Square roots equations

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

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 $\left[0,\dfrac\pi2\right]$

Answer 4

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 $x=100$.