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

If 4 (x^9 )^(1/(4 )) −9 (x^9 )^(1/(8 )) + 4 = 0 , then (x^9 )^(1/(4 )) + (x^(−9) )^(1/(4 )) =?

$${If}\:\mathrm{4}\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:−\mathrm{9}\:\sqrt[{\mathrm{8}\:}]{{x}^{\mathrm{9}} }\:+\:\mathrm{4}\:=\:\mathrm{0}\:,\:{then}\: \\ $$$$\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{−\mathrm{9}} }\:=? \\ $$

Answered by benjo_mathlover last updated on 18/Oct/20

4(x^(9/8) )^2 −9(x^(9/8) )+4 = 0 letting λ = x^(9/8) ⇒4λ^2 −9λ+4 = 0 ⇒λ^2 −(9/4)λ+1 = 0 , have the roots are λ_1 = x^(9/8) and λ_2 = x^(−(9/8)) ⇒we want to compute the value of x^(9/4) +x^(−(9/4)) it equals to λ_1 ^2 +λ_2 ^2 = (λ_1 +λ_2 )^2 −2λ_1 .λ_2 ⇒λ_1 ^2 + λ_2 ^2 = ((9/4))^2 −2.1 = ((81−32)/(16)) = ((49)/(16)) therefore : x^(9/4) + x^(−(9/4)) = ((49)/(16))

$$\:\mathrm{4}\left({x}^{\frac{\mathrm{9}}{\mathrm{8}}} \right)^{\mathrm{2}} −\mathrm{9}\left({x}^{\frac{\mathrm{9}}{\mathrm{8}}} \right)+\mathrm{4}\:=\:\mathrm{0} \\ $$$${letting}\:\lambda\:=\:{x}^{\frac{\mathrm{9}}{\mathrm{8}}} \Rightarrow\mathrm{4}\lambda^{\mathrm{2}} −\mathrm{9}\lambda+\mathrm{4}\:=\:\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\lambda+\mathrm{1}\:=\:\mathrm{0}\:,\:{have}\:{the}\:{roots} \\ $$$${are}\:\lambda_{\mathrm{1}} \:=\:{x}^{\frac{\mathrm{9}}{\mathrm{8}}} \:{and}\:\lambda_{\mathrm{2}} \:=\:{x}^{−\frac{\mathrm{9}}{\mathrm{8}}} \\ $$$$\Rightarrow{we}\:{want}\:{to}\:{compute}\:{the}\:{value}\:{of}\:{x}^{\frac{\mathrm{9}}{\mathrm{4}}} +{x}^{−\frac{\mathrm{9}}{\mathrm{4}}} \:{it} \\ $$$${equals}\:{to}\:\lambda_{\mathrm{1}} ^{\mathrm{2}} +\lambda_{\mathrm{2}} ^{\mathrm{2}} \:=\:\left(\lambda_{\mathrm{1}} +\lambda_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\lambda_{\mathrm{1}} .\lambda_{\mathrm{2}} \\ $$$$\Rightarrow\lambda_{\mathrm{1}} ^{\mathrm{2}} \:+\:\lambda_{\mathrm{2}} ^{\mathrm{2}} \:=\:\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{2}.\mathrm{1}\:=\:\frac{\mathrm{81}−\mathrm{32}}{\mathrm{16}}\:=\:\frac{\mathrm{49}}{\mathrm{16}} \\ $$$$\:{therefore}\::\:{x}^{\frac{\mathrm{9}}{\mathrm{4}}} \:+\:{x}^{−\frac{\mathrm{9}}{\mathrm{4}}} \:=\:\frac{\mathrm{49}}{\mathrm{16}} \\ $$

Commented by MJS_new last updated on 17/Oct/20

answer is ((49)/(16))

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

Commented by bramlexs22 last updated on 17/Oct/20

how sir?

$${how}\:{sir}? \\ $$

Commented by MJS_new last updated on 17/Oct/20

your answer is the same, just expand it!

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

Answered by MJS_new last updated on 17/Oct/20

we don′t have to solve for x let t=x^(9/8) ⇒ the problem turns to if 4t^2 −9t+4=0 then t^2 +t^(−2) =? 4t^2 −9t+4=0 t^2 −(9/4)t+1=0 the solutions of t^2 +pt+1 are t_1 =((−p−(√(p^2 −4)))/2) and t_2 =t_1 ^(−1) ⇒ t^2 +t^(−2) =t_1 ^2 +t_2 ^2 t_1 ^2 =((p^2 −2+p(√(p^2 −4)))/2)∧t_2 ^2 =((p^2 −2+p(√(p^2 −4)))/2) ⇒ t^2 +t^(−2) =t_1 ^2 +t_2 ^2 =p^2 −2 p=−(9/4) ⇒ answer is ((49)/(16))

$$\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$$\mathrm{let}\:{t}={x}^{\mathrm{9}/\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{problem}\:\mathrm{turns}\:\mathrm{to} \\ $$$$\mathrm{if}\:\mathrm{4}{t}^{\mathrm{2}} −\mathrm{9}{t}+\mathrm{4}=\mathrm{0}\:\mathrm{then}\:{t}^{\mathrm{2}} +{t}^{−\mathrm{2}} =? \\ $$$$\mathrm{4}{t}^{\mathrm{2}} −\mathrm{9}{t}+\mathrm{4}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:{t}^{\mathrm{2}} +{pt}+\mathrm{1}\:\mathrm{are} \\ $$$${t}_{\mathrm{1}} =\frac{−{p}−\sqrt{{p}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\:\mathrm{and}\:{t}_{\mathrm{2}} ={t}_{\mathrm{1}} ^{−\mathrm{1}} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} +{t}^{−\mathrm{2}} ={t}_{\mathrm{1}} ^{\mathrm{2}} +{t}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${t}_{\mathrm{1}} ^{\mathrm{2}} =\frac{{p}^{\mathrm{2}} −\mathrm{2}+{p}\sqrt{{p}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\wedge{t}_{\mathrm{2}} ^{\mathrm{2}} =\frac{{p}^{\mathrm{2}} −\mathrm{2}+{p}\sqrt{{p}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} +{t}^{−\mathrm{2}} ={t}_{\mathrm{1}} ^{\mathrm{2}} +{t}_{\mathrm{2}} ^{\mathrm{2}} ={p}^{\mathrm{2}} −\mathrm{2} \\ $$$${p}=−\frac{\mathrm{9}}{\mathrm{4}}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{49}}{\mathrm{16}} \\ $$

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