Question

for what condition 3x2+4mx+2=0 and 2x2+3x-2=0 having a common root

Answer 1

Answer :-

Splitting the middle term

$\sf\implies 2x^{2}+4x-1x-2=0$

$\sf\implies 2x(x+2)-1(x+2)=0$

$\sf\implies (x+2)(2x-1)=0$

$\sf\implies x+2=0\:Or\:2x-1=0$

$\sf\implies x=-2\:\: Or\:\:2x=1$

$\tt\implies x=-2\:Or\:x=\frac{1}{2}$

$\sf\:Now,\\substitute\:x \: values\\in \:3x^{2}+4mx+2=0,we\:get$

$Case\:1\\ \sf\:if\:x=-2,\\\implies 3\times \big(-2)^{2}+4m(-2)+2=0$

$\sf\implies 12-8x+2=0$

$\sf\implies 14-8x=0$

$\sf\implies -8x=-14$

$\sf\implies x = \frac{-14}{-8}$

$\sf\implies x = \frac{7}{4}$

$Case \:2\\ \sf\: If\:x=\frac{1}{2}$

$\implies 3\times \big(\frac{1}{2}\big)^{2}+4m\times \big(\frac{1}{2}\big)+2=0$

$\sf\implies \frac{3}{4}+2m+2=0$

$\sf\implies \frac{3+8m+8}{4}=0$

$\sf\implies 11+8m=0$

$\sf\implies 8m = -11$

$\sf\implies m = \frac{-11}{8}$

$\sf\implies m = \frac{-11}{8}$

Therefore,.

$value \:of \:m = \frac{7}{4}\:Or \:m =\frac{-11}{8}$