Question

Roots of the equation is $x^4-2x^3+x=380$.

• A. $5$
• B. $-4$
• C. $\frac{1\pm \sqrt{-75}}{2}$
• D. All of the above

Answer 1

The correct option is D All of the above

Given, $x^4-2x^3+x=380$

$\Rightarrow x^4-2x^3+x^2-x^2+x=380$

$\Rightarrow {(x^2-x)}^2-(x^2-x)=380$

Put $x^2-x=t$

$\Rightarrow t^2-t=380\Rightarrow t^2-t-380=0\Rightarrow t^2-20t+19t-380=0\Rightarrow t(t-20)+19(t-20)=0\Rightarrow (t+19)(t-20)=0\Rightarrow t=-19,20\Rightarrow x^2-x=t$

For $t=-19$, w ehave

$x^2-x=-19\Rightarrow x^2-x+19=0$

Using quadratic formula

$x=\frac{1\pm \sqrt{1-4\left(19\right)}}{2}=\frac{1\pm \sqrt{-75}}{2}$

For $t=20$, we have

$x^2-x=20\Rightarrow x^2-x-20=0\Rightarrow x^2-5x+4x-20=0\Rightarrow x(x-5)+4(x-5)=0\Rightarrow (x+4)(x-5)=0\Rightarrow x=-4,5$

So, the values of $x$ are $\frac{1+\sqrt{-75}}{2},\frac{1-\sqrt{-75}}{2},-4$ and $5$.