Question

If one zero of the polynomial $3x^2-8x+2k+1$ is seven times the other, find the value of $k$.

• A. $\frac{2}{3}$
  
• B. $\frac{3}{4}$
  
• C. $-1$
• D. $\frac{3}{5}$
  

Answer 1

The correct option is A $\frac{2}{3}$


Let $\alpha \text{and}\beta$ be the zeros of the polynomial. Then as per question $\beta =7\alpha$

Now sum of zeros $=\alpha +\beta =\alpha +7\alpha =-\left(\frac{-8}{3}\right)$

$\Rightarrow 8\alpha =\frac{8}{3}\text{or}\alpha =\frac{1}{3}$

and $\alpha \times \beta =\alpha \times 7\alpha =\frac{2k+1}{3}$

$\Rightarrow 7{\alpha }^2=\frac{2k+1}{3}\Rightarrow 7{\left(\frac{1}{3}\right)}^2=\frac{2k+1}{3}(\because \alpha =\frac{1}{3})$

$\Rightarrow \frac{7}{9}=\frac{2k+1}{3}\Rightarrow \frac{7}{3}=2k+1$

$\Rightarrow \frac{7}{3}-1=2k\Rightarrow k=\frac{2}{3}$