Question

find the value of k for which the equation x^2 + k(2x+k-1)+2=0 has real and equal roots.​

Answer 1

Solution :-

x² + k(2x + k - 1) + 2 = 0

⇒ x² + 2kx + k² - k + 2 = 0

Comparing the above equation with ax² + bx + c = 0 we get,

• a = 1
• b = 2k
• c = k² - k + 2

We know that

For a quadratic equation to have real and equal roots b² - 4ac = 0

⇒ b² - 4ac = 0

Substituting the values

⇒ ( 2k )² - 4( 1 )( k² - k + 2 ) = 0

⇒ 4k² - 4( k² - k + 2 )

⇒ 4k² - 4k² + 4k - 8 = 0

⇒ 4k - 8 = 0

⇒ 4k = 8

⇒ k = 8/4

⇒ k = 2

Therefore the value of k is 2.

Extra information :-

A quadratic equation ax² + bx + c = 0,

• to have real and distinct roots, b² - 4ac > 0
• to have real and equal roots, b² - 4ac = 0
• to have no real roots, b² - 4ac < 0