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 integral values of ${}^{\prime }{m}^{\prime }$ such that given quadratic equation has imaginary roots are,

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

By calculating the discriminant, we get

$4(1+3m{)}^2-4(1+8m\left)\right(1+m)\ge 0$ for real roots.

$(1+3m{)}^2-(1+9m+8m^2)\ge 0$

$1+9m^2+6m-1-9m-8m^2\ge 0$

$\Rightarrow m^2-3m\ge 0$

$\Rightarrow m(m-3)\ge 0$.

Hence, $m\le 0$ and $m\ge 3$.

Hence, $m\in (-\infty ,0]\cup [3,\infty )$.

Now for imaginary root,$m\in (0,3)$

Integral values of $m$ will be $1,2$

Hence, $2$ values.