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:

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 ratio is common.

So we have,

$\frac{x^2}{1}=\frac{6-x^2}{x^2}$

$x^4=6-x^2$

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

let x² = y

⇒ y² + y - 6 = 0

y² + 3y - 2y - 6 = 0

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

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

y = 2  and y = -3

⇒ x² = 2     and x² = -3

⇒ x = ± √2    and x = ± √3i  ( rejected due to imaginary number. )

Therefore, Value of x are √2 , - √2.

Answer 2

Answer:

Step-by-step explanation:

The answer is +-√2