Question

If one zero of the quadratic polynomial $2x^2-(3k+1)x-9$ is negative of the other find the value of 'k'.

• A. $-\frac{1}{6}$
• B. $3$
• C. $\frac{1}{4}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (D)

$-\frac{1}{3}$

Let $p\left(x\right)=2x^2-(3k+1)x-9$

Let $\alpha ,\beta$ be the roots of this polynomial.

We know that

$\alpha +\beta =\frac{3k+1}{2}=$ Sum of the roots

$\alpha \beta =\frac{-9}{2}=$ Products of the roots

Given that : One zero is negative of the other

$\Rightarrow \alpha =-\beta$

$\Rightarrow \alpha +\beta =0$

$\Rightarrow \alpha +\beta =\frac{3k+1}{2}=0$

$\Rightarrow 3k+1=0$

$\Rightarrow k=-\frac{1}{3}$