Question

if X + 1 upon X = √2 then the value of x square + 1 upon X is​

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a expression

• $\sf{x + \dfrac{1}{x} = \sqrt{2}}$

To Find:

• We have to find the value of another expression given below

• $\sf{x^2 + \dfrac{1}{x^2}}$

Solution:

We have been given given that

$\boxed{\sf{x + \dfrac{1}{x} = \sqrt{2}}}$

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Squaring Both Sides of Equation we get

$\implies \sf{ ( x + \dfrac{1}{x} ) ^2= (\sqrt{2})^2}$

$\boxed{\sf{(a+b)^2 = a^2 + b^2 + 2ab}}$

$\implies \sf{(x)^2 + { \left ( \dfrac{1}{x} \right ) }^2 + 2 \times x \times \dfrac{1}{x} = (\sqrt{2})^2}$

$\implies \sf{x^2 + \dfrac{1}{x^2} + 2 = 2}$

$\implies \sf{x^2 + \dfrac{1}{x^2} = 2 - 2}$

$\implies \sf{x^2 + \dfrac{1}{x^2} = 0 }$

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$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\large\boxed{\sf{\red{ x^2 + \dfrac{1}{x^2} = 0 }}}$

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$\purple{\underline{\underline{\sf{Some \: Important \: Indentities : }}}}$

$\sf{(a+b)^2 =a^2 +b^2 +2ab}$

$\sf{(a-b)^2 = a^2 + b^2 - 2ab}$

$\sf{(a+b)^3 = a^3 + b^3 + 3ab(a+b)}$

$\sf{(a-b)^3 = a^3 - b^3 - 3ab(a-b)}$

$\sf{a^3 + b^3 = (a+b)(a^2 + b^2 - ab)}$

$\sf{a^3 - b^3 = (a-b)(a^2 + b^2 + ab)}$