If solutions come in $\pm x$ pairs, what must the single solution be?

Try squaring $\sqrt{a+x}+\sqrt{a-x}$ and compare $\sqrt{a^2-x^2}$ to $a$.

Enter

• y = sqrt(a+x) + sqrt(a-x)
• y = 10

Frоm aniko.ai

Create a slider for a.

For a chosen value of a, look at how many intersection points the two graphs have. If you see two intersections, they will appear at opposite $x$-values (symmetric about $x=0$).

Move the slider for a until the line y=10 touches the curve at exactly one point (it should occur at $x=0$). Then read that a value from the slider and compute $4a$.

If $x$ is a solution, then replacing $x$ by $-x$ gives

so $-x$ is also a solution.

Therefore, if there is exactly one real solution, it cannot be a nonzero number (because then there would be at least two solutions, $x$ and $-x$). The only possible single solution is $x=0$.

Let

Square $F(x)$:

Since $a^2-x^2\le a^2$, we have $\sqrt{a^2-x^2}\le a$, with equality only when $x=0$. Thus

so $F(x)\le 2\sqrt{a}$, and the maximum value $2\sqrt{a}$ happens only at $x=0$.

To have exactly one real solution, the horizontal line $y=10$ must touch the graph of $y=F(x)$ at its single peak (otherwise there would be 0 intersections or 2 symmetric intersections). So

Therefore,

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