The given equation is
$ \sqrt{x} + \sqrt{x+16} = 3$
What is the range of the solution?
The given equation is
$ \sqrt{x} + \sqrt{x+16} = 3$
What is the range of the solution?
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.
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.
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 $\sqrt x+\sqrt {x+16}=3$