Question

Consider the quadratic equation $(1+m)x^2-2(1+3m)x+(1+8m)=0$, (where $m\in R-\{-1\})$, then the number of real values of $m$ such that the given quadratic equation has roots in the ratio $2:3$ are,

• A. $0$
• B. $2$
• C. $3$
• D. Infinitely many

Answer 1

Answer: (B)

$2$

Let the roots be $2\alpha$ and $3\alpha$,

$\therefore 2\alpha +3\alpha =\frac{2(1+3m)}{(1+m)}$

$\Rightarrow \alpha =\frac{2(1+3m)}{5(1+m)}.....\left(i\right)$

$\Rightarrow 2\alpha \cdot 3\alpha =\frac{(1+8m)}{1+m)}$

$\Rightarrow {\alpha }^2=\frac{(1+8m)}{6(1+m)}.....\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{4(1+3m{)}^2}{25(1+m{)}^2}=\frac{(1+8m)}{6(1+m)}$

$\Rightarrow 24(1+3m{)}^2(1+m)=25(1+m{)}^2(1+8m)$

$\Rightarrow (1+m)(16m^2-81m-1)=0$

$\therefore m=-1,m=\frac{81\pm \sqrt{(81{)}^2+16\times 4}}{2\times 16}=\frac{81\pm 5\sqrt{\left(265\right)}}{32}$

But $m=-1$ is not a solution. Therefore, there are two values of $m$