# Thread: Polynomial of degree 4 with 2 real roots

1. ## Polynomial of degree 4 with 2 real roots

If $ax^4 + (2a+b)x^2 + c + a = 0$ has only 2 real roots, which of the following is true? A. b=-2a B. b>-2a C. b<2a D. b<-2a E. none of them above.
Can someone pls explain how to solve this? Thanks in advance~

2. ## Re: Polynomial of degree 4 with 2 real roots

Originally Posted by shiny718
If $ax^4 + (2a+b)x^2 + c + a = 0$ has only 2 real roots, which of the following is true? A. b=-2a B. b>-2a C. b<2a D. b<-2a E. none of them above.
Can someone pls explain how to solve this? Thanks in advance~
This is a quadratic equation in \displaystyle \begin{align*} x^2 \end{align*}, so let \displaystyle \begin{align*} X = x^2 \end{align*} and the equation becomes

\displaystyle \begin{align*} a\,X^2 + \left( 2a + b \right) X + c + a &= 0 \\ X &= \frac{-(2a + b) \pm \sqrt{(2a + b)^2 - 4a(c + a)}}{2a} \\ x^2 &= \frac{-(2a + b) \pm \sqrt{(2a + b)^2 - 4a(c + a)}}{2a} \\ x &= \pm \sqrt{\frac{-(2a + b) \pm \sqrt{(2a + b)^2 - 4a(c + a)}}{2a}} \end{align*}

This will have exactly two solutions if \displaystyle \begin{align*} (2a + b)^2 - 4a(c + a) = 0 \end{align*} AND \displaystyle \begin{align*} \frac{-(2a + b)}{2a} > 0 \end{align*}.

3. ## Re: Polynomial of degree 4 with 2 real roots

if $ax^4 + (2a+b)x^2 + c + a = 0$ has but 2 real roots, then $au^2 + (2a+b)u + c + a = 0$ must have but 1 real root, so its discriminant:

$\Delta = (2a+b)^2 - 4(a)(c+a) = 0$

since $(2a+b)^2 - 4(a)(c+a) = 4a^2 + 4ab + b^2 - 4ac - 4a^2 = b^2 + 4(ab - ac)$ we have that:

$b^2 = 4a(c - b)$

EDIT: upon further reflection, and seeing ProveIt's answer, it's clear that although this condition is necessary, it is not sufficient, because u must be non-negative for x to be real.

this means that u = -(2a+b)/2a when Δ = 0. unfortunately, if we do not know if a > 0, we cannot decide if we should have 2a+b > 0, or 2a+b < 0.

one thing is certain, however, if b = -2a, then we only have one root for x.