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I don't know what can I substitute for $x$ so that equation becomes satisfied. Any assistance will be greatly valued.

Thanks!

Sufyan Naeem
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3 Answers3

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Use $A=4+\sqrt{15}$. That'll give you $A^{-1}=\dfrac{1}{4+\sqrt{15}}=\dfrac{4-\sqrt{15}}{16-15}=4-\sqrt{15}$

The equation then becomes,

$$A^{x/3}+A^{-x/3}=8\implies (A^{x/3})^2-8A^{x/3}+1=0$$

Solve this quadratic in terms of $A^{x/3}$ using the quadratic formula and you'll get,

$$A^{x/3}=\frac{8\pm\sqrt{64-4}}{2}=\frac{8\pm2\sqrt{15}}{2}=4\pm\sqrt{15}\\ \implies (4+\sqrt{15})^{x/3}=(4+\sqrt{15})^{\pm 1}\\ \implies \frac{x}{3}=\pm 1 \implies x=\pm 3$$

Prasun Biswas
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  • @AleksXPO, if my answer helped you, you can accept it by clicking on the check mark beside my answer. :) – Prasun Biswas Apr 16 '15 at 18:19
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    Yes, I'll do that, however I have to wait a few more minutes until the system let's me accept an answer. –  Apr 16 '15 at 18:19
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    Do you see how else you could see +-3 is the solution? – steedsnisps Apr 16 '15 at 18:21
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    Of course @PrasunBiswas has answered you greatly in a very short time. It is his right that you accept his answer :) – Sufyan Naeem Apr 16 '15 at 18:22
  • Yes, 1/3*3 = 1, so the root is gone in both parenthesis, and 4+4 = 8. –  Apr 16 '15 at 18:23
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    @wowlolbrommer, a nice approach to this problem is to note the symmetry of solutions. If $x=\alpha$ is a solution, $x=(-\alpha)$ is a solution too. Establishing the symmetry argument is trivial, I guess. And it is almost obvious that $x=3$ is a solution. By the argument of symmetry, $x=(-3)$ will also be a solution. – Prasun Biswas Apr 16 '15 at 18:24
  • it is the same approach as mine – Dr. Sonnhard Graubner Apr 16 '15 at 18:30
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Hint: $$\sqrt[3]{4-\sqrt{15}}\cdot \sqrt[3]{4+\sqrt{15}}=1$$ from here we get $$t^x+\frac{1}{t^x}=8$$

Dr. Sonnhard Graubner
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$$4-\sqrt{15}=t^3,\sqrt{15}=4-t^3$$ $$4+\sqrt{15}=8-t^3$$ $$(\sqrt[3]{8-t^3})^x=8-t^x$$ $$(8-t^3)^{\frac{x}{3}}=8-t^x$$ $$(8-t^3)^x=(8-t^x)^3$$ $x=3$ obviousley is a solution

Adi Dani
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