Prove a quartic equation ax^4+bx^3+cx^2+dx+e=0 has at least one real solution

[anemone]

If the equation $ax^2+(c-b)x+e-d=0$ has real solutions that are greater than 1, prove that the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real solution.

 

[Opalg]

If the equation $ax^2+(c-b)x+e-d=0$ has real solutions that are greater than 1, prove that the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real solution.

[sp]Let $s$ be a solution of $ax^2+(c-b)x+e-d=0$ with $s>1$. Then $as^2 + cs + e = bs+d.$

Let $f(x) = ax^4+bx^3+cx^2+dx+e$. Then $f(\sqrt s) = as^2 + cs + e + \sqrt s(bs+d) = (bs+d)(1 + \sqrt s).$ In the same way, $f(-\sqrt s) = (bs+d)(1- \sqrt s).$

If $bs+d = 0$ then $\sqrt s$ and $ - \sqrt s$ are zeros of $f(x)$. If $bs+d \ne 0$ then $f(\sqrt s)$ and $f(-\sqrt s)$ have opposite signs (because $\sqrt s > 1$). So by the intermediate value theorem $f(x)$ must take the value zero somewhere between $-\sqrt s$ and $\sqrt s$. In either case, the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real solution.

[/sp]

 

[anemone]

[sp]Let $s$ be a solution of $ax^2+(c-b)x+e-d=0$ with $s>1$. Then $as^2 + cs + e = bs+d.$

Let $f(x) = ax^4+bx^3+cx^2+dx+e$. Then $f(\sqrt s) = as^2 + cs + e + \sqrt s(bs+d) = (bs+d)(1 + \sqrt s).$ In the same way, $f(-\sqrt s) = (bs+d)(1- \sqrt s).$

If $bs+d = 0$ then $\sqrt s$ and $ - \sqrt s$ are zeros of $f(x)$. If $bs+d \ne 0$ then $f(\sqrt s)$ and $f(-\sqrt s)$ have opposite signs (because $\sqrt s > 1$). So by the intermediate value theorem $f(x)$ must take the value zero somewhere between $-\sqrt s$ and $\sqrt s$. In either case, the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real solution.

[/sp]

Awesome, Opalg!

Spoiler

My attempt revolved around the intermediate value theorem as well, but I tried to work with letting the roots be -1, 0 and 1 and those didn't help at all...therefore thank you so much for the intelligent approach!