Question

If 2 is a root of the equation x2+kx+12=0 and the equation x2+kx+q=0 has equal roots, find the value of q.

Answer 1

x2 + kx +12=0.
2 is a root of the equation.
Then, 4+2k+12=0.
Then, 2k+16=0.
So, 2k= -16.
Therefore, k= -8.
x2+kx+q=0.
x2-8k+q=0.
Equal roots, so b2 =4ac.
64=4*1*q
q=16.

Answer 2

Answer : The value of q = 16

Solution :

Given, 2 is a root of the equation x² + kx + 12 = 0,

Then Substituting x = 2 should make the equation hold good.

So,

⇒ x² + kx + 12 = 0

⇒(2)² + k(2) + 12 = 0

⇒ 4 + 2k + 12 = 0

⇒2k + 16 = 0

⇒2k = - 16

⇒k = - 16/2

⇒k = - 8

Also, Equation x²+kx+q=0 has equal roots, So discriminant should be 0

⇒Δ = 0

⇒b² - 4ac = 0

⇒ k² - 4(1)(q)= 0

⇒(-8)² - 4q = 0

⇒ 64 - 4q = 0

⇒ 64 = 4q

⇒64/4 = q

⇒ q = 16

Therefore, Value of q = 16