Question

The roots of the equation 12x2+4kx+3=0 are real and equal. Then value of K

is

a)±2 b) ±9 c) 4 d) ±3​

Answer 1

Correct option is

A=±3

⇒ The given quadratic equation is 12x^2 +4kx+3=0, comparing it with ax^2 +bx+c=0.

⇒ We get, a=12,b=4k and c=3

⇒ It is given that roots are real and equal.

∴ b^2 −4ac=0

⇒ (4k)^2 −4(12)(3)=0

⇒ 16k^2 −144=0

⇒ 16k 2 =144

⇒ k^2 = 144/16

⇒ k^2 =9

∴ k=±3