Quadratic inequations

A quadratic inequation is an inequation in which we find numbers, a variable (which we call $x$) that can be multiplied by itself, and an inequality symbol.

An example of a quadratic inequation might be: $$ 2x^2-x < 2x-1$$

where we can observe that the term $2x^2$ is the quadratic term, characteristic of the quadratic inequations, because if there is no quadratic term, we would have a first degree inequation.

To solve a quadratic inequation we will use a method consisting of a series of steps to be followed.

To apply this method we will need the quadratic formula to solve quadratic equations, so let's remember this:

Given a quadratic equation: $ax^2+bx+c=0$, the solutions are given by the formula: $$ x_{+}=\dfrac{-b+\sqrt{b^2-4ac}}{2a} \qquad x_{-}=\dfrac{-b-\sqrt{b^2-4ac}}{2a}$$

Two, one or no solution may be found, depending on the value of $\sqrt{b^2-4ac}$ (for more information take a look at the unit Quadratic Equation).

Step by step method to find the solution:

• Given the inequation, operate until obtaining a zero on one of the sides of the inequation, obtaining an expression of the type: $ax^2+bx+c < 0$ or $ax^2+bx+c > 0$ where the values $b$ and $c$ are real numbers that can be positive or negative or even zero, and $a$ is a positive value. In case of finding a negative value of $a$, we will multiply the entire inequation by $(-1)$ tota la inequació, changing in this way the sign of $a$ (and, consequently, the sign of the other terms and the order of the inequality).

• We will look for the solutions of the equation $ax^2+bx+c= 0$, induced by the inequation $ax^2+bx+c < 0$ or $ax^2+bx+c > 0$.

• At this point, we have three options:

  • If we do not have solutions of the equation, we must consider two cases:

    • If $ax^2+bx+c > 0$: The solution is any real value: all the numbers satisfy the inequation.
    • If $ax^2+bx+c < 0$: No value of $x$ satisfies the inequation, therefore, the inequation has no solution.

    If we observe the graph of $y=ax^2+bx+c$ we will see that it does not cut the X axis since the equation has no solutions. Being also the value of a positive, the entire graph is over the X axis, with positive values of $y$, therefore if the inequation has sign greater than (or greater than or equal to), then any point is the solution to the inequation, and if it has a sign less than (or less than or equal to), then no point will be a solution.

  • If we have a solution, we will do:

    • If we had the inequation $ax^2+bx+c > 0$, and we proceed: $$ ax^2+bx+c > 0 \Rightarrow (x-x_1)^2 > 0 \Rightarrow (x-x_1)(x-x_1) > 0$$ $$\Rightarrow \left\{ \begin{array}{l} (x-x_1) < 0 \Rightarrow x < x_1 \\ (x-x_1) > 0 \Rightarrow x > x_1 \end{array} \right. $$

      We have to consider the last two valid cases since a product of two numbers is positive if these two numbers are both positive or both negative.

      So the solution of the inequation will be $x$ whereby $x$ satisfies $x < x_1$ and $x > x_1$ in which $x_1$ is the solution to the equation $ax^2+bx+c=0$.

      In case that we have an inequality of the type $ax^2+bx+c \geqslant 0$, apart from the same solutions that we were considering earlier, we would add the solution $x_1$ and the result would be to have region solution the entire real line.

    • If we have the inequation $ax^2+bx+c < 0$, we will do: $$ ax^2+bx+c < 0 \Rightarrow (x-x_1)^2 < 0 \Rightarrow \text{ There is no solution} $$ since a squared number will always be positive, and we are demanding that it should be negative.

      In the event that we have an inequality of the type $ax^2+bx+c \leqslant 0$, so we would have an exact solution, the solution of the equation $x_1$.

  • If we have two solutions, $x_1$ and $x_2$, considering as well $x_1 < x_2$, fwe will follow the following procedure:

    (let's remember that the value of $a$ is always positive)

    • If $ax^2+bx+c > 0$: $$ ax^2+bx+c > 0 \Rightarrow (x-x_1)(x-x_2) > 0 \Rightarrow$$ $$\Rightarrow \left\{ \begin{array}{l} \text{a) } \ (x-x_1) > 0 \ \text{ and } \ (x-x_2) > 0 \\ \text{b) } \ (x-x_1) < 0 \ \text{ and } \ (x-x_2) < 0 \end{array} \right. $$ $$\Rightarrow \left\{ \begin{array}{l} \text{a) } \ x > x_1 \ \text{ and } \ x > x_2 \\ \text{b) } \ x < x_1 \ \text{ and } \ x < x_2 \end{array} \right. $$

      and as we have supposed that $x_1 < x_2$, we remain with the inequalities $x < x_2$ and $ x < x_1$.

    • If $ax^2+bx+c < 0$: $$ ax^2+bx+c < 0 \Rightarrow (x-x_1)(x-x_2) < 0 \Rightarrow$$ $$\Rightarrow \left\{ \begin{array}{l} \text{a) } \ (x-x_1) > 0 \ \text{ and } \ (x-x_2) < 0 \\ \text{b) } \ (x-x_1) < 0 \ \text{ and } \ (x-x_2) > 0 \end{array} \right. $$ $$\Rightarrow \left\{ \begin{array}{l} \text{a) } \ x > x_1 \ \text{ and } \ x < x_2 \\ \text{b) } \ x < x_1 \ \text{ and } \ x > x_2 \end{array} \right. $$ and as we have supposed that $x_1 < x_2$, we remain with the inequalities $x < x_2$ and $ x < x_1$.

• Once we have found the region where the inequation is satisfied, we are done.

Remember that in the resolution algorithm we have just used strict inequalities (less than, greater than), but the same reasoning can be applied to inequalities of the type 'less than or equal to' or 'greater than or equal to'.

Next we will see an example of each type:

$$ x^2+x+2 > -1-x $$

Resolution: $$ x^2+x+2 > -1-x \Rightarrow x^2+2x +1 > 0 $$

We find the solutions of the equation $x^2+2x+1=0$: $$ x=\dfrac{-2\pm \sqrt{4-4}}{2}=-1$$ There is only one solution.

Using the given outline, the solution is $x < -1$ and $x > -1$, that is, every point except $-1$.

$$ x^2+2 < -1-2x $$

Resolution: $$ x^2+2 < -1-2x \Rightarrow x^2+2x +1 < 0 $$

We find the solutions of the equation $x^2+2x+1=0$: $$ x=\dfrac{-2\pm \sqrt{4-4}}{2}=-1$$ There is only one solution.

Using the given outline, we see that there are not any possible solutions.

$$ -x(x-1)-x < -1 $$

Resolution: $$ -x(x-1)-x < -1 \Rightarrow -x^2+x-x +1 < 0 \Rightarrow -x^2 +1 < 0 \Rightarrow x^2 -1 > 0 $$

We find the solutions of the equation $x^2-1=0$: $x=\pm 1$

As we have two solutions, the solution of the problem is (using the outline) $x < -1$ and $x > 1$.