Question

Find the value of k for which x =3 is a solution of the quadratic equation. (k+2)x^2-kx+6=0.thus find the other root of the equation

Answer 1

(k+2)x²-kx+6
(k+2)3²-k(3)+6=0
(k+2)9-3k+6=0
9k+18-3k+6=0
6k+24=0
6k=-24
K=-24/6
k=-4

Answer 2

Hey

Here is your answer,

Substitute the value of x to find the value of k,

(k+2)x^2-kx+6=0
(K+2)9 - 3k +6=0
9k + 18 -3k +6=0
6k + 24 =0
6k = -24
K= -24/6
K= -4

(k+2)x^2-kx+6=0
(-4+2)x^2 +4x +6=0
2x^2-4x -6=0
X^2 - 2x -3=0

X^2 +x -3x -3=0
X(x+1)-3(x+1)=0
(X+1)(x-3)=0
X=3.-1

Hope it helps you!