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Yanka [14]
3 years ago
9

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

Mathematics
1 answer:
RideAnS [48]3 years ago
3 0

<u>ANSWER</u>

Extraneous solution: x=6

Real solution: x=11

<u>EXPLANATION</u>

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.

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