Question

Square root of the quantity x minus 2 end quantity plus 8 equals x

Answer 1

ANSWER

x=11

EXPLANATION

We want to solve the equation:

$\sqrt{x - 2} + 8 = x$

Add -8 to both sides:

$\sqrt{x - 2} = x - 8$

Square both sides of the equation to get:

$(\sqrt{x - 2} )^{2} =( x - 8)^{2}$

$x - 2 = {x}^{2} - 16x + 64$

Rewrite in the form ax² +bx+c=0

${x}^{2} - 16x - x + 64 + 2 = 0$

${x}^{2} - 17x + 66 = 0$

Factor to get,

$(x - 11)(x - 6) = 0$

This implies that,

$x = 11 \: or \: x = 6$

Checking for extraneous solutions.

When x=6, the original equation becomes,

$\sqrt{6- 2} + 8 = 6$

$\sqrt{4 } + 8 = 6$

2+8= 6 is not true.

x=6 is an extraneous solution.

When x=11, the original equation becomes,

$\sqrt{11- 2} + 8 = 11$

$\sqrt{9} + 8 = 11$

$3 + 8 = 11$

This is true, hence x=11 is the only solution.