Question

If x = √7 +√6 then find value of ( x²+ 1/x² )​

Answer 1

Answer:

27

Step-by-step explanation:

Given,

x = √7 + √6

To Find :-

Value of ' x² + 1/x²'

How To Do :-

As they gave the value of 'x' we need find the value of '1/x' using that and we need to do Rationalizing the denominator of it. After getting that value we need to substitute the values of 'x' and '1/x' in the formula of x²+ 1/x².

Formula Required :-

(x + 1/x)² = x² + 1/x² + 2×x×1/x

→1)  x²+ 1/x² = (x + 1/x)² - 2 × x × 1/x

2) a+b)(a-b) = a² - b²

3)  (ab)² = a²b²

Solution :-

x = √7 + √6

→ 1/x = 1/√7 + √6

Rationalizing the denominator :-

$=\dfrac{1}{\sqrt{7}+\sqrt{6}}\times \dfrac{\sqrt{7}-\sqrt{6}}{\sqrt{7}-\sqrt{6}}$

$=\dfrac{\sqrt{7}-\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2}$

[ ∴ (a+b)(a-b) = a² - b²]

$=\dfrac{\sqrt{7}-\sqrt{6}}{7-6}$

= √7 - √6/1

∴ 1/x = √7 - √6

∴ x²+ 1/x² = (x + 1/x)² - 2 × x × 1/x

= (x + 1/x)² - 2

Substituting the values of both 'x' and '1/x' :-

$=\left(\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6})^2-2$

= (2√7)² - 1

= (2)²(√7)² - 1

[ ∴ (ab)² = a²b²]

= 4(7) - 1

= 28 - 1

= 27

∴ x²+ 1/x² = 27

Answer 2

Given:

$\tt{x=\sqrt7 + \sqrt6 }$

To Find :

$\tt{x^2 + \dfrac{1}{{x^2}}$

Solution :

$\tt{x = \sqrt7 + \sqrt6}\\ \\ \implies \tt{\dfrac{1}{x} = \dfrac{1}{\sqrt7 + \sqrt6}} \\ \\ \tt{Now\ \ \ we\ \ rationalize\ \ it:}\\ \\ \implies \tt{\dfrac{1}{x} = \dfrac{1}{\sqrt7 + \sqrt6}\ \times \ \dfrac{\sqrt7 - \sqrt6}{\sqrt7 - \sqrt6} }\\ \\ \\ \implies \tt{\dfrac{1}{x} = \dfrac{\sqrt7 - \sqrt6}{7-6} = \sqrt7 - \sqrt6 }\\ \\ Therefore:\ \tt{\Big(x + \dfrac{1}{x}\Big) = (\sqrt7 + \sqrt6) + (\sqrt7 - \sqrt6) = 2\sqrt7}$

$\tt{We\ know\ that: } \\ \\ \tt{x^2 + \dfrac{1}{x^2} = \Big( x + \dfrac{1}{x} \Big)^2 - 2 } \\ \\ \implies \tt{x^2 + \dfrac{1}{x^2} = \big(2\sqrt7\big)^2 - 2 = (4\times7) - 2 = 26} \\ \\ Hence, \\ \\ \boxed{\tt{x^2 + \dfrac{1}{x^2} = 26}}$

Hope this help : )