Quadratic Inequalities

Normally, quadratic inequalities would involve intervals as solution sets. However, we can investigate the cases that could possibly lead to a unique solution.

First, let us get the degenerate cases out of the way where one of the expressions become linear. This may or may not be necessary but it is an easy enough exercise. By checking $c=1$ and $c=0$ , we find that neither satisfy our needs.

Next, we consider the cases where either of the quadratic expressions have only one (repeated) real root by using the discriminant:

$0=(2c)^2-4(c-1)(c+4)\Rightarrow c=\frac{4}{3}$
$0=[2(c+1)]^2-4c(c+1)\Rightarrow c=-1$
Substituting these values into our system, we find that only $c=\frac{4}{3}$ would yield a unique solution for the system.

Lastly, we check the cases where the two quadratic expressions share a common root by using the quadratic formula: $\frac{-2c\pm\sqrt{(2c)^2-4(c-1)(c+4)}}{2(c-1)}=\frac{-2(c+1)\pm\sqrt{[2(c+1)]^2-4c(c+1)}}{2c}$

These equations only yield two candidate values, $-\frac{3}{4}$ and $\frac{5}{4}$ . Once more, we substitute these into our system and find that only $-\frac{3}{4}$ will yield a unique solution.

Hence, the sum of all possible $c$ 's that would make the system have a unique solution is $\frac{4}{3}-\frac{3}{4}=\frac{7}{12}$ and our answer, $19$ , follows immediately.

For brevity, label the polynomials as follows:

$L_1(x) = (c-1)x^2 + 2cx + c + 4 \leq 0$ $L_2(x) = cx^2 + 2(c+1)x + c + 1 \geq 0$

We will first discuss when solutions are unique, and then find the solutions.

When does the system of inequalities have a unique solution? If a solution is unique it must occur when either $L_1(x) = 0$ or $L_2(x) = 0$ . Further, we can require that a unique solution is (1) the only solution for either $L_1(x) = 0$ or $L_2(x) = 0$ , or (2) a point $x_0$ for which both $L_1(x_0) = 0$ and $L_2(x_0) = 0$ .

For (1) suppose that $a$ is a non-unique solution to $L_1(x) = 0$ , and $L_2(a) > 0$ . If there are two solutions to $L_1(x) = 0$ , then $L_1(x)$ crosses the x-axis twice, so clearly there exists some $\epsilon$ (positive or negative) for which $L_1(a + \epsilon) < 0$ . Pick $\epsilon$ small enough; then $L_2(a + \epsilon) > 0$ still. Then $(a + \epsilon )$ is also a solution to the system of inequalities, and $a$ is not a unique solution. Cases where $L_1(a) < 0$ and $L_2(a) \geq 0$ , or $L_1(a) < 0$ and $L_2(a) > 0$ , or where the quadratic term of $L_1(x)$ or $L_2(x)$ vanishes follow by similar arguments.

For (2), it remains to show that if $a$ is a point such that both $L_1(a) = 0$ and $L_2(a) = 0$ , then $a$ does not have to be a unique solution to either $L_1(x) = 0$ nor $L_2(x) = 0$ . It may be the case that for a small $\epsilon > 0$ , we have $L_1(a - \epsilon)$ and $L_2(a - \epsilon)$ greater than 0, and both $L_1(a + \epsilon)$ , and $L_2(a + \epsilon)$ less than 0. So it is possible that $a$ is a unique solution, then, since $L_1(x)$ is not satisfied before $a$ and $L_2(x)$ is not satisfied after, but both are satisfied at $a$ .

What remains now is to compute the solutions. The natural thing to do is to look for unique solutions to either $L_1(x) = 0$ or $L_2(x) = 0$ , and solutions to $L_1(x_0) = L_2(x_0) = 0$ . We will first rewrite the system of inequalities by factoring:

$L_1(x) = (c-1)(x+1)^2 + 2x + 5 \leq 0$ and $L_2(x) = c(x+1)^2 + 2x + 1 \geq 0$ .

We can simplify this further if we substitute $u = x + 1$ :

$F_1(u, c) = L_1(x) = (c-1)u^2 + 2u + 3 \leq 0$ $F_2(u, c) = L_2(x) = cu^2 + 2u - 1 \geq 0$

Now we will examine unique solutions of $F_1(u, c)$ and $F_2(u, c)$ case by case. We know that a solution to a quadratic equation is unique if and only if the determinant (number under the square root of the quadratic formula) is 0. Plugging in and solving for c, we get $\det F_1(u, c_1) = 0 \implies c_1 = 4/3$ . Since $F_1(u, 4/3)$ bends upwards (since $4/3 - 1 > 0$ ), there is only one solution to $F_1(u, 4/3) \leq 0$ . Solve $F_1(u, 4/3) = 0$ for $u$ to get $u = -3$ . Check that $F_2(-3, 4/3)$ is indeed greater than 0 to confirm our first unique solution, $c_1 = 4/3$ . Repeating the same process for $F_2(u, c)$ does not yield a unique solution to the system of inequalities.

Lastly we solve $F_1(u, c) = F_2(u, c)$ :

$cu^2 - u^2 + 2u + 3 = cu^2 +2u -1$ , so $u^2 = 4$

This means is that a solution $F_1(u, c) = F_2(u, c)$ does not depend on $c$ at all, it is true whenever $u^2 = 4$ . Solve $F_2(2, c_2) = 0$ for $c_2$ to find $c_2 = - 3/4$ . Note that nearby $u=2$ , both $F_1(u, -3/4)$ and $F_2(u, -3/4)$ are greater than 0 on the left, and less than 0 on the right, since both $F_1(u, -3/4)$ and $F_2(u, -3/4)$ curve downwards and both $\max F_1(u, -3/4)$ and $\max F_2(u, -3/4)$ occur for $u < 2$ . Additionally, since $F_1(u, -3/4)$ crosses the x-axis to the left of $F_2(u, -3/4)$ (besides $u = 2$ , of course), we have confirmed our second unique solution, $c_2 = -3/4$ . Repeating the same process for $u = -2$ does not yield a unique solution.

