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Solve for the sqrt(x-2)+8=x

User MESepehr
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1 Answer

7 votes

Answer: x = 11

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Work Shown:


√(x-2)+8=x\\√(x-2)=x-8\\\left(√(x-2)\right)^2=(x-8)^2 \text{ square both sides}\\x-2=x^2-16x+64\\0=x^2-16x+64-x+2\\0=x^2-17x+66\\x^2-17x+66 = 0\\(x-11)(x-6) = 0\\x-11 = 0 \text{ or } x-6 = 0\\x = 11 \text{ or } x = 6\\

Those are the two possible solutions . We need to check each possible solution.

Plug in x = 11 then simplify. If we get the same number on both sides, then x = 11 is confirmed.


√(x-2)+8=x\\√(11-2)+8=11\\√(9)+8=11\\3+8=11\\11=11\\

We get 11 on both sides, so the solution x = 11 is confirmed.

Repeat for x = 6 as well


√(x-2)+8=x\\√(6-2)+8=6\\√(4)+8=6\\2+8=6\\10=6\\

We do not get the same thing on both sides, so x = 6 is not a solution. We consider this extraneous.

User Shua
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