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The given equation is

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

What is the range of the solution?

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    How do you usually solve equations that have square roots in them? Try that and see what happens.2012-08-10

3 Answers 3

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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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    You are right. I noticed David Mitra's comment above.2012-08-10
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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.2012-08-10
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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 $\sqrt x+\sqrt {x+16}=3$