Determine x^4 − (1/x^4 ) , if x^2 + (1/x^2 )=34 .


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Question Number 8148 by Rasheed Soomro last updated on 02/Oct/16

$Determine x^{4} - \frac{1}{x^{4}}, if x^{2} + \frac{1}{x^{2}} = 34 .$
Commented by sou1618 last updated on 02/Oct/16

$s e t X = x^{2} ⩾ 0$ $X + \frac{1}{X} = 34$ ${(X - \frac{1}{X})}^{2} = {(X + \frac{1}{X})}^{2} - 4 = 34^{2} - 2^{2} = 36 \times 32$ $\Rightarrow$ $X - \frac{1}{X} = \pm 24 \sqrt{2}$ $X^{2} - \frac{1}{X^{2}} = (X + \frac{1}{X}) (X - \frac{1}{X}) = 34 \times (\pm 24 \sqrt{2})$ $= \pm 816 \sqrt{2}$
Commented by Rasheed Soomro last updated on 02/Oct/16

$Thank S! G \underset{\overset{}{⌣}}{\overset{⌢}{∙}} \overset{⌢}{∙} D approach!$
	Answered by trapti rathaur@ gmail.com last updated on 02/Oct/16

	$Missing \left or extra \right$ $34 \times 34 = x^{4} + \frac{1}{x^{4}} + 2$ $x^{4} + \frac{1}{x^{4}} = 1154$ $Missing \left or extra \right$ $Missing \left or extra \right$ ${(1154)}^{2} = {(x^{4} - \frac{1}{x^{4}})}^{2} + 4$ ${(x^{4} - \frac{1}{x^{4}})}^{2} = {(1154)}^{2} - {(2)}^{2}$ $(x^{4} - \frac{1}{x^{4}}) = \sqrt{1331712}$ $= 816 \sqrt{2}$ $A N S W E R -$ $(x^{4} - \frac{1}{x^{4}}) = 816 \sqrt{2}$
	Commented by Rasheed Soomro last updated on 02/Oct/16

	$Thank S!$ $You lost one answer by taking only + ve$ $squareroot of 1331712 : (x^{4} - \frac{1}{x^{4}}) = \sqrt{1331712}$ $You should have written$ $(x^{4} - \frac{1}{x^{4}}) = \pm \sqrt{1331712} = \pm 816 \sqrt{2}$
	Commented by trapti rathaur@ gmail.com last updated on 02/Oct/16

	$t h a n k s . i f o r g e t i t$
	Answered by Rasheed Soomro last updated on 02/Oct/16
	$x^2 + (1/x^2 )=34 , x^4 − (1/x^4 )=? Approach: x^2 + (1/x^2 )→ { ((x+ (1/x))),((x− (1/x))) :} →x^2 − (1/x^2 )∧x^2 + (1/x^2 )→x^4 − (1/x^4 ) x^2 + (1/x^2 )=34 x^2 +2+ (1/x^2 )=34+2 ∧ x^2 −2+ (1/x^2 )=34−2 (x+(1/x))^2 =(±6)^2 ∧ (x−(1/x))^2 =(±4(√2) )^2 x+(1/x)=±6 ∧ x−(1/x)=±4(√2) ⇒x^2 − (1/x^2 )=±24(√2) {: ((x^2 − (1/x^2 )=±24(√2))),(( × )),((x^2 + (1/x^2 )=34 (given))) }⇒x^4 −(1/x^4 )=±816(√2)$
	$x^{2} + \frac{1}{x^{2}} = 34, x^{4} - \frac{1}{x^{4}} = ?$ $Approach :$ $x^{2} + \frac{1}{x^{2}} \to {\begin{cases} x + \frac{1}{x} \\ x - \frac{1}{x} \end{cases} \to x^{2} - \frac{1}{x^{2}} \land x^{2} + \frac{1}{x^{2}} \to x^{4} - \frac{1}{x^{4}}$ $x^{2} + \frac{1}{x^{2}} = 34$ $x^{2} + 2 + \frac{1}{x^{2}} = 34 + 2 \land x^{2} - 2 + \frac{1}{x^{2}} = 34 - 2$ ${(x + \frac{1}{x})}^{2} = {(\pm 6)}^{2} \land {(x - \frac{1}{x})}^{2} = {(\pm 4 \sqrt{2})}^{2}$ $x + \frac{1}{x} = \pm 6 \land x - \frac{1}{x} = \pm 4 \sqrt{2}$ $\Rightarrow x^{2} - \frac{1}{x^{2}} = \pm 24 \sqrt{2}$ $\begin{matrix} x^{2} - \frac{1}{x^{2}} = \pm 24 \sqrt{2} \\ \times \\ x^{2} + \frac{1}{x^{2}} = 34 (given) \end{matrix}} \Rightarrow x^{4} - \frac{1}{x^{4}} = \pm 816 \sqrt{2}$
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