Question

plz solve this equation​

Answer 1

The given equation is

$mx^2+4x+n=0$

$x=\frac{2}{3}\:is\:one\:of\:its\:roots\\\implies m(\frac{2}{3})^2+4\times\frac{2}{3}+n=0\\\implies \frac{4m}{9}+\frac{8}{3}+n=0\\\implies\frac{4m+8\times3+9n}{9}=0\\\implies4m+9n=-24\:\:\:\:(eqtn.\:1)$

Also,

$x=2\:is\:the\:other\:root\\\implies 2^2m+4\times 2+n=0\\\implies 4m+8+n=0\\\implies 4m+n=-8\:\:\:\:(eqtn.\:2)$

Subtracting equation 1 from 2, we get

$8n=-24+8\\\implies n=-\frac{16}{8}=-2$

Putting value of n in equation 2, we get

$4m-2=-8\\\implies 4m=-6\\\implies m=-\frac{3}{2}$