Question

if x-1/x=10 find the value of (x+1/x)^2

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Answer 1

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$x - \frac{1}{x} = 10 \\ = > (x - \frac{1}{x} ) {}^{2} = 100 \\ = > x {}^{2} - 2 + \frac{1}{x {}^{2} } = 100 \\ = > x {}^{2} + \frac{1}{x {}^{2} } = 102 \\ now \\ (x + \frac{1}{x} ) {}^{2} = x {}^{2} + 2 + \frac{1}{x {}^{2} } {}^{} \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 \\ = 102 + 2 \\ = 104$

Answer 2

Question :

If $\bf{x - \dfrac{1}{x} = 10}$ , find the value of $\bf{\bigg(x + \dfrac{1}{x}\bigg)^{2}}$

Given :

$\bf{x - \dfrac{1}{x} = 10}$

To find :

The value of $\bf{\bigg(x + \dfrac{1}{x}\bigg)^{2}}$.

Solution :

On squaring the Equation $\bf{x - \dfrac{1}{x} = 10}$ , we get :

$:\implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = 10^{2}}$

$:\implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{2} = 100}$

Using the identity,

⠀⠀⠀⠀⠀⠀(a - b)² = a² + b² - 2ab

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀We get :

$:\implies \bf{x^{2} + \dfrac{1}{x^{2}} - 2 \times x \times \dfrac{1}{x} = 100}$

$:\implies \bf{x^{2} + \dfrac{1}{x^{2}} - 2 \times \not{x} \times \dfrac{1}{\not{x}} = 100}$

$:\implies \bf{x^{2} + \dfrac{1}{x^{2}} - 2 = 100}$

$:\implies \bf{x^{2} + \dfrac{1}{x^{2}} = 100 + 2}$

$:\implies \bf{x^{2} + \dfrac{1}{x^{2}} = 102}$

Now using the identity ,

⠀⠀⠀⠀⠀⠀a² + b² = (a + b)² - 2ab

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀We get :

$:\implies \bf{\bigg(x^{2} + \dfrac{1}{x}\bigg)^{2} - 2 \times x \times \dfrac{1}{x} = 102}$

$:\implies \bf{\bigg(x^{2} + \dfrac{1}{x}\bigg)^{2} - 2 \times \not{x} \times \dfrac{1}{\not{x}} = 102}$

$:\implies \bf{\bigg(x^{2} + \dfrac{1}{x}\bigg)^{2} - 2 = 102}$

$:\implies \bf{\bigg(x^{2} + \dfrac{1}{x}\bigg)^{2} = 102 + 2}$

$:\implies \bf{\bigg(x^{2} + \dfrac{1}{x}\bigg)^{2} = 104}$

$\boxed{\therefore \bf{\bigg(x^{2} + \dfrac{1}{x}\bigg)^{2} = 104}}$

Hence, the value of $\bf{\bigg(x + \dfrac{1}{x}\bigg)^{2}}$ is 104.