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This the Pre-Calculus Problem:

$x-7= \sqrt{x-5}$

So far I did it like this and I'm not understanding If I did it wrong.

$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:

$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.









Now this is where I'm stuck because when I tried to see if I got the right numbers in the parentheses I got this....



As you may see I'm doing something bad because I don't get $x^2-15x+19$

Could anyone please help me and tell me what I'm doing wrong?

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$7^2=49\text{}$ – J. M. Sep 21 '11 at 2:21
note: $-7\times -7 \neq 14$ – Deven Ware Sep 21 '11 at 2:22
(+1) For showing work. If there weren't a mistake somewhere along the way, then you wouldn't have needed to ask, I guess! – Altar Ego Sep 21 '11 at 2:33
As others have pointed out, you made a mistake in expanding $(x-7)^2$. It should be $(x^2 - 14x + 49)$. Can you now redo the problem and post the progress. (If it is correct, you can also post it as an answer and accept it :)) – Srivatsan Sep 21 '11 at 2:50
Very Very Detail Error Solution :) – Xiang Sep 21 '11 at 2:51

We can avoid squaring both sides. Let $x-5=u^2$, where $u \ge 0$. Then $\sqrt{x-5}=u$. Also, $x=u^2+5$, so $x-7=u^2-2$. Thus our equation can be rewritten as $$u^2-2=u, \quad\text{or equivalently}\quad u^2-u-2=0.$$ But $$u^2-u-2=(u-2)(u+1).$$ Thus the solutions of $u^2-u-2=0$ are $u=2$ and $u=-1$. Since $u \ge 0$, we reject the solution $u=-1$.

We conclude that $u=2$, and therefore $x=u^2+5=9$.

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$x-7= \sqrt{x-5}$





$x^2 - 15x + 54 = 0$

$(x - 9)(x - 6) = 0$

$x - 9 = 0 $ or $x - 6 = 0$

$x = 9$ or $x = 6$

Now check for extraneous solutions...

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See this question for much on extraneous roots. – Bill Dubuque Sep 21 '11 at 3:15






$x=6$, or $x=9$

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Now check for extraneous solutions... – Xiang Sep 21 '11 at 3:10
Couldn't have said it better myself, @kso ! – Altar Ego Sep 21 '11 at 3:13

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