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

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In summary: You summarized the conversation perfectly. In summary, if the equation $ax^2+(c-b)x+e-d=0$ has real solutions that are greater than 1, then the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real solution due to the properties of real numbers and the intermediate value theorem.
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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]
 
  • #3
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!
 

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

1. How do you prove that a quartic equation has at least one real solution?

The most common method is to use the fundamental theorem of algebra, which states that a polynomial of degree n has n complex roots. Since a quartic equation has degree 4, it must have at least 1 real solution. This is because complex roots always come in pairs, so if there are 4 complex roots, there must be 2 real roots.

2. Can you provide an example of a quartic equation with one real solution?

One example is the equation x4 + 3x3 + 2x2 + 4x + 3 = 0. This equation has one real solution, which is approximately -0.835.

3. Are there any other methods to prove the existence of a real solution for a quartic equation?

Yes, another method is to use the Descartes' rule of signs. This rule states that the number of positive real roots of a polynomial is equal to the number of sign changes in the coefficients or less by an even number. Similarly, the number of negative real roots is equal to the number of sign changes in the coefficients or less by an odd number.

4. Is it possible for a quartic equation to have more than one real solution?

Yes, it is possible for a quartic equation to have up to 4 real solutions. However, it is also possible for it to have 2 or 3 real solutions. The number of real solutions depends on the coefficients of the equation.

5. Can a quartic equation have no real solutions?

Yes, a quartic equation can have no real solutions. This is because the fundamental theorem of algebra and Descartes' rule of signs only guarantee the existence of at least one real solution, but not necessarily more. In rare cases, all 4 roots of a quartic equation may be complex, meaning there are no real solutions.

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