Question

If(x2+1/x2)=79 then find the value of x+1/x​

Answer 1

Answer:

We know that, (x + 1/x)^2 = x^2 + 1/x^2 + 2x + 1/x

Given , x^2 + 1/x^2 = 79

So, (x + 1/x)^2 = 79 + 2

       (x + 1/x)^2 = 81

       (x + 1/x ) = 9

Step-by-step explanation:

Hope it helped u dear frnd!!

Answer 2

Given, x+1/x = 9 , Squaring both sides

(x+1/x)² = 81 ……………………………………………………..(1)

Now we have

(x- 1/x)² = x² - 2.x.1/x + (1/x)² = x² - 2+ 1/x²

Or, (x- 1/x)² = x² + 2 -2 - 2+ 1/x² (Adding and subtracting 2)

Or, (x- 1/x)² = x² + 2.1 - 4+ 1/x² = x² + 2.x.1/x - 4+ 1/x²

= (x² + 2.x.1/x + 1/x²) - 4 = (x+1/x)² - 4

Substituting for (x+1/x)² from (1),

(x- 1/x)² = 81 - 4 = 77 , Taking square root of both sides

⇒ x- 1/x = ±77½ or ±8.775 (rounded to third decimal) (Answer)