$$ \dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$
How can I solve this equation in the easiest way?
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1Did you tried something? Can you post your attempts? – Tomás Feb 14 '13 at 17:50
3 Answers
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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.
user7530
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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.
Hagen von Eitzen
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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
Norbert
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