Question

If x-1/x=√5 find the value of x+1/x

Answer 1

$\: \: \: \: \: (x - \frac{1}{x} ) = \sqrt{5} \\ = > {(x - \frac{1}{x}) }^{2} = {( \sqrt{5} )}^{2} \\ = > {x}^{2} + \frac{1}{ {x}^{2} } - 2.x. \frac{1}{x} = 5 \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 5 + 2 \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 7$

Now,

$\: \: \: \: \: {(x + \frac{1}{x} )}^{2} \\ = {x}^{2} + \frac{1}{ {x}^{2} } + 2.x. \frac{1}{x} \\ =7 + 2 \\ = 9$

Since ,

${(x + \frac{1}{x} )}^{2} = 9 \\ = > (x + \frac{1}{x} ) = \sqrt{9} = 3$

Answer 2

Answer:

±3

Step-by-step explanation:

Given

x-1/x=√5

Let us square on both the sides

e.i., (x-1/x)² = (√5)²

=> x²+1/x²-2x•1/x = √25

=> x²+1/x² -2 = 5

=> x²+1/x² = 5+2

=> x²+1/x² = 7

Now,

We have, x²+1/x²

So, (x+1/x)²

=> (x+1/x)² = x²+1/x² +2x•1/x

Since, we know that the value of x²+1/x² = 7

therefore,

=> (x+1/x)² = 7+2

=> (x+1/x)² = 9

=> x+1/x = √9

=> x+1/x = 3

Hence,

x+1/x = ±3