Question

If one root of `sqrt2 x^2-kx-2sqrt2 = 0  "is"  2sqrt2`, find the value of k and also find other root.

Answer 1

Given:

`sqrt2 x^2-kx-2sqrt2 = 0`

`x = 2sqrt2`

Substitute:

`sqrt2(2sqrt2)^2 - k(2sqrt2) - 2sqrt2 = 0`

`(2sqrt2)^2`

= 4 × 2

= 8

`sqrt2 xx 8`

`= 8sqrt2`

`k (2sqrt2)`

`= 2ksqrt2`

`8sqrt2 - 2ksqrt2 - 2sqrt2 = 0`

Combine:

`(8sqrt2 - 2sqrt2)-2ksqrt2 = 0`

`=> 6sqrt2 - 2ksqrt2 = 0`

6 − 2k = 0

⇒ 2k = 6 

⇒ k = 3

​`sqrt2 x^2 - 3x - 2sqrt2 = 0`

`x = (-(-3) +-sqrt(-3)^3 - 4(sqrt2)(-2sqrt2))/(2sqrt2)`

`x = (3+-sqrt25)/(2sqrt2)`

`= (3+-5)/(2sqrt2)`

`x = (3+5)/(2sqrt2) = 8/(2sqrt2) = 4/sqrt2 = 2sqrt2`

`x = (3-5)/(2sqrt2) = -2/(2sqrt2) = (-1)/sqrt2 = -sqrt2/2`

Value of k = 3

Other root = `-sqrt2/2`