Question

Find the value pf k for which the equation x² + k (2k + k - 1 ) + 2= 0 has real and equal roots

Answer 1

For the following question the answer is k = 2/√3

Answer 2

Given ::- x² + k( 2k + k -1) + 2 = 0

Roots are real and equal

To find ::- value of k

Solution ::-

Comparing the given equation with

ax² + bx + c= 0

So, a = 1 b = k ( 2k + k -1) ⇒ k ( 3k -1) c = 2

b²-4ac = 0

[( 3k² - k)]² - 4(1)(2) = 0

9k⁴ - k² - 8 = 0

9 k⁴ - k² = 8

But if we assume that the value of k is 1 ,

9 (1)⁴ - ( 1)²

= 9 -1

= 8

We get LHS = RHS

Or even if we take the value of k as -1 ,

9 (-1)⁴ - (-1)²

= 9 -1

= 8

We get LHS = RHS

SO, the value of k is 1 or -1

HOPE IT HELPS YOU!!