Find the value of x-x2-1-x2-4x-1-x-6

( x^2+1/x^2 ) 4 ( x+1/x )+6 ( x+1/x )^2=x^2+1/x^2+2x.1/x we need to subtract 2 in order to combasate ( x+1/x )^2 4 ( x+1/x ) 2+6=6 x+1/xt t^2 4t+4=0 (t 2)^2=0 t ...

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Byju's Answer

Standard XII

Mathematics

Definition of Sets

Find the valu...

Question

Find the value of x: $(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$

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Solution

$(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$
${(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2 x . \frac{1}{x}$
we need to subtract 2 in order to combasate
${(x + \frac{1}{x})}^{2} - 4 (x + \frac{1}{x}) - 2 + 6 = 6$
$x + \frac{1}{x} t$
$t^{2} - 4 t + 4 = 0$
$(t - 2)^{2} = 0$
$t = 1$
$x + \frac{1}{x} = 2$
$x^{2} + 1 - 2 x = 0$
$(x - 1)^{2} = 0$
x =1

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Find the value of x: (x2+1x2)−4(x+1x)+6

(x2+1x2)−4(x+1x)+6 (x+1x)2=x2+1x2+2x.1x we need to subtract 2 in order to combasate (x+1x)2−4(x+1x)−2+6=6 x+1xt t2−4t+4=0 (t−2)2=0 t=1 x+1x=2 x2+1−2x=0 (x−1)2=0 x =1

Find the value of x: $(x^{2} + \frac{1}{x^{2}}) - 4 (x + \frac{1}{x}) + 6$