Question

The number of imaginary roots of the equation $4x(x^2+x+3)+5x(x^2-5x+3)=-\frac{3}{2}$ is

• A. $4$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$4x(x^2+x+3)+5x(x^2-5x+3)=\frac{-3}{2}$

$9x^3-21x^2+27x=\frac{-3}{2}$

$6x^3-14x^2+18x+1$

Let $f\left(x\right)=6x^3-14x^2+18x+1$

$f^{\prime }\left(x\right)=18x^2-18x+18$

$△={\left(28\right)}^2-4\left(18\right)\left(18\right)<0$

$\therefore f^{\prime }\left(x\right)>0\forall xϵR$

$\therefore f\left(x\right)$ is monotonically increasing graph

i.e., if $x_1>x_2$ then $f\left(x_1\right)>f\left(x_2\right)$

$f\left(0\right)=1>0$

$f(-1)=-37<0$

$\therefore f\left(x\right)$ has only one real root

That only real root is in $(-1,0)$