Question

If one zero of $p\left(x\right)=3x^2-8x+(2k+1)$ is 7 times the other,find k.

Answer 1

Given,

$p\left(x\right)=3x^2-8x+(2k+1)$

$\beta =7\alpha$

$\alpha +\beta =\alpha +7\alpha =8\alpha =-\frac{b}{a}=\frac{8}{3}$

$\therefore \alpha =\frac{1}{3}$

$\alpha \beta =\alpha \left(7\alpha \right)=7{\alpha }^2=\frac{c}{a}=\frac{2k+1}{3}$

$\Rightarrow {\alpha }^2=\frac{2k+1}{21}$

$\frac{1}{9}=\frac{2k+1}{21}$

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

$\therefore k=\frac{2}{3}$