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Question Number 8148 by Rasheed Soomro last updated on 02/Oct/16
Determine x^4 − (1/x^4 ) , if x^2 + (1/x^2 )=34 .
$$\mathrm{Determine}\:\mathrm{x}^{\mathrm{4}} −\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:,\:\mathrm{if}\:\mathrm{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{34}\:. \\ $$
Commented by sou1618 last updated on 02/Oct/16
set X=x^2 ≥0 X+(1/X)=34 (X−(1/X))^2 =(X+(1/X))^2 −4=34^2 −2^2 =36×32 ⇒ X−(1/X)=±24(√2) X^2 −(1/X^2 )=(X+(1/X))(X−(1/X))=34×(±24(√2)) =±816(√2)
$${set}\:{X}={x}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${X}+\frac{\mathrm{1}}{{X}}=\mathrm{34} \\ $$$$ \\ $$$$\left({X}−\frac{\mathrm{1}}{{X}}\right)^{\mathrm{2}} =\left({X}+\frac{\mathrm{1}}{{X}}\right)^{\mathrm{2}} −\mathrm{4}=\mathrm{34}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} =\mathrm{36}×\mathrm{32} \\ $$$$\Rightarrow \\ $$$${X}−\frac{\mathrm{1}}{{X}}=\pm\mathrm{24}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${X}^{\mathrm{2}} −\frac{\mathrm{1}}{{X}^{\mathrm{2}} }=\left({X}+\frac{\mathrm{1}}{{X}}\right)\left({X}−\frac{\mathrm{1}}{{X}}\right)=\mathrm{34}×\left(\pm\mathrm{24}\sqrt{\mathrm{2}}\right) \\ $$$$=\pm\mathrm{816}\sqrt{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 02/Oct/16
ThankS! G •^(⌢) •_(⌣^ ) ^(⌢) D approach!
$$\mathrm{Thank}\mathcal{S}!\:\:\:\mathcal{G}\:\underset{\overset{} {\smile}} {\overset{\frown} {\bullet}}\:\:\overset{\frown} {\bullet}\mathcal{D}\:\mathrm{approach}!\:\: \: \\ $$
Answered by trapti rathaur@ gmail.com last updated on 02/Oct/16
(x^2 +(1/x^2 ))^2 =x^4 +(1/x^4 )+2×x^2 ×(1/(x2)) 34×34=x^4 +(1/x^4 )+2 x^4 +(1/x^4 )=1154 using (a+b)^2 =(a−b)^2 +4ab (x^4 +(1/x^4 ))^2 =(x^4 −(1/x^4 ))^2 +4×x^4 ×(1/x^4 ) (1154)^2 =(x^4 −(1/x^4 ))^2 +4 (x^4 −(1/x^4 ))^2 =(1154)^2 −(2)^2 (x^4 −(1/x^4 ))=(√(1331712)) =816(√2) ANSWER− (x^4 −(1/x^4 ))=816(√2)
$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\overset{\mathrm{2}} {\right)}={x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2}×{x}^{\mathrm{2}} ×\frac{\mathrm{1}}{{x}\mathrm{2}} \\ $$$$\:\:\:\mathrm{34}×\mathrm{34}={x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2} \\ $$$$\:\:\:\:\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{1154} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{using}\:\left({a}+{b}\overset{\mathrm{2}} {\right)}=\left({a}−{b}\overset{\mathrm{2}} {\right)}+\mathrm{4}{ab} \\ $$$$\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\overset{\mathrm{2}} {\right)}=\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\overset{\mathrm{2}} {\right)}+\mathrm{4}×{x}^{\mathrm{4}} ×\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\ $$$$\left(\mathrm{1154}\right)^{\mathrm{2}} =\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} +\mathrm{4} \\ $$$$\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} =\left(\mathrm{1154}\right)^{\mathrm{2}} −\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)=\sqrt{\mathrm{1331712}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{816}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{ANSWER}− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)=\mathrm{816}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 02/Oct/16
ThankS! You lost one answer by taking only +ve squareroot of 1331712 : (x^4 −(1/x^4 ))=(√(1331712)) You should have written (x^4 −(1/x^4 ))=±(√(1331712))=±816(√2)
$$\mathrm{Thank}\mathcal{S}! \\ $$$$\mathrm{You}\:\mathrm{lost}\:\mathrm{one}\:\mathrm{answer}\:\mathrm{by}\:\mathrm{taking}\:\mathrm{only}\:+\mathrm{ve} \\ $$$$\mathrm{squareroot}\:\mathrm{of}\:\mathrm{1331712}\::\:\:\:\:\:\:\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)=\sqrt{\mathrm{1331712}} \\ $$$$\mathrm{You}\:\mathrm{should}\:\mathrm{have}\:\mathrm{written}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)=\pm\sqrt{\mathrm{1331712}}=\pm\mathrm{816}\sqrt{\mathrm{2}} \\ $$
Commented by trapti rathaur@ gmail.com last updated on 02/Oct/16
thanks.i forget it
$${thanks}.{i}\:{forget}\:{it} \\ $$
Answered by Rasheed Soomro last updated on 02/Oct/16
x^2 + (1/x^2 )=34 , x^4 − (1/x^4 )=? Approach: x^2 + (1/x^2 )→ { ((x+ (1/x))),((x− (1/x))) :} →x^2 − (1/x^2 )∧x^2 + (1/x^2 )→x^4 − (1/x^4 ) x^2 + (1/x^2 )=34 x^2 +2+ (1/x^2 )=34+2 ∧ x^2 −2+ (1/x^2 )=34−2 (x+(1/x))^2 =(±6)^2 ∧ (x−(1/x))^2 =(±4(√2) )^2 x+(1/x)=±6 ∧ x−(1/x)=±4(√2) ⇒x^2 − (1/x^2 )=±24(√2) {: ((x^2 − (1/x^2 )=±24(√2))),(( × )),((x^2 + (1/x^2 )=34 (given))) }⇒x^4 −(1/x^4 )=±816(√2)
$$\mathrm{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{34}\:\:,\:\:\mathrm{x}^{\mathrm{4}} −\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=? \\ $$$$\mathrm{Approach}: \\ $$$$\mathrm{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\rightarrow\begin{cases}{\mathrm{x}+\:\frac{\mathrm{1}}{\mathrm{x}}}\\{\mathrm{x}−\:\frac{\mathrm{1}}{\mathrm{x}}}\end{cases}\:\:\:\:\rightarrow\mathrm{x}^{\mathrm{2}} −\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\wedge\mathrm{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\rightarrow\mathrm{x}^{\mathrm{4}} −\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} } \\ $$$$\mathrm{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{34} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{34}+\mathrm{2}\:\:\:\wedge\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{2}+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{34}−\mathrm{2} \\ $$$$\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} =\left(\pm\mathrm{6}\right)^{\mathrm{2}} \:\:\:\:\:\wedge\:\:\:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} =\left(\pm\mathrm{4}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\pm\mathrm{6}\:\:\wedge\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\pm\mathrm{4}\sqrt{\mathrm{2}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\pm\mathrm{24}\sqrt{\mathrm{2}} \\ $$$$\left.\begin{matrix}{\mathrm{x}^{\mathrm{2}} −\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\pm\mathrm{24}\sqrt{\mathrm{2}}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{\mathrm{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{34}\:\left(\mathrm{given}\right)}\end{matrix}\right\}\Rightarrow\mathrm{x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=\pm\mathrm{816}\sqrt{\mathrm{2}} \\ $$

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