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Question Number 118435 by bramlexs22 last updated on 17/Oct/20

$$If4\sqrt[4]{x^9}-9\sqrt[8]{x^9}+4=0,then$$$$\sqrt[4]{x^9}+\sqrt[4]{x^{-9}}=?$$

Answered by benjo_mathlover last updated on 18/Oct/20

$$4{\left(x^{\frac{9}{8}}\right)}^2-9\left(x^{\frac{9}{8}}\right)+4=0$$$$letting\lambda =x^{\frac{9}{8}}\Rightarrow 4{\lambda }^2-9\lambda +4=0$$$$\Rightarrow {\lambda }^2-\frac{9}{4}\lambda +1=0,havetheroots$$$$are{\lambda }_1=x^{\frac{9}{8}}and{\lambda }_2=x^{-\frac{9}{8}}$$$$\Rightarrow wewanttocomputethevalueof{x}^{\frac{9}{4}}+x^{-\frac{9}{4}}it$$$$equalsto{\lambda }_1^2+{\lambda }_2^2={({\lambda }_1+{\lambda }_2)}^2-2{\lambda }_1.{\lambda }_2$$$$\Rightarrow {\lambda }_1^2+{\lambda }_2^2={\left(\frac{9}{4}\right)}^2-2.1=\frac{81-32}{16}=\frac{49}{16}$$$$therefore:x^{\frac{9}{4}}+x^{-\frac{9}{4}}=\frac{49}{16}$$

Commented by MJS_new last updated on 17/Oct/20

$$\mathrm{answer}\mathrm{is}\frac{49}{16}$$

Commented by bramlexs22 last updated on 17/Oct/20

$$howsir?$$

Commented by MJS_new last updated on 17/Oct/20

$$\mathrm{your}\mathrm{answer}\mathrm{is}\mathrm{the}\mathrm{same},\mathrm{just}\mathrm{expand}\mathrm{it}!$$

Answered by MJS_new last updated on 17/Oct/20

$$\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{to}\mathrm{solve}\mathrm{for}x$$$$\mathrm{let}t=x^{9/8}$$$$\Rightarrow \mathrm{the}\mathrm{problem}\mathrm{turns}\mathrm{to}$$$$\mathrm{if}4t^2-9t+4=0\mathrm{then}{t}^2+t^{-2}=?$$$$4t^2-9t+4=0$$$$t^2-\frac{9}{4}t+1=0$$$$\mathrm{the}\mathrm{solutions}\mathrm{of}{t}^2+pt+1\mathrm{are}$$$$t_1=\frac{-p-\sqrt{p^2-4}}{2}\mathrm{and}{t}_2=t_1^{-1}$$$$\Rightarrow t^2+t^{-2}=t_1^2+t_2^2$$$$t_1^2=\frac{p^2-2+p\sqrt{p^2-4}}{2}\wedge t_2^2=\frac{p^2-2+p\sqrt{p^2-4}}{2}$$$$\Rightarrow t^2+t^{-2}=t_1^2+t_2^2=p^2-2$$$$p=-\frac{9}{4}\Rightarrow \mathrm{answer}\mathrm{is}\frac{49}{16}$$

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