Question

Find the value of $k$ so that the zero of the quadratic polynomial $3x^2-kx+14$ are in the ratio $7:6$

Answer 1

Let $\alpha$ and $\beta$ be the zeros $\left(roots\right)$ of $p\left(x\right)=3x^2+kx+14$

$\alpha +\beta =\frac{k}{3}$ and $\alpha \beta =\frac{14}{3}$

Given ,$\alpha :\beta =7:6$

Or, $\frac{\alpha }{\beta }=\frac{7}{6}\Rightarrow \alpha =\frac{7\beta }{6}-----\left(1\right)$

Now, $\alpha +\beta =\frac{k}{3}$

Or, $\frac{7\beta }{6}+\beta =\frac{k}{3}$

Or, $\frac{7\beta +6\beta }{6}=\frac{k}{3}\Rightarrow 13\beta =2k-----\left(2\right)$

Again,$\alpha \beta =\frac{14}{3}$

Or, $\frac{7\beta }{6}\times \beta =\frac{14}{3}\left[from\left(1\right)\right]$

Or, ${\beta }^2=4\Rightarrow \beta =\pm 2$

Putting $\beta =2$ in $\left(2\right),13\times 2=2k\Rightarrow k=13$

Putting $\beta =-2$ in $\left(2\right),13\times (-2)=2k\Rightarrow k=-13$

$\therefore$ value of $k=\pm 13$ Ans.