Range of the solutions for $\sqrt{x} + \sqrt{x+16} = 3$

Question

The given equation is

$$ \sqrt{x} + \sqrt{x+16} = 3$$

What is the range of the solution?

Answer 1

Clearly existence of $\sqrt x\implies x\geq 0$ for $x\in \Bbb R$.Thus, $\sqrt {x+16}$ is atleast $4\implies \sqrt x+\sqrt {x+16}\geq 4$ for $x\in \Bbb R\implies$ no real solution for $\sqrt x+\sqrt {x+16}=3$

Answer 2

Try this. Be aware $x\ge 0$ or we are dead on the spot. We begin with this.

$$\sqrt{x + 16} = 3 -\sqrt{x}$$ Squaring gives $$x + 16 = 9 - 6\sqrt{x} + x. $$ Now cancel to get $$6\sqrt{x} = -7. $$ This does not look so good. I think it's devoid of real solutions.

Answer 3

Take square of both sides and get $$ 2x+16+2\sqrt{x(x+16)}=9 $$ We thus get $$ \sqrt{x(x+16)}=-x-\frac{7}{2}. $$ Take square of both sides and get $$ x(x+16)=x^2+7x+\frac{49}{4}. $$ This is a quadratic equation in $x$, so the rest is easy.