Find $a$ if $(a+3)x^2-4x+2<0$ $\forall x\in [-2,1]$

Question

Find $a$ if $(a+3)x^2-4x+2<0 \qquad \forall x\in [-2,1]$

My Attempt: $a+3=0$ $\implies -4x+2<0$ $$2x-1>0$$ $$x>\dfrac {1}{2}$$ $\implies a \neq -3$.

Answer 1

Let $f(x)=(a+3)x^2-4x+2$

1) Let $a+3>0 \Leftrightarrow a>-3$.

Then $f(-2)<0$ and $f(1)<0$ $f(-2)<0 \Leftrightarrow 4(a+3)+10<0$, then $a<-\frac{11}{2}$

$f(-1)<0 \Leftrightarrow a+3-2<0$, then $a<-1$

2) Let $a+3>0$ and let $x_0=\frac{4}{2(a+3)}$

Then $f(-2)<0$ and $f(1)<0$ and $x_0\not\in[-2;1]$

Answer 2

First we calculate values of $a$ at boundaries for which it is negative $$ (a+3)(-2)^2-4(-2)+2<0 \rightarrow a<-11/2$$ $$(a+3)(1)^2-4(1)+2<0 \rightarrow a<-1$$

Thus for boundary pointwe have $$a<-11/2$$

Now dividing into two cases, namely positive and negative leading coefficient

1) $a+3>0 \rightarrow a>-3$

Now this is not possible as from the values found at boundaries for $a$ this does not satisfy the conditions

2) $a+3<0 \rightarrow a<-3$

Now to make sure all values are negative, we impose the boundary points being negative. On top of that we say that sign of derivative is same for both then only they will be lying on same side of the roots. $$f(x)=(a+3)x^2-4x+2$$ $$f'(x)=2(a+3)x-4$$ now to have same sign we can say $$f'(-2)\cdot f'(1)>0$$ $$(-4(a+3)-4)\cdot(2a-2)>0$$

Find the values of $a$ that satisfy this condition and take an intersection with values of $a$ for which function is negative at $-2,1$

Answer 3

HINT

Let $f(x)= (a+3)x^2-4x+2$

Firstly, we discuss for $a+3$ is positive, negative, or zero.

Then for each case, we check for $x=-\frac{-4}{2(a+3)}=\frac{2}{a+3}$, is it on the left of $[-2,1]$, on the right, or in between?

For example, if $a+3 < 0$:

If $\frac{2}{a+3} \le -2$: it is decreasing, and we just want $f(-2) <0$

If $-2 < \frac{2}{a+3} <1$: we need its local maximal to be negative, $f(\frac{2}{a+3}) < 0$

If $\frac{2}{a+3} \ge 1$: it is increasing, $f(1) <0$

Answer 4

$a\ne -3$ is indeed a necessary condition, but it is in no way sufficient.

Actually, there's a theorem on the sign of a quadratic function:

A quadratic polynomial $ax^2+bx+c$ $(a\ne0)$ has the sign of its leading coefficient $a$, except between its roots, if any.

The (reduced) discriminant of the given polynomial $p(x)$ is $\Delta'=4-2(a+3)=-2(a+1)$. We see that, if there is no real root ($a>-1)$, the leading coefficient $a+3$ is positive, i.e. $p(x)>0$ for all $x$.

If $a<-1$, $p(x)$ has two real roots, and it is asked that $-2$ and $1$ separate these roots, which by the cited theorem means that $$\begin{cases} (a+3)p(-2)<0\\(a+3)p(1)<0 \end{cases}\iff\begin{cases} 2(2a+11)<0\\ a+1<0 \end{cases} $$ Thus the set of solutions is the interval $\bigl(-\infty,-\frac{11}2\bigr) $.