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

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The discussion establishes that if the quadratic equation $ax^2+(c-b)x+e-d=0$ has real solutions greater than 1, then the quartic equation $ax^4+bx^3+cx^2+dx+e=0$ must have at least one real solution. By defining the function $f(x) = ax^4+bx^3+cx^2+dx+e$ and evaluating it at $\sqrt{s}$ and $-\sqrt{s}$, the discussion demonstrates that either $f(\sqrt{s})$ and $f(-\sqrt{s})$ have opposite signs or both are zeros of $f(x)$. This conclusion is supported by the intermediate value theorem, confirming the existence of a real solution for the quartic equation.

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anemone
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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.
 
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anemone said:
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]
 
Opalg said:
[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!
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!
 

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