finding sum of roots of equation

Question

The sum of all the real number $x$ such that $5x^4 - 10x^3 + 10x^2 - 5x - 11=0$, without aid of Calculus is there any way I tried to find one root inspection but was not getting anywhere, and tried recombining to get a solvable equation. Any feasible approach.

Answer 1

Note that\begin{align*}5x^4-10x^3+10x^2-5x-11&=5\left(x^4-2x^3+2x^2-x-\frac{11}5\right)\\&=5\left(\left(x^2-x+\frac12\right)^2-\frac{49}{20}\right)\\&=5\left(x^2-x+\frac12-\frac7{2\sqrt5}\right)\left(x^2-x+\frac12+\frac7{2\sqrt5}\right).\end{align*}The equation $x^2-x+\frac12-\frac7{2\sqrt5}=0$ has two real roots, whereas the equation $x^2-x+\frac12+\frac7{2\sqrt5}=0$ has none. So, the sum of the real roots of your polynomial is the some of the real roots of the polynomial $x^2-x+\frac12-\frac7{2\sqrt5}$, which is $1$.

Answer 2

Rewrite our equation in the following form. $$x^5+(1-x)^5=12.$$ Now, let $1-x=y$.

Thus, we need to solve $$x^5+y^5=12$$ or $$x^4+y^4+x^2y^2-xy(x^2+y^2)=12$$ or $$x^4+2x^2y^2+y^4-x^2y^2-xy(1-2xy)=12$$ or $$(x^2+y^2+xy)(x^2+y^2-xy)-xy(1-2xy)=12$$ or $$(1-xy)(1-3xy)-xy(1-2xy)=12$$ or $$5x^2y^2-5xy-11=0,$$ which gives $$xy=\frac{5-\sqrt{245}}{10}$$ because $$xy\leq\left(\frac{x+y}{2}\right)^2=\frac{1}{4}.$$

Thus, we need to solve $$x^2-x-\frac{7\sqrt{5}-5}{10}=0,$$ which gives two roots: $$\frac{1+\sqrt{\frac{14}{\sqrt{5}}-1}}{2}$$ and $$\frac{1-\sqrt{\frac{14}{\sqrt{5}}-1}}{2}.$$ Done!

Answer 3

If you have an equation like this $$P (x) = (x - a) (x - b) (x - c) (x - d)=0$$ then $a,b,c,d$ are the solutions. Thus expanding polynomial $P(x)$ we get $$x^4 -(a+b+c+d)+x^2 (a b+a c+a d+b c+b d+c d)-x (a b c+a b d+a c d+b c d)+a b c d=0$$ therefore the sum of the roots of your equation can be easily obtained dividing the equation by the leading coefficient $$x^4-2 x^3+2 x^2-x-\frac{11}{5}=0$$ the coefficient of the term having degree $n-1$ is the opposite of the sum of the roots, so $x_1+x_2+x_3+x_4=2$

Hope it is clear