Question

★ STANDARD QUESTION 8 ★

★ CONTENT QUALITY SUPPORT REQUIRED ★

Solve for " x "
1.) x ∈ N
2.)x ∈ R

$\frac{x + \sqrt{x {}^{2} - 1} }{x - \sqrt{x {}^{2} - 1} } + \frac{x - \sqrt{x {}^{2} - 1} }{x + \sqrt{x {}^{2} - 1} } = 98$

Answer 1

Solution :

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Given :

$\frac{x+ \sqrt{x^2 -1 } }{x - \sqrt{x^2 - 1} } + \frac{x- \sqrt{x^2 -1 } }{x + \sqrt{x^2 - 1} } = 98$

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To Find :

Value of x,.

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By cross - Multiplication,.

We  get,

⇒ $\frac{(x+ \sqrt{x^2 -1 } )(x + \sqrt{x^2 - 1} )+( x- \sqrt{x^2 -1 } )(x - \sqrt{x^2 - 1} )}{(x + \sqrt{x^2 - 1} )(x - \sqrt{x^2 - 1} )} = 98$

⇒ $\frac{x^2 + ( \sqrt{x^2 -1})^2 - 2x( \sqrt{x^2 - 1} )+ x^2 + ( \sqrt{x^2 - 1} )^2 +2x( \sqrt{x^2 - 1} ) }{x^2 - ( \sqrt{x^2 - 1} )}  = 98$

⇒ $\frac{x^2 + x^2 - 1 + x^2 + x^2 - 1}{x^2 - x^2 + 1} = 98$

⇒ $\frac{4x^2 - 2}{1} = 98$

⇒ $4x^2 - 2 = 98$

⇒ $4x^2 = 98 + 2$

⇒ $4x^ 2 = 100$

⇒ $x^2 = \frac{100}{4}$

⇒ $x^2 = 25$

⇒ $x = \sqrt{25}$

∴ x = 5 (or) x = - 5

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Answer 2

Step-by-step explanation:

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