Question

(I)Find the value of m for which the roots of the quadratic equation $mx(6x+10)+25=0,$ are equal.

(II)Solve for x, $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}$$,x\ne -1,-2,-4$

Answer 1

$\left(I\right)$ $mx(6x+10)+25=0$

$6x^{2}m+10mx+25=0$

Roots are equal

$\therefore {\alpha }^2=\frac{25}{6m}$

$\alpha =\frac{-5}{6}$

$\to {\alpha }^2=\frac{25}{36}$

$\therefore \frac{25}{36}=\frac{25}{6m}$

$\therefore m=6$

$\left(II\right)$ $\frac{1}{x+1}+\frac{2}{x+2}=\frac{4}{x+4}$

$⟹[x+2+2x+2][x+4]=4[x^2+3x+2]$

$⟹(3x+4)(x+4)=4(x^2+3x+2)$

$⟹3x^2+16x+16=4x^2+12x+8$

$⟹x^2-4x-8=0$

$⟹x=\frac{4\pm \sqrt{16-4\left(8\right)}}{2}=\frac{4\pm i16}{2}=2+i8$