Question

The number of real roots of the equation $3x^4+25x^3+6x^2+25x+3=0$ is

• A. $1$
• B. $0$
• C. $4$
• D. $2$

Answer 1

The correct option is D $2$

Given equation is $3x^4+25x^3+6x^2+25x+3=0$
This is a type $1$ biquadratic equation.
On dividing by $x^2$ and rearranging we get,
$3(x^2+\frac{1}{x^2})+25(x+\frac{1}{x})+6=0$
Substituting, $x+\frac{1}{x}=t,$ we get:
$x^2+\frac{1}{x^2}=t^2-2$
We can write the above equation as,
$\Rightarrow 3(t^2-2)+25t+6=0$
$\Rightarrow 3t^2+25t=0$
$\Rightarrow t(3t+25)=0$
$\Rightarrow t=-\frac{25}{3},0$
But $t\ne 0$ as $t\in (-\infty ,-2]\cup [2,\infty )$

$x+\frac{1}{x}=-\frac{25}{3}$

$\Rightarrow 3x^2+3=-25x$
$\Rightarrow 3x^2+25x+3=0$
Using discriminant method to solve this equation we get,
$x=\frac{-25\pm \sqrt{625-36}}{6}$
$\Rightarrow x=\frac{-25\pm \sqrt{589}}{6}$
Hence, there are $2$ real roots of the equation.