Question

if x +1/x =sqrt6, find the value of x^2+1/2

Answer 1

Correct question: If x + 1/x = √6, find the value of x² + 1/x².

Answer:

$\boxed{ \bf \therefore \: {x}^{2} + \frac{1}{x {}^{2} } = 4}$

Step-by-step explanation:

Given that ;

$\implies \sf x + \frac{1}{x} = \sqrt{6} \\ \\ \bf On \: squaring \: both \: sides : \\ \\ \implies \sf (x + \frac{1}{x} ) {}^{2} = ( \sqrt{6} ) {}^{2} \\ \\ \implies \sf {x}^{2} + \frac{1}{x {}^{2} } + 2 \times x \times \frac{1}{x} = 6 \\ \\ \implies \sf {x}^{2} + \frac{1}{x {}^{2} } + 2 = 6 \\ \\ \implies \sf {x}^{2} + \frac{1}{x {}^{2} } = 6 - 2 \\ \\ \boxed{ \bf \therefore \: {x}^{2} + \frac{1}{x {}^{2} } = 4}$

Answer 2

$\bf{\large{\underline{\underline{Correct \:Question:-}}}}$

If x + 1/x = √6 , Find the value of x² + 1/x²

$\bf{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bf{{x}^{2} + \dfrac{1}{ {x}^{2}} = 4}}$

$\bf{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{x + \dfrac{1}{x} = \sqrt{6}}$

To find :- $\sf{x^2 + \dfrac{1}{x^2}}$

Solution :-

$x + \dfrac{1}{x} = \sqrt{6}$

By squaring on both the sides :-

${(x + \dfrac{1}{x})}^{2} = {( \sqrt{6})}^{2}$

We know that, (a + b)² = a² + b² + 2ab

Here a = x, b = 1/x

By substituting the values in the identity we have :-

${x}^{2} + \dfrac{1}{ {x}^{2} } + 2(x)( \dfrac{1}{x}) = 6$

${x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 6$

${x}^{2} + \dfrac{1}{ {x}^{2} } = 6 - 2$

${x}^{2} + \dfrac{1}{ {x}^{2}} = 4$

$\boxed{\bf{{x}^{2} + \dfrac{1}{ {x}^{2}} = 4}}$

$\bf{\large{\underline{\underline{Identity\:Used:-}}}}$

(a + b)² = a² + b² + 2ab

$\bf{\large{\underline{\underline{Some\:Important\:Identities:-}}}}$

[1] (a + b)² = a² + b² + 2ab

[2] (a - b)² = a² + b² - 2ab

[3] (a + b)(a - b) = a² - b²

[4] (x + a)(x + b) = x² + (a + b)x + ab