How to solve equation $\frac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$?

$$\dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$

How can I solve this equation in the easiest way?

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Did you tried something? Can you post your attempts? – Tomás Feb 14 '13 at 17:50

You can solve it algebraically by isolating one of the square roots, squaring both sides, solving for the other square root, and squaring both sides again. This will give you a quadratic equation in $x^2$.

But you can also argue more cleverly directly from the function. First, notice that the LHS is undefined for $|x|<4$. For $|x|\geq 4$, $$\frac{\sqrt{x^2-16}+\sqrt{x^2-9}}{2} \geq \frac{\sqrt{x^2-9}}{2}\geq \frac{\sqrt{7}}{2} > 1,$$ so your equation has no (real) solutions.

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Very nice way, +1. – 1015 Feb 14 '13 at 18:05
That is, has no real solutions. – Lubin Feb 14 '13 at 18:28
@Lubin I've made an edit. – user7530 Feb 14 '13 at 19:06

Multiply by $2$ to obtain $$\tag1\sqrt{x^2-16}+\sqrt{x^2-9}=2$$ and multiply by the conjugate $\sqrt{x^2-16}-\sqrt{x^2-9}$ to obtain $$\tag2 -\frac72=\frac12((x^2-16)-(x^2-9))=\sqrt{x^2-16}-\sqrt{x^2-9}.$$ Add $(1)$ and $(2)$ and divide by $2$ to obtain $$\sqrt {x^2-16}=-\frac34$$ Which has no real solution.

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possible duplicate – Norbert Feb 14 '13 at 18:07
Apologies. Norbert's solution is correct. – zaarcis Feb 14 '13 at 18:38

We have the equation $$\frac{1}{2}(\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$ Let's multiply it by $\sqrt{x^2-16} - \sqrt{x^2-9}$ to get $$-\frac{7}{2}=\sqrt{x^2-16} - \sqrt{x^2-9}$$ Hence $$2\sqrt{x^2-16} = (\sqrt{x^2-16} + \sqrt{x^2-9}) + (\sqrt{x^2-16} - \sqrt{x^2-9})=2-\frac{7}{2}<0$$ This is imossible so there is no real solution for this equation

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Square both sides, isolate the square root and square again.

Do not forget to verify the results ;)

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