Freddy Friedman

Answered

2022-07-16

Suppose that I have a polynomial of degree 4 like ${x}^{4}+4x+c=0$ where $c\in \mathbb{R}$.

What is the difference when the question says

Show that it has at most two real roots

and

Show that it has exactly two real roots

?

Are approaches to solve the problem in these two cases different?

What is the difference when the question says

Show that it has at most two real roots

and

Show that it has exactly two real roots

?

Are approaches to solve the problem in these two cases different?

Answer & Explanation

kitskjeja

Expert

2022-07-17Added 13 answers

The polynomial ${x}^{4}+4x+c=0$ has only one minimum (or maximum). Taking the derivative and setting it equal to 0 we get

${x}^{3}=-1.$

Among the reals there is only one number that satisfies this condition, namely −1. At −1 the value of the polynomial is $c-3$. Note that if $|x|\to \mathrm{\infty}$ then the value of the polynomial tends to $+\mathrm{\infty}$. (So the extremum is a minimum.)

Then there are three possibilities: (1) $c=3$ then we have exactly one real solution, (2) $c<3$ then we have exactly two real solutions. (3) $c>3$ then there are no real solutions.

So there are at most two real roots; and there are exactly two real roots if $c<3$.

Our polynomial is a continuous function so we can refer to the intermediate value theorem. The intermediate value theorem (in the second special case) says that since at −1 the value of the polynomial is negative and for some $x<-1$ the polynomial's value will be positive, there must be a root in between. Similarly, since for some $x>-1$ there will be positive values again, so in between there must be another zero crossing again. However the intermediate value theorem does not say that there will not be further such zero crossings.

We know (in the second special case) that there are only two zero crossings because we know that our polynomial is strictly decreasing for all $x<-1$ and it is strictly increasing for all $x>-1.$

${x}^{3}=-1.$

Among the reals there is only one number that satisfies this condition, namely −1. At −1 the value of the polynomial is $c-3$. Note that if $|x|\to \mathrm{\infty}$ then the value of the polynomial tends to $+\mathrm{\infty}$. (So the extremum is a minimum.)

Then there are three possibilities: (1) $c=3$ then we have exactly one real solution, (2) $c<3$ then we have exactly two real solutions. (3) $c>3$ then there are no real solutions.

So there are at most two real roots; and there are exactly two real roots if $c<3$.

Our polynomial is a continuous function so we can refer to the intermediate value theorem. The intermediate value theorem (in the second special case) says that since at −1 the value of the polynomial is negative and for some $x<-1$ the polynomial's value will be positive, there must be a root in between. Similarly, since for some $x>-1$ there will be positive values again, so in between there must be another zero crossing again. However the intermediate value theorem does not say that there will not be further such zero crossings.

We know (in the second special case) that there are only two zero crossings because we know that our polynomial is strictly decreasing for all $x<-1$ and it is strictly increasing for all $x>-1.$

Intomathymnma

Expert

2022-07-18Added 5 answers

The first question says, it might have no real roots, or one repeated real root, or two real roots.

You can use the same approach to solve it.

You can use the same approach to solve it.

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