Show that $x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4=0$ has at least two solutions

Question

Let $$p(x)= x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4$$ be a polynomial with coefficients $a_1, a_2, a_3, a_4$.

If $a_4 < 0$, then show $p(x)$ has at least 2 solutions.

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So far I thought, about using the Intermediate Value Theorem but this only shows that there must exist a solution. I know it is obvious that it must have 2 solutions since it behaves like a parabola and the less term being negative obligates the function to cross the x-axis at least twice.

So, my problem is how can I prove this? is it possible to do it by contradiction? can I assume there is less than 2 solutions?

I know since $a_4 < 0$ then is impossible to not have solutions, since not all x in the function p are positive indeed we have $p(0)<0$.

But now how can I show that cannot have only one solution? I know is like a parabola so it needs to have at least 2 but is there a more mathematically way to do this?

Answer 1

Method 1

Note that $$\lim_{x\to\infty}p(x)=\lim_{x\to-\infty}p(x)=\infty$$ Now, since the value it takes at zero is negative, it has to cross the $x$-axis twice by intermediate value theorem.

Method 2

Since the non-real complex roots occur in conjugate pairs, it cannot occur that it has one real and three non-real complex roots. Suppose that it has no real root. Then the imaginary roots are $a+ib,a-ib,c+id,c-id$ for some $a,b,c,d\in \mathbb R$. Then the product of the roots is $$a_4 = (a^2+b^2)(c^2+d^2)>0$$ which is a contradiction. Hence, it must have at least two real roots.

Answer 2

$f(-\infty)=f(\infty)>0 \implies f(0)$ has $0$ or $2$ or $4$ real solutions.$f(0)=a_4$ If $f(0)<0 \implies a_4<0$, it will have two real roots one in $(-\infty, 0)$ and one in (0, \infty)$$. So if $a_4<0$, $F(x)=0$ has at least two real roots.