If [tex]x-12\sqrt{x} +36=0[/tex], what is the value of x?

A. [tex]6[/tex]

B. [tex]6^{2}[/tex]

C. [tex]6^{3}[/tex]

D. [tex]6^{4}[/tex]

Respuesta :

Answer:

x = 36

Step-by-step explanation:

[tex] x - 12\sqrt{x} + 36 = 0 [/tex]

Subtract x and 36 from both sides.

[tex] -12\sqrt{x} = -x - 36 [/tex]

Divide both sides by -1.

[tex] 12\sqrt{x} = x + 36 [/tex]

Square both sides.

[tex] 144x = x^2 + 72x + 1296 [/tex]

Subtract 144x from both sides.

[tex] 0 = x^2 - 72x + 1296 [/tex]

Factor the right side.

[tex] 0 = (x - 36)^2 [/tex]

[tex] x - 36 = 0 [/tex]

[tex] x = 36 [/tex]

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

Check x = 36:

[tex] x - 12\sqrt{x} + 36 = 0 [/tex]

[tex] 36 - 12\sqrt{36} + 36 = 0 [/tex]

[tex] 36 - 12\times 6 + 36 = 0 [/tex]

[tex] 36 - 72 + 36 = 0 [/tex]

[tex] 0 = 0 [/tex]

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