Question

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

• A. $4$
• B. $2$
• C. $0$

Answer 1

The correct option is B $2$

Given: $2x^4-4x^3-4x+2=0$
Now, the equation is of the form $a{x}^4+b{x}^3+c{x}^2+bx+a=0$
Comparing which, we get: $a=2;b=-2;c=0$
Now, dividing the equation with $x^2,$ we get the transformed equation as:
$2x^2-4x-\frac{4}{x}+\frac{2}{x^2}=0\Rightarrow 2(x^2+\frac{1}{x^2})-4(x+\frac{1}{x})=0$
Now, let $x+\frac{1}{x}=t\text{where}t\in (-\infty ,-2]\cup [2,\infty )$
Thus, we get the transformed equation as:
$2(t^2-2)-4t=0\Rightarrow 2t^2-4t-4=0\Rightarrow t^2-2t-2=0$
Now, using the quadratic formula:
$t=\frac{2\pm \sqrt{4+8}}{2}\Rightarrow t=\frac{2\pm 2\sqrt{3}}{2}\Rightarrow t=1\pm \sqrt{3}$
Now, $1-\sqrt{3}\in (-2,2)$
Thus, $t=1-\sqrt{3}$ is ignored.
$\Rightarrow t=1+\sqrt{3}=x+\frac{1}{x}\Rightarrow x+\frac{1}{x}=1+\sqrt{3}\Rightarrow x^2-(1+\sqrt{3})x+1=0$
Now, the discriminant of this equation:
$D=(1+\sqrt{3}{)}^2-4=1+3+2\sqrt{3}-4=2\sqrt{3}>0$
Thus, $D>0$ thus the equation will give two real values of $x$