Question

If one solution of $3x^2=8x+2k+1$ is seven times the other. Find the solutions and the value of $k$.

Answer 1

if one of the root is a then other is 7a

Rearranging the given equation,

$3x^2-8x-2k-1=0$

Sum of the roots must be -(-8)/3

=8/3

So,

$\begin{array}{c}a+7a=\frac{8}{3}\\ 8a=\frac{8}{3}\\ a=\frac{1}{3}\end{array}$

and product of the roots should be (2k+1)/3

So,

$\begin{array}{c}a\times 7a=\frac{-2k-1}{3}\\ 7a^2=\frac{-2k-1}{3}\\ 7{\left(\frac{1}{3}\right)}^2=\frac{-2k-1}{3}\\ -2k-1=\frac{7}{3}\\ -6k-3=7\\ -6k=10\\ k=\frac{-5}{3}\end{array}$