Question

a) Find the value of 'K' for which $x$ = 3 is a solution of the quadratic equation.
$(K+2)x^2-Kx+6=0.$
b) Thus find the other root of the equation.
[6 MARKS]

Answer 1

Each subpart: 3 Marks

a) $x$ = 3 is solution of equation
$(K+2)x^2-Kx+6=0$
Hence, it will satisfy the equation
$(K+2)3^2-K\times 3+6=0$
$\Rightarrow 9(K+2)-3K+6=0\Rightarrow 9K+18-3K+6=0\Rightarrow 6K+24=0\Rightarrow K=-4$

b) Substituting $K=-4$, the given equation is changed to,
$-2x^2+4x+6=0\Rightarrow x^2-2x-3=0\Rightarrow x^2-3x+x-3=0\Rightarrow x(x-3)+1(x-3)=0\Rightarrow (x-3)(x+1)=0$
Either $x=3orx=-1$
Hence, other root is -1.