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

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 $x^2-15x+19$

• $7^2=49\text{}$ Commented Sep 21, 2011 at 2:21
• note: $-7\times -7 \neq 14$ Commented Sep 21, 2011 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! Commented Sep 21, 2011 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 :)) Commented Sep 21, 2011 at 2:50
• Very Very Detail Error Solution :) Commented Sep 21, 2011 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 $x=u^2+5=9$.

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

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

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

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

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

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

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

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

$x = 9$ or $x = 6$

Now check for extraneous solutions...

• See this question for much on extraneous roots. Commented Sep 21, 2011 at 3:15

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

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

• 6 is not an extraneous root lol Commented Apr 8, 2017 at 7:02
• Root(x) - can give only positive results
– Sid
Commented Apr 8, 2017 at 8:01

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

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

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

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

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

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

• Now check for extraneous solutions... Commented Sep 21, 2011 at 3:10
• Couldn't have said it better myself, @kso ! Commented Sep 21, 2011 at 3:13