Quadratic inequalities: the parabola and the sign

If you are in Year 3 at a liceo delle scienze umane (the Italian high school focused on psychology, pedagogy and social sciences), it is early October: quadratic equations have just been explained or are about to be, and straight after them come quadratic inequalities. The chapter usually closes in November, and the test that includes it falls in five to eight weeks. There is plenty of time to learn it properly, and it is worth it: the same technique comes back with rational inequalities, with the study of domains and, later on, with limits.

The idea is a single one and you can learn it in an afternoon: you do not solve the inequality, you look at a parabola. I have written about quadratic equations themselves elsewhere (the version for the art high school and for the vocational track): here I take it for granted that you can calculate the discriminant and the roots. For inequalities with fractions there is a separate article, and for tangents and families of parabolas this one. Here I stay on the sign.

The idea: an inequality is a question about the sign

Take a trinomial $a{x}^2+bx+c$ with $a\ne 0$. Its graph is a parabola. The inequality $a{x}^2+bx+c>0$ asks: for which $x$ is the parabola above the x-axis? With $<0$ it asks where it is below. With $\ge$ and $\le$ you also include the points where the parabola touches the axis.

You do not need to draw it precisely: you only need three pieces of information.

1. The concavity, from the sign of $a$: upwards if $a>0$, downwards if $a<0$.
2. The intersections with the $x$-axis, that is, the roots, if they exist.
3. The direction of the inequality ($>$ or $<$, with or without the equals sign).

The method in four steps

1. Bring everything to one side and rewrite in the form $a{x}^2+bx+c\gtrless 0$. If $a$ is negative it helps to multiply by $-1$, remembering to flip the direction.
2. Calculate the discriminant $\Delta =b^2-4ac$.
3. Find the roots (if $\Delta \ge 0$) and draw a sketch: a parabola with the right concavity, meeting the axis at the roots.
4. Read the sign from the sketch and write the solution.

The three cases, with $a>0$

With $a>0$ the parabola is U-shaped. The behaviour depends on the discriminant, and it is worth knowing by heart like a times table.

The practical rule for $\Delta >0$ is: values outside the roots when you look for the positive, values between the roots when you look for the negative (always with $a>0$). With $a<0$ the parabola is upside down and everything swaps: the sketch shows you the result straight away without having to learn two tables.

With $\ge$ and $\le$ the roots are part of the solution. For $\Delta =0$ this is decisive: $a{x}^2+bx+c\le 0$ has one solution, $x=x_1$, whereas with $<0$ it has none.

Inequalities with more than one term, and systems

If the inequality is not in standard form, you first bring it to standard form: for example $x^2\le 4x-3$ becomes $x^2-4x+3\le 0$. Never leave $x$ on one side and a number on the other.

A system of inequalities is solved in three moves: solve each inequality on its own, draw the solutions on parallel lines (one under the other, same abscissas) and take the intersection, that is, the values where all the lines are “solid”. If one solution is empty, so is the whole system.

The mistakes I see every year

Wrong direction. After multiplying by $-1$ you forget to flip the sign: from $-x^2+4x-3\ge 0$ you go to $x^2-4x+3\le 0$, not to $\ge 0$. Quick check: try a number (for example $x=0$) in the starting inequality.

Forgetting the case $\Delta <0$. If the discriminant is negative there are no roots and many people get stuck or invent two intervals. The solution is always “all the reals” or “no value”, depending on the direction and the sign of $a$.

Dividing by $x$. Faced with $x^2>3x$ the temptation is to cancel $x$ and write $x>3$. But you do not know whether $x$ is positive or negative: if it is negative the direction would change, and you lose all the solutions with $x<0$. You bring everything to one side, $x^2-3x>0$, and study the sign: $x(x-3)>0$ gives $x<0$ or $x>3$.

Saying that $\Delta =0$ gives “all the reals” or “none” without looking. With $(x-3{)}^2\ge 0$ the solution is all the reals, but with $(x-3{)}^2>0$ it is $x\ne 3$: the excluded value has to be written down.

Badly written intervals. Writing $2>x>5$ makes no sense: the “outside” solution is $x<2$ or $x>5$, never a single chain. The word “or” (union) and the word “and” (intersection) have to be used with care.

Three worked examples

1. Two inequalities with $\Delta >0$

Solve $x^2-5x+6>0$ and then $-x^2+4x-3\ge 0$.

First case: $\Delta =25-24=1>0$, roots

A parabola with $a=1>0$ (U-shaped) and the inequality “greater than zero”: outside values. Solution: $x<2$ or $x>3$.

Second case: I multiply by $-1$ and flip the direction, getting $x^2-4x+3\le 0$. Here $\Delta =16-12=4$, roots

U-shaped parabola, “less than or equal to zero”: values between the roots, endpoints included. Solution: $1\le x\le 3$.

Check on the second: $x=2$ in the original inequality gives $-4+8-3=1\ge 0$, true; $x=0$ gives $-3\ge 0$, false. Consistent with the interval.

2. Zero discriminant and negative discriminant

Solve $x^2-6x+9\le 0$, then $x^2-6x+9>0$, then $2x^2-3x+4<0$.

The trinomial $x^2-6x+9$ is $(x-3{)}^2$, with $\Delta =36-36=0$. A square is never negative, so:

• $(x-3{)}^2\le 0$ holds only if $(x-3{)}^2=0$: solution $x=3$;
• $(x-3{)}^2>0$ holds everywhere except at $x=3$: solution $x\ne 3$.

For $2x^2-3x+4$ the discriminant is

The parabola has $a=2>0$ and does not touch the $x$-axis: it is always above it, so $2x^2-3x+4$ is always positive. The request "$<0$" has no solutions.

3. A system

Solve

First inequality: $\Delta =1+24=25$, roots $x=\frac{1\pm 5}{2}$, that is $x_1=-2$ and $x_2=3$. With “less than zero” and a U-shaped parabola, values between the roots: $-2<x<3$.

Second inequality: $x^2-4x+4=(x-2{)}^2$, with $\Delta =0$. Saying “greater than zero” means $x\ne 2$.

Now the intersection: from the interval $(-2,3)$ I remove the point $x=2$, which belongs to the interval but is excluded by the second inequality. Solution:

Check with $x=2$: the second inequality becomes $0>0$, false. With $x=0$: $-6<0$ true and $4>0$ true. With $x=4$: $16-4-6=6<0$ false, so 4 is outside, as expected.

How to prepare over the coming weeks

You do not need to do lots of exercises, you need to do them in the right order: first the three cases of the discriminant with simple numbers, then the direction (how often do you get it wrong?), then the systems. Redo the three exercises above without looking, changing the coefficients. If you always go wrong at the same point, that point is the problem, not the exercise.

When the chapter is finished in class, you can try the simulator of tests for high school: it offers a practice test per topic. More resources on the subject are on the maths page.

How we work in lessons

I give one-to-one online maths lessons. With anyone who comes to me on this topic we always draw the sketch of the parabola, even when “you could do it in your head”: the sketch removes almost all the errors of direction, and after a few exercises it becomes automatic.

If you want a hand

If the test is getting close and you want to see where you stand, you can book a free introductory call: we look together at where the gaps are and whether it makes sense to work on them.