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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.

Thomas Andrews
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Math-Learner
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3 Answers3

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Direct substitution with $\alpha=x$ and $c=x^2-x$ leads to.

$x^4-x^3+x^2-x(x^2-x+1)-((x^2-x)^2+x^2-x)$

$=(1-1)x^4+(2-2)x^3+(2-2)x^2+(1-1)x=0$

herb steinberg
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    This does not prove that $c = \alpha^2 - \alpha$ is the only solution. For one thing, it does not use either of the conditions that $\alpha \in \mathbb R$ and $c \gt 0$, which should be a red flag. – dxiv May 30 '21 at 04:13
  • @dxiv The question asks only if it is a solution, not what all solutions may be. – herb steinberg May 30 '21 at 16:25
  • The way I read it, the question is to prove that $c$ must be $\alpha ^2 - \alpha$, not that it can be that. – dxiv May 30 '21 at 16:48
  • @dxiv I'll leave it to the poster to clear this up. – herb steinberg May 30 '21 at 16:52
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As per reading your comments, you have got that $x^2-x-c=0$.

$p(\alpha)=\alpha^2-\alpha-c=0$

$$ \alpha^2-\alpha-c=0 $$

$$ c=\alpha^2-\alpha $$

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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.

Toothpick Anemone
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    You know what you expect to be under the radical, so it should be easy to verify that in fact$4\alpha^4-4\alpha^3+5\alpha^2 -2 \alpha + 1 = \left(2 \alpha^2 - \alpha + 1\right)^2$. Then you still have to show that one, and only one, of the two roots is eligible. – dxiv May 30 '21 at 04:17
  • So This approach gives two real $c$! – Math-Learner May 30 '21 at 04:24
  • Two real $c$, but one is $c=-\alpha^2-1$ which is always negative. – Peter May 30 '21 at 06:22