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

How can I solve this equation in the easiest way?

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1  
Did you tried something? Can you post your attempts? –  Tomás Feb 14 '13 at 17:50

4 Answers 4

up vote 15 down vote accepted

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
2  
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

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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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

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

Do not forget to verify the results ;)

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