Therefore, unique solutions exist for $c_1 = 4/3$ and $c_2 = -3/4$ with $c_1 + c_2 = 7/12 = a/b$ , so $a + b = 19$ .

When dealing with inequalities of the form $ax^2 + bx + c$ ,we usually have two types of solutions $[ d ,e ]$ or $( -\infty , d] U [e,\infty)$ where $d$ and $e$ are roots of the quadratic and $d<=e$ . We are looking for unique solutions for the system and one type of unique solution is when the solution of one of the quadratics is of the form $( -\infty , d] U [e,\infty)$ and the other has one solution or $[ f , f]$ . (The intersection of the two sets yields one distinct solution). . A quadratic inequality has one solution( $[f,f]$ ) whenever the discriminant is 0. So we can find one type of c by setting the discriminant of $(c-1)x^2 + 2cx + c + 4 <=0$ to 0,which becomes $4c^2 - 4(c-1)(c+4)=0$ which gives us $3c=4$ or $c=4/3$ before concluding this is a solution we check it by replacing it in the system and confirming it is correct. We can also do this for the second inequality which gives us $c=-1$ but we see that this is incorrect as we check by plugging c into the system. We then realize that the only solutions are not only the ones that nullify discriminant,but $c$ 's that make the two equations equivalent could also be valid if they satisfy the constraints. So we equate the two quadratics $(c-1)x^2 + 2cx + c + 4 = cx^2 + 2(c+1)x + c+1$ after simplifying,the above reduces to $x^2+2x-3$ ,this gives us $x=1$ and $x=-3$ . We replace these $x$ 's back in the system and for $x=1$ we get $c=- 3/4$ ( $x>=-3/4$ and $x<=-3/4$ ) and for $x=-3$ we get $c=5/4$ ( $x>=5/4$ and $x<=5/4$ ) we then filter out the values that satisfy the required criterion. Only $c=-3/4$ gives us a unique solution so we sum up our two $c$ 's $-3/4 + 4/3$ and we get $7/12$ .

Firstly, we solve the following equation : $(c-1)x^2+2cx+c+4=0$ and $cx^2+2(c+1)x+(c+1=0$ . We have the sets of equation are $\left\{ \frac{-c+\sqrt{4-3c}}{c-1}, -\frac{c+\sqrt{4-3c}}{c-1} \right\}$ and $\left\{ \frac{-c-1+\sqrt{c+1}}{c}, -\frac{c+1+\sqrt{c+1}}{c} \right\}$ In order to satisfy the condition such that the system has a unique solution, the necessary condition is 2 of the 4 roots above are equal to each other. By solving the corresponding equations, we can find that there are 3 values of $c$ satisfying the necessary condition are $\frac{4}{3}, \frac{5}{4}, -\frac{3}{4}$ . By substituting each obtained value of $c$ to the given system and solving it, we have only 2 values of $c$ satisfying the given condition : $\frac{4}{3}, -\frac{3}{4}$ . Then the sum of all values of $c$ is $\frac{4}{3} - \frac{3}{4} = \frac{7}{12}$ . The value of $a+b$ is $7+12=19$ .

If $0< c <1$ , then both inequalities are satisfied for all $x$ large enough in absolute value. Also, if $c$ is $0$ or $1$ then one of the inequalities is quadratic that is satisfied for all $x$ with large enough absolute value, and the other is linear (and not constant), so again they have infinitely many common solutions. So either $c>1$ or $c<0$ .

There are two possibilities for two quadratic inequalities to have exactly one solution.

Case 1. The quadratic polynomials have a common root, which happens to be the only common solution.

Case 2. One of the quadratic polynomials has a double root, and the inequality is such that this is the only solution. It must also satisfy the other inequality.

We will first address Case 1. If for some $x$ both inequalities are equalities, then, subtracting the first from the second, we get $x^2+2x-3=0$ . This means that $x=1$ or $x=-3$ . If $x=1$ , then $(c-1)+2c+c+4=0,$ so $c=-\frac{3}{4}$ . One can check that this value of $c$ does work. If $x=-3,$ then $9(c-1)-6c+c+4=0,$ so $c=\frac{5}{4}$ . In this case, the first quadratic inequality is satisfied for $x\leq -3$ or $x\geq -\frac{3}{5}$ , while the second one is satisfied for $-7\leq x\leq -3,$ which gives infinitely many common solutions. So in Case 1 we get $c=-\frac{3}{4}$ .

In Case 2, if the first polynomial has a double root, then by considering the discriminant, we get $c^2-(c-1)(c+4)=0,$ which gives $c=\frac{4}{3}$ . This yields $\frac {1}{3} x^2 + \frac {8}{3} x + \frac {16}{3} = 0$ , so $x=-4$ . Check that $\frac {4}{3} (-4)^2 + 2(\frac {4}{3}+1)(-4)+(\frac {4}{3} + 1) = 5 \geq 0$ satisfies the second inequality, so $c = \frac {4}{3}$ is valid. If the second polynomial has a double root, then the discriminant gives $(c+1)^2-c(c+1)=0$ , so $c=-1$ . The corresponding double root is $x=0,$ for which the first inequality is not satisfied since $(-1-1)(0)^2 + 2 (-1) (0) + (-1 +4) = 3 > 0$ .

Putting this all together, the sum of the possible values of $c$ is $-\frac{3}{4}+\frac{4}{3}=\frac{7}{12}$ . Hence $a+b=19$ .