Question

For what value of x do the numbers 1, x^2, and 6-x^2, taken in that order, form a geometric progression?

Answer 1

Answer:

please refer to the attachment

Answer 2

Answer:

Value of x are √2 , - √2 .

Step-by-step explanation:

Given:

Terms of a GP = 1 , x² and 6 - x²

To find: Value of x.

We know that in GP

( GM )² = a × b

GM = x² , a = 1  and b = 6 - x²

$(x^2)^2=(1)(6-x^2)$

$x^4+x^2-6=0$

let x² = z

⇒ z² + z - 6 = 0

z² + 3z - 2z - 6 = 0

z( z + 3 ) - 2 ( z + 3 ) = 0

( z - 2 )( z + 3 ) = 0

z = 2  and z = -3

⇒ x² = 2     and x² = -3 ( This value rejected because it give imaginary no. )

⇒ x = ± √2    

Therefore, Value of x are √2 , - √2.