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

The given equation is

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

What is the range of the solution?

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Note $x$ cannot be negative, so $\sqrt{x+16}\ge4$. – David Mitra Aug 10 '12 at 1:30
How do you usually solve equations that have square roots in them? Try that and see what happens. – Francis Adams Aug 10 '12 at 1:33

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

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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.

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There aren't any solutions. – ncmathsadist Aug 10 '12 at 1:36

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.

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But then you have to plug the solutions you find into the original equation to see if they are real or were introduced by the squaring. – Ross Millikan Aug 10 '12 at 1:42
You are right. I noticed David Mitra's comment above. – Michel Aug 10 '12 at 4:52