Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

$(x-7)(x-7)=x-5$

$x^2-7x-7x+14=x-5$

$x^2-14x+14=x-5$

$x^2-14x-x+14=x-x-5$

$x^2-15x+14=-5$

$x^2-15x+14+5=-5+5$

$x^2-15x+19=0$

$(x-1)(x-19)=0$

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....

$x^2-19x-1x+19=0$

$x^2-20x+19=0$

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?

share|improve this question
    
$7^2=49\text{}$ –  J. M. Sep 21 '11 at 2:21
    
note: $-7\times -7 \neq 14$ –  Deven Ware Sep 21 '11 at 2:22
2  
(+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 :) –  4545454545SI Sep 21 '11 at 2:51

3 Answers 3

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 $x=u^2+5=9$.

share|improve this answer

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

$(x-7)^2=\sqrt{x-5}^2$

$(x-7)^2=x-5$

$(x-7)(x-7)=x-5$

$x^2-7x-7x+49=x-5$

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

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

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

$x = 9$ or $x = 6$

Now check for extraneous solutions...

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

$x^2-14x+49=x-5$

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

$x^2-9x-6x+54=0$

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

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

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

share|improve this answer
1  
Now check for extraneous solutions... –  4545454545SI Sep 21 '11 at 3:10
1  
Couldn't have said it better myself, @kso ! –  Altar Ego Sep 21 '11 at 3:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.