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]
Accepted Solution
A:
Answer:x = 36Step-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.