Question

If k and 2k are zeros of $f\left(x\right)=x^3+4x^2+9kx=90$, find k and all three zeros of f(x).

• A. K = 1, roots = -3, -6, -5
• B. K = -3, roots = -3, -6, 5
• C. K = -3, roots = -3, 6, -5
• D. K = 3, roots = -3, -6, 5

Answer 1

The correct option is B

K = -3, roots = -3, -6, 5

Since K and 2 K are zeros of f(x)
F(K) = 0
$\Rightarrow K^3+4K^2+9K^2-90=0\Rightarrow K^3+13K^2-90=0....\left(i\right)$
And F(2K) = 0
$\Rightarrow 8K^3+16K^2+18K^2-90=0\Rightarrow 8K^3+34K^2-90=0\Rightarrow 4K^3+17K^2-45=0....\left(ii\right)\left(ii\right)-4\times \left(i\right)\Rightarrow -35K^2+315=0\Rightarrow K^2=9\Rightarrow K=\pm 3$
K = 3 does not satisfy the given
Polynomial.
$\therefore k=-3$
The two roots are – 3 and – 6
Sum of roots = - 4
$\Rightarrow -3-6$+third root =-4
$\Rightarrow$ third root =5