Question

Solve for real x $x^4-2x^3+x-380=0$

Answer 1

Considering the quesion to be,

to find the roots of the equation $x^4-2x^3+x-380=0$

Now by trial and error method we find that $x=5$ is a solution to the equation.

$5^4–2\times 5^3+5–380=0$

$=625–250+5–380=0$

$380–380=0$

Thus $x=5$ is a solution.

We can say that $x-5$ divides : $x^4–2x^3+x-380$

We also find that $x=-4$ is a solution

$-4^4–2\times -4^3–4–380=0$

$=256+128–4–380$

$380–380=0$

We can say that $x+4$ divides:$-x^4–2x^3+x-380$

In a polynomial:

sum of roots $=\frac{-b}{a}$

Let the other roots be: $\alpha ,\beta$

$\alpha +\beta +5-4=-\left(\frac{-2}{1}\right)$

$\Rightarrow \alpha +\beta =2-1=1$

$\Rightarrow \alpha =1-\beta$ ........(1)

product = $=da$

$5\times -4\times \alpha \times \beta =-380$

$\Rightarrow \alpha \beta =19$

$\Rightarrow \beta \times (1-\beta )=19$

$\Rightarrow \beta -{\beta }^2-19=0$

Edit:

$\Rightarrow \beta -{\beta }^2-19=0$

$\Rightarrow \beta =\frac{1\pm \sqrt{1-76}}{2}$

$\Rightarrow \beta =\frac{1\pm \sqrt{-75}}{2}$

$\frac{1\pm i5\sqrt{3}}{2}$

Now we can find,

$\alpha =1-\beta$

$\Rightarrow 1-\frac{1\pm i5\sqrt{3}}{2}$

$\Rightarrow \frac{2-1\pm i5\sqrt{3}}{2}$

$\Rightarrow \alpha =\frac{1\pm i5\sqrt{3}}{2}$

We see that alpha and beta are equal.

Hence the roots of the equation are:

$\frac{1+i5\sqrt{3}}{2}$

$\frac{1-i5\sqrt{3}}{2}$

The real roots are : $5$ and $-4$