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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.

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

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

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

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

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

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

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

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

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?

  • 0
    Oh, BTW: +1 for posting your solution. How I wish more askers did that.2011-09-21

4 Answers 4

2

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

0

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

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

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

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

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

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

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

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

$x = 9$ or $x = 6$

Now check for extraneous solutions...

  • 4
    See [this question](http://math.stackexchange.com/q/61252/242) for much on extraneous roots.2011-09-21
0

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

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

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

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

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

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

  • 1
    Couldn't have said it better myself, @kso !2011-09-21
0

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

Check the domain first

$x-5 \geq0 \cap x-7 \geq0$

$x \geq7$

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

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

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

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

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

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

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

$x = 9$ or $x = 6$

checking the domain x=6 is an extraneous root.

x = 9

  • 0
    Root(x) - can give only positive results2017-04-08