Solving Radical Equations $x-7= \sqrt{x-5}$

Question

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?

(+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 02:33
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 02:50
Oh, BTW: +1 for posting your solution. How I wish more askers did that. — J. M. ain't a mathematician, Sep 21 '11 at 02:52

André Nicolas · Answer 1 · 2011-09-21T05:23:49.913

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

edited Sep 21 '11 at 05:23

answered Sep 21 '11 at 05:16

André Nicolas

507,029

score 1 · Answer 2 · edited Jun 12 '20 at 10:38

1

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

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

edited Jun 12 '20 at 10:38

Community

1

answered Apr 08 '17 at 06:45

Sid

1,234

1

6 is not an extraneous root lol – Saketh Malyala Apr 08 '17 at 07:02
Root(x) - can give only positive results – Sid Apr 08 '17 at 08:01

score 1 · Answer 3 · answered Sep 21 '11 at 02:50

1

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

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

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

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

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

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

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

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

$x = 9$ or $x = 6$

Now check for extraneous solutions...

answered Sep 21 '11 at 02:50

Altar Ego

5,282

4

See this question for much on extraneous roots. – Bill Dubuque Sep 21 '11 at 03:15

score 0 · Answer 4 · answered Sep 21 '11 at 02:58

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 $x=9$

answered Sep 21 '11 at 02:58

Pedja

12,883

1

Now check for extraneous solutions... – Xiang Sep 21 '11 at 03:10
1

Couldn't have said it better myself, @kso ! – Altar Ego Sep 21 '11 at 03:13

Solving Radical Equations $x-7= \sqrt{x-5}$

4 Answers4

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