\pm\sqrt{\frac{275}{36}}=\pm\frac{\sqrt{275}}{\sqrt{36}}$.
Simplify $\sqrt{275}=\sqrt{25\times11}=5\sqrt{11}$ and $\sqrt{36} = 6$.
Thus, $x=\pm\frac{5\sqrt{11}}{6}$.

Answer:

C. $\pm\frac{5\sqrt{11}}{6}$