Introductory Olympiad Algebra - Nested Square Roots Disappear

In the first square root, the inner expression is equal to $( \sqrt{x+2} - 2)^2$ , and hence, it is equal to $| \sqrt{x+2} - 2|$ .

In the second square root, the inner expression is equal to $\sqrt{x+2} - 3) ^2$ , hence it is equal $| \sqrt{x+2} - 3|$ .

We want to find the domain where $| \sqrt{ x+2} -2 | + |\sqrt{ x+2} - 3| = 1 .$

The triangle inequality states that $|a| +|b| \geq | a + b |$ , where equality holds if and only if $a$ and $b$ have the same sign. Applying this with $a = \sqrt{x+2}-2$ and $b = 3 - \sqrt{x+2}$ , we get that

$| \sqrt{ x+2} -2 | + |\sqrt{ x+2} - 3| \geq | (3 - \sqrt{x+2} ) + ( \sqrt{x+2} - 2) | = 1 .$

Hence, for solutions, we must have equality hold throughout, which implies that $a$ and $b$ have the same sign. Since $a+b = 1 > 0$ , they must both be positive. Thus, we have

$a = \sqrt{x+2} - 2 \geq 0, b = 3 - \sqrt{x+2} \geq 0 \Rightarrow$

$2 \leq \sqrt{ x+2} \leq 3 \Rightarrow 4 \leq x +2 \leq 9 \Rightarrow 2 \leq x \leq 7 .$

Let’s look at the equation $\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1$

We can replace $x-2$ by $u^2$ so the equation becomes: $\sqrt{u^2+4-4u}+\sqrt{u^2+9-6u}=1$

This is equal to: $\sqrt{(u-2)^2}+\sqrt{(u-3)^2}=1$ or $|u-2|+|u-3|=1$ .

When solving for $3\geq u \geq 2$ we get: $u-2-u+3=1 \to 1=1$ so every number in the set $[2,3]$ is a solution for $u$ .

We now have:

$3 \geq \sqrt{x+2} \geq 2$ we square this ( $\sqrt{x+2}=u$ ) and substract 2:

$9-2 \geq x+2-2 \geq 4-2$

$7 \geq x \geq 2 \to x \in [2,7]$

Suppose x+2=b^2,

sol the equation be sqrt(b^2-4b+4)+sqrt(b^2-6b+9)=1

i.e |b-2|+|b-3|=1

i.e b=3 or b=2

Replacing b with sqrt(x+2), we get x=7 or x=2

i.e. [2,7]

because by keeping the 2and 7 the under rot comes to negative