Question

for what value of k, the given equation has real roots
$x {}^{2} - k(4x - k + 1) + 2 = 0$

Answer 1

Answer:

x² - k ( 4x - k + 1 ) + 2 = 0

=> x² -4kx + k² - k + 2 = 0

=> a = 1, b = -4k, c = k² - k + 2

Discriminant = b² - 4ac

= ( -4k )² - 4 ( 1 ) ( k² - k + 2 )

= 16k² - 4k² + 4k - 8

= 12k² + 4k - 8

Since the solutions are real, the discriminant will be greater than 0.

=> 12k² + 4k - 8 > 0

Dividing by 4 throughout, we get,

=> 3k² + k - 2 > 0

=> 3k² + 3k - 2k - 2 > 0

=> 3k ( k + 1 ) -2 ( k + 1 ) > 0

=> ( 3k - 2 ) ( k + 1 ) > 0

=> ( k + 1 ) > 0 / ( 3k - 2 )

=> k + 1 > 0

=> k > ( -1 )

Also,

=> ( 3k - 2 ) > 0 / ( k + 1 )

=> 3k - 2 > 0

=> k > 2 / 3

So the interval at which k can take different values is : ( 2 / 3, ∞ )

This is the required answer.