Question

If x} +36=0" align="absmiddle" class="latex-formula">, what is the value of x?
A. $6$

B. $6^{2}$

C. $6^{3}$

D. $6^{4}$

Answer 1

Answer:

x = 36

Step-by-step explanation:

$x - 12\sqrt{x} + 36 = 0$

Subtract x and 36 from both sides.

$-12\sqrt{x} = -x - 36$

Divide both sides by -1.

$12\sqrt{x} = x + 36$

Square both sides.

$144x = x^2 + 72x + 1296$

Subtract 144x from both sides.

$0 = x^2 - 72x + 1296$

Factor the right side.

$0 = (x - 36)^2$

$x - 36 = 0$

$x = 36$

Since the solution of the equation involved squaring both sides, we musty check the answer for possible extraneous solutions.

Check x = 36:

$x - 12\sqrt{x} + 36 = 0$

$36 - 12\sqrt{36} + 36 = 0$

$36 - 12\times 6 + 36 = 0$

$36 - 72 + 36 = 0$

$0 = 0$

Since 0 = 0 is a true statement, the solution x = 36 is a valid solution.