Question

If x^2 − 6x + 1 = 0, then the value of (x^4 + 1/x^2) (x^2 + 1) is

Answer 1

Answer:

Given,

x2−6x+1=0

Since x=0 is not a solution of the given equation, we can divide the equation by x

x−6+1x=0

x+1x=6(1)

Given expression whose value is required

y=x4+1x2x2+1

Taking out 1x2 common from numerator

1x2 x6+1x2+1

We can simplify the above expression

By identify, a3+b3=(a+b)(a2−ab+b2)

Thus,

x6+1=(x2)3+(12)3

=(x2+1)(x4−x2+1)

Dividing by (x2+1) on both sides,

x6+1x2+1=(x4−x2+1)

Hence, the required expression becomes,

⟹1x2(x4−x2+1)

⟹x2−1+1x2

Note that

(x+1x)2=x2+2+1x2

Hence, the simplified expression may be written as

⟹(x+1x)2−3

By (1) , we know x+1x=6

Hence, value of required expression

=62−3

=33

Hope that helps :)

Answer 2

The answer is $(x^{4} + \frac{1}{x^{2} }) (x^{2} +1)$ = $6x (36 - 2x )$

Given: x² - 6x + 1 = 0

To find: The value of  $(x^{4} + \frac{1}{x^{2} }) (x^{2} +1)$

Solution: x² - 6x + 1 = 0  can be written as

⇒ x² + 1 = 6x _(1)  

⇒ $\frac{x^{2}+1 }{x} = \frac{6x}{x}$              [ divide both sides with x ]

⇒ $x^{2} + \frac{1}{x} = 6$  

⇒ $(x^{2} + \frac{1}{x} )^{2} = 6^{2}$           [ do squaring on both sides ]

⇒ $(x^{2} )^{2} + (\frac{1}{x} )^{2} + 2(x^{2})(\frac{1}{x}) = 6^{2}$      [ ∵ (a+b) = a²+b² + 2ab ]

⇒ $x^{4} + \frac{1}{x^{2} } + 2(x) = 36$

⇒ $x^{4} + \frac{1}{x^{2} } = (36 - 2x )$  

Therefore, $(x^{4} + \frac{1}{x^{2} }) (x^{2} +1)$ =  $6x (36 - 2x )$

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