Question

if one solution of the equation 3x2 =8x+2K+1 is seven times the other. Find the solution and value of K​

Answer 1

» Given:

equation = 3x²- 8x +2k +1

$if\:\alpha\:is \:a\:root,then\\ \\ \\ solutions\:are\:\alpha \:and \:7\alpha$

» Solution -

if $\alpha \:and\: 7\alpha\:roots\:then\:f(\alpha)= 0 \:and\:f(7\alpha) = 0$

$f(\alpha) = 3(\alpha)^2 -8(\alpha)+2K+1 = 0\\ \\ \\ 2K+1 = -3\alpha^2 +8\alpha...(1)\\ \\ \\ f(7\alpha) = 3(7\:alpha)^2 -8(7\alpha) +2K +1 = 0 \\ \\ \\ 147\alpha^2 -56\alpha+2K+1 = 0 \\ \\ \\ 2K+1 = -147\alpha^2+56\alpha....(2)\\ \\ \\ (1) = (2)\\ \\ \\ -3\alpha^2+8\alpha = -147\alpha^2+56\alpha\\ \\ \\-3\alpha^2+147\alpha^2 = 56\alpha-8\alpha\\ \\ \\144\alpha^2 = 48\alpha\\ \\ \\by\: dividing\:\alpha\:on\:both\:sides\\ \\ \\ 144\alpha = 48\\ \\ \\ \alpha = \frac{48}{144} \\ \\ \\ \alpha = \frac{1}{3}\\ \\ \\{\pink{\boxed{\therefore\:roots \:are\:\alpha= \frac{1}{3}\:and\7\alpha = \frac{7}{3}}}}$

Value of K

$f(\frac{1}{3}) = 3\times(\frac{1}{3} \times\frac{1}{3})-8(\frac{1}{3})+2K+1 \\ \\ \\ = \frac{1}{3}-\frac{8}{3}+2k+1 \\ \\ \\ = \frac{-7}{3}+1+2K \\ \\ \\ = \frac{-4}{3}+2K = 0 \\ \\ \\ 2K = \frac{4}{3} \\ \\ \\ k = \frac{4}{3} \times\frac{1}{2} = \frac{2}{3} \\ \\ \\ {\purple{\boxed{\therefore\:K = \frac{2}{3}}}}$