Question

Plz help me in this question of class 9th linear equation in one variable

Answer 1

Step-by-step explanation:

We have,

$x + \sqrt{15} = 4$

$\implies \: x = 4 - \sqrt{15}$

$\implies \: \frac{1}{x} = \frac{1}{4 - \sqrt{15}}$

$\implies \: \frac{1}{x} = \frac{(4 + \sqrt{15}) }{(4 - \sqrt{15})(4 + \sqrt{15} )}$

$\implies \: \frac{1}{x} = \frac{(4 + \sqrt{15}) }{16 - 15}$

$\implies \: \frac{1}{x} = \frac{4 + \sqrt{15} }{1}$

$\implies \: \frac{1}{x} = 4 + \sqrt{15}$

So,

$x + \frac{1}{x} = 4 - \sqrt{15} + 4 + \sqrt{15}$

$\implies \: x + \frac{1}{x} = 8$