Question

If x=3 is a solution of the quadratic equation (k+2)x²+kx+6=0 then, find the value of k

Answer 1

x = 3

⇒ (k+2)(x²+kx+6) = 0

⇒ (k+2)(3²+3k+6) = 0

⇒ (k+2)(9+6+3k) = 0

⇒ (k+2)(15+3k) = 0

⇒ 15k+3k²+30+6k = 0

⇒3k²+21k+30 = 0

⇒ k²+7k+10 = 0

by splitting the middle term

k²+5k+2k+10 = 0

k(k+5)+2(k+5) = 0

k = -2 , -5