# Square roots equations

I had to solve this problem: $$\sqrt{x} + \sqrt{x-36} = 2$$ So I rearranged the equation this way: $$\sqrt{x-36} = 2 - \sqrt{x}$$ Then I squared both sides to get: $$x-36 = 4 - 4\sqrt{x} + x$$ Then I did my simple algebra: $$4\sqrt{x} = 40$$ $$\sqrt{x} = 10$$ $$x = 100$$ The problem is that when I go back and plug my $x$-value into the equation, it doesn't work. $$\sqrt{100} + \sqrt{100-36} = 2$$ $$10+8 = 2$$ Which is obviously wrong.

• there is no real solution. One needs $x \geq 36$ to get $x-36 \geq 0,$ but then the other $\sqrt x \geq 6$ which is bigger than $2$ Jul 6, 2015 at 3:40
• So why did I get x=100? Jul 6, 2015 at 3:43
• If I asked you to solve $\sqrt t = -7$ and you squared both sides, what would happen? Jul 6, 2015 at 3:45
• You would get t=49 whereas there is actually no real solution. Right? Jul 6, 2015 at 3:49

Your argument shows that if there is a real root, that root must be $100$. But there is no real root. For $\sqrt{x-36}$ exists only if $x\ge 36$, and in that case $$\sqrt{x}+\sqrt{x-36}\ge 6.$$

Remark: When you squared both sides of $\sqrt{x-36}=2-\sqrt{x}$, you were introducing the additional possibility $\sqrt{x-36}=-(2-\sqrt{x})$. And indeed $x=100$ is a solution of that equation. The $x=100$ is an extraneous root that comes from the fact that the equations $\sqrt{x-36}=2-\sqrt{x}$ and $(\sqrt{x-36})^2=(2-\sqrt{x})^2$ are not equivalent.

• And thats why root equations might have extraneous solutions? If so, I think I got it. Jul 6, 2015 at 3:50
• Oh ok. I wrote my comment before the last edit you made. Thanks this really makes sense now. Jul 6, 2015 at 3:52
• Yes, the process of squaring can produce "extraneous" solutions, since the process of squaring is not (uniquely) reversible. Jul 6, 2015 at 3:53
• Is that the same idea that f(x)=x² is somehow Not a one-to-one function? Jul 6, 2015 at 4:03
• Exactly. Squaring erases information. If we are told that a number was squared and the result was $16$, we do not know whether the number was $4$ or $-4$. So whenever one squares, one should, as you did, check at the end whether the answer(s) obtained work. Checking is anyway not a bad idea in general, it is usually cheap. Jul 6, 2015 at 4:07

Avoid squaring whenever possible as it immediately introduces extraneous root(s) which demand(s) exclusion.

$$(\sqrt x-\sqrt{x-36})(\sqrt x+\sqrt{x-36})=x-(x-36)=36$$

$$\implies\sqrt x+\sqrt{x-36}=2\ \ \ \ (1)\iff \sqrt x-\sqrt{x-36}=\dfrac{36}2=18\ \ \ \ (2)$$

But $\sqrt{x-36}\ge0\implies \sqrt x+\sqrt{x-36}\ge\sqrt x-\sqrt{x-36}$

Can you take it from here?

Method $\#1:$

As for real $a,\sqrt a\ge0\ \ \ \ (1)$

$(\sqrt x+\sqrt{36-x})^2=36+2\sqrt{x(36-x)}\ge36$

$\implies\sqrt x+\sqrt{36-x}\ge6\ \ \ \ (2)$ or $\sqrt x+\sqrt{36-x}\le-6\ \ \ \ (3)$

Finally $(1)$ nullifies $(3)$

Method $\#2:$

WLOG let $\sqrt x=6\csc2y$ where $0<2y\le\dfrac\pi2\implies\sqrt{x-36}=+6\cot2y$

$\sqrt x+\sqrt{36-x}=6\cdot\dfrac{1+\cos2y}{\sin2y}=6\cot y$

Now $0<2y\le\dfrac\pi2\implies0<y\le\dfrac\pi4\implies\cot0>\cot y\ge\dfrac\pi4=1$ as $\cot y$ is decreasing in $\left[0,\dfrac\pi2\right]$

Surprised this hasn’t been mentioned yet. This method is helpful in general.

Let $$u=\sqrt{x}$$ and $$v=\sqrt{x-36}$$. Then $$u^2-v^2=36$$ and $$u+v=2$$. With $$u^2-v^2=(u+v)(u-v)$$, $$u-v=18$$. Thus $$u=10$$ and $$x=100$$.