Non negativity condition for polynomials?

Question

Let $p(x)=3ax^2+2bx+(1-a-b), \,\,\,\,0<x<1$

What are the conditions on a and b to $p(x)>0$

Could someone give me a hint ?

Answer 1

Hint: complete the square. This means writing $p(x)$ in the form $$p(x)=A(x+B)^2+C $$ where $A$, $B$, $C$ are constants possibly depending on $a$ and $b$, but not on $x$. When you're done you will have expressed $p(x)$ as a nonnegative piece $A(x+B)^2$ (this requires $A\ge0$) and a constant $C$. So if we require $p(x)>0$ for all $x$ we must have $A\ge0$ and $C>0$.

In your case you'd complete the square by starting out $$ P(x)=3a(x^2+\frac{2b}{3a}x)+(1-a-b) . $$ To get this first section in the form $A(x+B)^2$ you'd want to complete the square with $(\frac b{3a})^2$. To repair what you've just done, you need to subtract it off again: $$P(x)=3a\left(x^2+\frac{2b}{3a}x + (\frac b{3a})^2 - (\frac b{3a})^2)\right) + (1-a-b).$$ Next, regroup: $$\begin{align} P(x)&=3a\left(x^2+\frac{2b}{3a}x + (\frac b{3a})^2\right) -3a(\frac b{3a})^2 + (1-a-b).\\ &=3a\left(x+\frac b{3a}\right)^2 -3a(\frac b{3a})^2 + (1-a-b).\\ \end{align} $$ From this last expression you can read off $A$, $B$, $C$. (You can simplify $C$ down to a more tidy expression.)

Answer 2

Hint:

1. You must have $3a>0$

2. The derivative is $6ax+2b$ vanishes at $x = -{{b}\over{3a}}$ which is the minimum if $a>0$.

3. Find $a,b$ such that $p(-{b\over{3a}})>0$.