Quadratic equations in Year 3 of the art high school

Mid-September 2026: lessons started a few days ago, and in Year 3 at the Italian liceo artistico (art-focused high school) the teacher has just finished revising the line and systems in order to begin quadratic equations. The first test arrives in four to seven weeks, so between the end of October and early November: little time for a topic that on its own is worth a good part of the first term.

Why it is different here from the science high school

At the art high school maths has 2 hours a week in Year 3, against 4 at the liceo scientifico (science-focused high school). The programme, however, is not cut in half: with the “azzurro” programme typical of this track, what at the science high school is spread between Year 2 and Year 3 arrives here concentrated into a single year. The practical result is that in class there is no time for exercises: you see the theory, do one example, and the rest, which is the part that really makes the difference in a test, has to be done at home.

And at home, in this track, time is what is left after the 12 weekly hours of specialist subjects: graphic and painting subjects, geometry subjects, sculpture subjects, art workshop. Drawings and projects have fixed deadlines, and maths, which has a fixed deadline only on the day of the test, is the first to slip. That is why the study plan needed here is not “two hours of exercises a day”: it is 30-40 targeted minutes, every day, on the single block arriving in class that week. Twenty minutes done with method are better than a whole afternoon put off to the weekend before the test.

The Year 3 “azzurro” programme, on quadratic equations, asks for: normal form, discriminant, incomplete equations, factorising the trinomial, quadratic problems. It does not normally ask for parametric equations or the relations between roots and coefficients written out in full: if your textbook covers them, ask the teacher whether they are part of this year’s actual programme before studying them in depth.

Normal form, and why the first coefficient can’t be zero

A quadratic equation is written in normal form like this:

with $a\ne 0$: if $a$ were zero, the $x^2$ term would vanish and the equation would become first-degree, not second-degree. It is the first thing to check when an equation arrives already “awkward”, with terms to move or brackets to expand: first bring everything to normal form, then look at what $a$, $b$ and $c$ are.

Incomplete equations: the commonest mistake is using the formula

When the first-degree term or the constant term is missing, the equation is incomplete, and has a solution method shorter than the general formula. Using the general formula on an incomplete one isn’t wrong in the result, but it wastes time that you don’t have in a test, and it is mistake no. 1 on this topic.

Pure (missing $b$, the first-degree term): $a{x}^2+c=0\Rightarrow x^2=-\frac{c}{a}$. It has two opposite solutions if $-\frac{c}{a}>0$, no real solution if it is negative.

Two steps, no discriminant: whoever applies the quadratic formula here gets to the same result but takes three times as long.

Spurious (missing $c$, the constant term): you take $x$ out as a common factor.

One solution is always $x=0$: forgetting it, stopping only at the second, is the commonest mistake on this type.

Monomial (both $b$ and $c$ missing): $a{x}^2=0\Rightarrow x=0$, a double solution.

The quadratic formula and the discriminant

When the equation is complete, that is no coefficient is zero, you use the quadratic formula:

The quantity under the root is called the discriminant, $\Delta =b^2-4ac$, and its sign decides how many real solutions exist:

• $\Delta >0$: two distinct real solutions;
• $\Delta =0$: one real solution, called a double one;
• $\Delta <0$: no real solution.

When the coefficient $b$ is even, the reduced formula is convenient, with $b=2b^{\prime }$: it avoids carrying a factor 4 and a larger root than necessary.

There is a way of reading the discriminant that isn’t in the track’s programme but helps to understand what you are calculating: the left-hand side of the equation, $y=a{x}^2+bx+c$, is a parabola, and the solutions of the equation are the points where that curve crosses the $x$-axis. $\Delta >0$ means the parabola crosses the axis twice, $\Delta =0$ that it touches it at a single point (it is tangent), $\Delta <0$ that it never touches it. If in geometry subjects you have already met the parabola as a curve (in arches, in conic sections, in certain ruler-and-compass constructions), it is the same curve: here it is looked at from the equation side instead of the drawing side.

Factorising the trinomial with the solutions

Once you have found $x_1$ and $x_2$, the trinomial can be rewritten as a product:

It serves to check the result (multiplying out you must get back to the starting trinomial) and comes in useful later, with quadratic inequalities.

Four worked exercises step by step

Exercise 1: spurious. Solve $3x^2-9x=0$.

You take $3x$ out as a common factor:

Check: if I took out only $x$ I would get $x(3x-9)=0$, which still gives $x=0$ or $x=3$: the extra numerical factor doesn’t change the solutions, it only changes how clean the step is.

Exercise 2: complete, $\Delta$ a perfect square. Solve $x^2-5x+6=0$.

Here $a=1$, $b=-5$, $c=6$. The discriminant:

$\Delta =1$ is a perfect square ($\sqrt{1}=1$), so the solutions are whole numbers:

Check by factorising: $(x-2)(x-3)=x^2-5x+6$. It works.

Exercise 3: complete, $\Delta <0$. Solve $x^2+2x+5=0$.

$\Delta$ is negative: the equation has no real solutions. It is not a calculation error to be checked twice, it is the correct result, and it must be written as such: “no real solution”, not left blank.

Exercise 4: a quadratic problem. A rectangular painting has a base that exceeds the height by 3 cm, and the area of the canvas is $108{\text{cm}}^2$. Find the dimensions.

If the height is $x$, the base is $x+3$, and the area is their product:

The negative solution is discarded: a length can’t be negative, not because “negative numbers don’t exist” but because here $x$ represents a real measurement. That leaves height $9\text{cm}$ and base $12\text{cm}$: check, $9\cdot 12=108$.

You can practise more exercises of this type in the maths test: quadratic equations: the correct answer is free, the step-by-step guided solution unlocks with the first lesson. An honest warning: that practice test is catalogued under “Year 2”, because at the science high school quadratic equations are studied in Year 2; at the art high school the same topic arrives a year later, in Year 3, but the exercises are identical.

Typical mistakes, and how to spot them

How we work in lessons

With 2 hours a week and 12 hours of specialist subjects competing for your time, the point is not “doing more exercises”: it is doing the right ones. In lessons I start from a mini-test to see whether the fragile block is calculating the discriminant, handling signs, or recognising the incomplete ones, and I work only on that, with immediate correction at every step. If we haven’t met yet, there is a free 30-minute introductory call to work out where to start, with no commitment.

If you want to revisit last year’s topics in the same track, I wrote maths in Year 1 at the art high school: numbers and fractions. For physics in the same year there is physics at the art high school: measurements and significant figures. And if you want to see which other algebraic shortcuts look like real rules and aren’t, there is five maths mistakes that look right. The full programme of the year is on the maths page. The same topic but in another track, with a more applied approach and a rhythm by terms: quadratic equations in Year 3 at a vocational school.

If you want a hand before the end-of-October test, write to me: on a precise block like this, one well-aimed hour recovers much more than a whole evening of disorganised exercises.