Inequalities – 1st part – Digitális matematika

What are inequalities?

Inequalities are counterpart of equations. Is it true? Well, not really, but there is some truth in such a statement.

Inequalities are counterpart of equations. Is it true? Well, not really, but there is some truth in such a statement. In fact, if we have an equation, then replacing the $=$ sign by any of the inequality signs $<,>,\leq,\geq$ we obtain a relation which we call an inequality. It makes sense to look for solutions of inequalities obtained in this way from famous equations.

Let’s start with linear equations: $a x+b=0$ where we suppose that $a$ is a nonzero real number, $b$ is an arbitrary real number and we are looking for those real numbers $x$ satisfying this equation. Clearly, the only $x$ with this property is $-\frac{b}{a}$ , this is the unique solution of our equation. We can interpret this problem geometrically: writing the equation in the form $ax+b=0$ we know that the left side represents the $y$ coordinates of the points on a straight line in the plane with slope $a$ as $x$ runs through the real numbers and we are looking for the $x$ coordinates of those points on the line where the corresponding $y$ coordinate is $0$ . In other words, we are looking for the point of intersection with the $x$ axis. From geometry, it is clear that there is exactly one point — thanks to the assumption $a\ne 0$ .

We can apply this argument for the intersection of any two non-parallel lines which leads to the problem of solving the equation

$ax+b=cx+d,$

where $a\ne c$ . The $=$ sign expresses the intersection property. What if we replace $=$ by $<$ in this example? The inequality

$ax+b<cx+d$

expresses that given an $x$ coordinate the $y$ coordinate of the corresponding point on the left line is smaller than that of the point on the right line. In other words, we are looking for those points lying on the same vertical line, that is, having equal $x$ coordinates, and that on the left line has smaller $y$ coordinate than the other on the right line. This means that we want to describe the set of all those $x$ values for which the left line is below the right line.

Geometrically it is clear, that the two lines have exactly one intersection point, and on the left of this point one of the two lines is above the other, and on the right of the point we have the converse. So the unique $x$ , which is the solution of the corresponding equation, splits the $x$ -axis into two half lines: one of them represents the set of points we are looking for. Clearly, our inequality can be reduced to the simpler one $A x+B<0$ when denoting $a-c$ by $A$ and $b-d$ by $B$ .

Our final answer, however, depends on the sign of the slope $A=a-c$ : if $A>0$ , then $x<-\frac{B}{A}$ , but if $A<0$ , then $x>-\frac{B}{A}$ . Observe that this depends on if the line $y=Ax+B$ is increasing or decreasing. And this shows that we cannot copy the method we follow when solving the corresponding equation: in fact, when multiplying, or dividing both sides with some nonzero number then we keep the inequality sign if that number is positive, but we change it for the reverse if that number is negative.

Roughly speaking, we can do the same things when solving inequalities as we do with equations: the only difference is that we have to interchange $<$ by $>$ and $\leq$ by $\geq$ when multiplying with negative numbers. This is easy, right? And geometry helps a lot.

For instance, when solving a quadratic inequality, like $x^2+x-6\leq 0$ we think of the graph of the function on the left side: it is a parabola which opens upward and we are looking for those point of that parabola which are below the $x$ -axis, or just lying on it. Clearly, they are the two solutions of the corresponding equation $x^2+x-6$ , which are $2$ and $-3$ , and all values in between: $-3\leq x\leq 2$ .

It seems that – in some cases – for the solution of an inequality the solution of the corresponding equation can be utilized: replacing the inequality sign by equality the solutions of the equation represent, in some sense, the „boundary” of the solution set of the original inequality. Of course, this is a very rough statement but the idea works in certain cases.

Let’s take an example: solve the inequality

$x^5+3x^4-23 x^3-51 x^2+94 x+120\geq 0$

in the set of real numbers. The corresponding equation

$x^5+3x^4-23 x^3-51 x^2+94 x+120= 0$

is a very complicated quintic equation, and we know — thanks to Galois — that there is no solution formula for such equations. So, how to find the solutions? If we are not smart enough to solve this equation how could we solve the more complicated inequality? Maybe we should factorize our polynomial, but the coefficients look really awful — trial and error seems to be hopeless.

Don’t worry: there is a simple method how to find at least the rational roots of an equation with integer coefficients. Indeed, if $x=\frac{m}{n}$ is a rational solution of the above equation with some integers $m,n$ , where $n> 0$ and $m,n$ are coprime, then substitution into the equation gives

$m^5+3 m^4 n-23 m^3 n^2-51 m^2 n^3+ 94 m n^4+120 n^5=0.$

From this equation it is clear that $n$ divides $m$ , hence, by assumption, $n=1$ . This implies that every rational solution is an integer and $x=m$ divides $120$ . Bad luck that $120$ has a lot of divisors: $\pm 1$ , $\pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12$ , $\pm 15$ , $\pm 20$ , $\pm 24$ , $\pm 30$ , $\pm 40$ , $\pm 60$ , $\pm 120$ . We have so many possibilities — this point only trial and error helps. After some minutes we’ll have the roots: $-5,-3,-1,2,4$ — we can stop here as a quintic polynomial cannot have more than five roots.

Now we have the desired factorization and our inequality has the form:

$x^5+3x^4-23 x^3-51 x^2+94 x+120=(x+5)(x+3)(x+1)(x-2)(x-4)\geq 0.$

How to solve this inequality? We have a product of five factors which is positive if and only if an even number of the factors is negative: zero, two, or four. But there is a simpler way to find out the solutions: between two consecutive roots our polynomial does not change its sign, hence we can take some test values between any two consecutive roots to find those intervals where the graph is above the $x$ -axis: the values $-6, -4, -2, 0, 3,5$ do well, as the following figure shows:

We can see that the solutions are: $-5\leq x\leq -3$ , or $-1\leq x\leq 2$ , or $x\geq 4$ .

So, the method works! Nevertheless, don’t expect miracles: in more complicated situations it can be really hard to find the solutions of the equation itself — but don’t give it up: proper investment brings profit.