Let $c$ be a positive real number for which the equation $x^4-x^3+x^2-(c+1)x-(c^2+c)=0$ has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$ Let $c$ be a positive real number for which the equation
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$
I tried to to solve using relation between roots and coefficients but unable to progress much. Please help. Thanks in advance.
 A: Direct substitution with $\alpha=x$ and $c=x^2-x$ leads to.
$x^4-x^3+x^2-x(x^2-x+1)-((x^2-x)^2+x^2-x)$
$=(1-1)x^4+(2-2)x^3+(2-2)x^2+(1-1)x=0$
A: As per reading your comments, you have got that $x^2-x-c=0$.
$p(\alpha)=\alpha^2-\alpha-c=0$
$$ \alpha^2-\alpha-c=0 $$
$$ c=\alpha^2-\alpha $$
A: Suppose that $\alpha$ is a root to the equation:
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$

Okay, then we have:
$\alpha^4-\alpha^3+\alpha^2-(c+1)\alpha-(c^2+c)=0$

$\begin{align}
\text{You can} & \text{put it in standard form as:} \\
& \\
& c^2+(1+\alpha)c^1 + (-\alpha^4+\alpha^3-\alpha^2 + \alpha)c^0=0 \\ 
\end{align}$

$\begin{align}
\text{Use the quadratic formula} & \\  
& \\
 \forall b,c, x \in \mathbb{R},&\\
& x^2 + bx^1+cx^{0} = 0 \text{ if and only if } x = \dfrac{-b\pm\sqrt{b^2-4c}}{2} \\
\end{align}$

$c=\dfrac{-(1+\alpha)\pm\sqrt{(1+\alpha)^2-4(-\alpha^4+\alpha^3-\alpha^2 + \alpha)}}{2}$
$c=\dfrac{-(1+\alpha)\pm\sqrt{(1+2\alpha+\alpha^2)+(4\alpha^4-4\alpha^3+4\alpha^2 -4 \alpha)}}{2}$
$c=\dfrac{-(1+\alpha)\pm\sqrt{4\alpha^4-4\alpha^3+5\alpha^2 -2 \alpha + 1}}{2}$
Maybe, after some additional work, We have that that $c=\alpha ^2 - \alpha$
I don't know.
