Question

if x + 1 upon x =9 find the value of (i) x2 +1 upon x​

Answer 1

Answer:

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)

Answer 2

Answer:

$x + 1 \\ 9(1) = 9 \\ x2 + 1 \\ 9 \times 2 + 1 = 19$