Question

Solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

Square root of x-2+8=x

Answer 1

ANSWER

Extraneous solution: x=6

Real solution: x=11

EXPLANATION

The given expression is

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

Add -8 to both sides:

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

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

Square both sides.

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

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

We expand the to get

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

Write in standard quadratic form.

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

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

Factor to get:

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

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

We check for extraneous solutions by substituting each value of x into the original equation.

When x=6

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

$\sqrt{4} + 8 =6$

$2 + 8 = 10 \ne8$

Hence x=6 is an extraneous solution.

When x=11

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

$\sqrt{9} + 8 = 11$

$3 + 8 = 11$

This statement is true.

Hence x=11 is the only solution.