Find a and b in equation given range of x

Question

I have the problem to find $a$ and $b$ given

$-ax^2+bx+4\geqslant0$, $-1/3\leqslant x\leqslant4$

and have they key with the answer $a=3,b=11$, but which steps do I take to get to that answer?

Answer 1

What is given says that the roots of the equation $-ax^2+bx+4=0$ are $x_1=-1/3, x_2=4$ and $a>0$. Therefore

-a(-1/3)^2+b(-1/3)+4=0

and

$-a(4)^2+b.4+4=0$

This is the same as

$-a-3b+36=0$

$-16a+4b+4=0$

Solve the system of these equations to get $a=3$, $b=11$.

Answer 2

We have to find a two degree polynomial function which is positive when $-\dfrac13\le x\le 4$.

So, actually have to find a polynomial which is downward facing. So, our co-efficient of $x^2$ should be negative. So, we set up a polynomial function as follows:

$f(x)=-(x+\dfrac13)(x-4)=-x^2+\dfrac{11}{3}x+\dfrac43$.

Now, if we eradicate the fraction part, we can get the polynomial to be of the form, $f(x)=-3x^2+11x+4$.

So, we find $a=3,b=11$.

Hence, done.

Answer 3

Hints: let $\;\alpha\,,\,\beta\;,\;\;\alpha\le\beta\;$, be the roots of the given quadratic, then

$$-ax^2+bx+4=-a(x-\alpha)(x-\beta)\ge 0\iff\begin{cases}x\le\alpha\;\;or\;\;x\ge\beta&,\;\;\text{if}\;\;a<0\\{}\\\alpha\le x\le\beta&,\;\;\text{if}\;\;a>0\end{cases}$$

Since it is given that the solution set is$\;-\frac13\le x\le 4\;$ we're in the second case. The solution now follows from Viete's formulas:

$$\begin{cases}\alpha\beta=-\frac4a\\{}\\\alpha+\beta=-\frac ba\end{cases}$$