# Quartic Equation Inequality

I am trying to find the range of b for which the following quartic inequality is satisfied where $0<a\leq\frac{1}{8}$. $$b^4+(2a-1)b^2+a^2+a<0.$$ Let $b^2=x$. Then, we get: $$x^2+(2a-1)x+(a^2+a)<0.$$ I am not sure how to explicitly solve the inequality, so I tried to find the roots with the hope of being able to divide $b$ into subintervals and seeing where the inequality above is less than 0. Using the quadratic formula, I get that: $$x = \frac{1-2a \pm \sqrt{1-8a}}{2}.$$ Does it immediately follow that if $x<0$, then the inequality holds? I feel like this is not true. This would hold in the case the inequality was linear, but otherwise it should fail. From here, I am not sure how to check whether the inequality is less than or greater than 0.

Any and all help is greatly appreciated! Some one advised that I plug this into Maple but I hope to find the method of solving equations like this for the future!

Thanks!

Let's make a functional approach.

We have the inequality : $$b^4+(2a-1)b^2+a^2+a<0$$

By letting $b^2 = x$ we get : $$x^2+(2a-1)x+(a^2+a)<0$$

Now, let's name a function $$f:\mathbb R \to \mathbb R$$ $$f(x) = x^2+(2a-1)x+(a^2+a)$$

Differentiating : $$f'(x) = 2x + 2a-1$$

Taking the derivative equal to zero : $$f'(x) = 0 \Leftrightarrow 2x + 2a-1 = 0 \Leftrightarrow x = \frac{1-2a}{2}$$

Pretty elementary to see that the function is decreasing until $x = \frac{1-2a}{2}$ and increasing after it.

Solving the equation : $$x^2+(2a-1)x+(a^2+a) = 0\Leftrightarrow$$

$$x = \frac{1-2a \pm \sqrt{1-8a}}{2}$$

So, now it's obvious that : $f(x) < 0$ when $x \in ( \frac{1-2a - \sqrt{1-8a}}{2},\frac{1-2a + \sqrt{1-8a}}{2})$

• This is an approach if you do not know how to handle quadratic inequalities. Otherwise, there is a method form/way that you approach such staff which is pretty straight forward. – Rebellos Apr 26 '17 at 9:36