Question

if y=3+√8, find the value of y2+1y2​

Answer 1

Solution :-

→ y = 3 + √8 => √9 + √8

→ 1/y = 1/(√9 + √8) => 1/(√9 + √8) * {(√9 - √8)/(√9 - √8)} => (√9 - √8) / (9 - 8) => √9 - √8 => 3 - √8

[ using (a+b)(a-b) = a² - b² ] .

So,

→ y + 1/y = (3 + √8) + (3 - √8) => 6

Squaring Both sides ,

→ (y + 1/y)² = 6²

using (a+b)² = a² + b² + 2ab , we get,

→ y² + 1/y² + 2 * y * 1/y = 36

→ y² + 1/y² + 2 = 36

→ y² + 1/y² = 36 - 2 => 34 (Ans.)

Answer 2

Question:-

If y=3+√8, find the value of y² + 1/y²

Formula Used:-

(a + b)(a - b) = a² - b²

(a + b)² = a² + b² + 2ab

Answer:-

Finding 1/y:

y = 3 + √8

=> 1/y = 1/(3 + √8)

Rationalising the denominator

=> 1/y = [1(3 - √8)]/[(3 + √8)(3 - √8)]

=> 1/y = [3 - √8]/[3² - (√8)²]

=> 1/y = (3 - √8)/(9 - 8)

=> 1/y = 3 - √8

Finding y + 1/y:

y + 1/y = 3 + √8 + 3 - √8

=> y + 1/y = 6

Finding y² + 1/y²:

(y + 1/y)² = 6²

=> y² + 1/y² + 2*y*1/y = 36

=> y² + 1/y² = 36 - 2

=> y² + 1/y² = 34

Ans. y² + 1/y² = 34