Question 514·Hard·Math

Here $h(x)=2$ gives the output. Replace $h(x)$ with the rule from the question: $\sqrt{x+21}-\sqrt{x+5}=2$. It’s the difference of the roots that equals $2$, not either root by itself.

Define $h(x)$ in Desmos, then type $h(x_1)\sim2\{x_1\ge-5\}$. The subscript makes $x_1$ the input Desmos finds; $\sim$ tells it to match the output, and the braces enforce the given domain. Under PARAMETERS, Desmos shows $x_1=4$. So the input that gives an output of $2$ is $x=4$. Grid in 4.

Let $A=\sqrt{x+21}$ and $B=\sqrt{x+5}$. Replacing $h(x)$ with its rule turns the given output into $A-B=2$. Naming the roots lets you keep track of their difference while finding the input $x$.

Squaring a square root gives the expression inside it, so subtract the squares: $A^2-B^2=(x+21)-(x+5)=16$. The $x$ terms cancel. You now know this difference even though you don’t know $x$.

A difference of squares factors as $A^2-B^2=(A-B)(A+B)$: multiplying out makes the middle terms cancel. Use the two differences you know: $(A-B)(A+B)=16$. Substitute $A-B=2$: $2(A+B)=16$. Divide by $2$: $A+B=8$. The known difference and the difference of squares give you the sum of the roots.

Subtract the difference $A-B=2$ from the sum $A+B=8$: $(A+B)-(A-B)=8-2$. Simplify, since the $A$ terms cancel: $2B=6$.

Divide by $2$: $B=3$. This is the smaller square root, $\sqrt{x+5}$, not the input $x$.

Since $B=\sqrt{x+5}$, square both sides of $\sqrt{x+5}=3$ to remove the radical: $x+5=9$. The $9$ is $x+5$, so there’s one more move.

Subtract $5$: $x=4$. This input is allowed because it satisfies the stated domain $x\ge-5$. Grid in 4.