Quadratic equations, Year 3 vocational

End of September 2026: in Year 3 at Italian vocational schools (istituti professionali) the maths programme opens with quadratic equations, after finishing the revision of first-degree equations and inequalities. Vocational schools almost always run by terms (quadrimestri), with the school report at the end of January: the first serious test on this block falls between five and eight weeks from now, so between the end of October and mid-November, and it weighs directly on that report.

This chapter has a feature that none of the topics seen so far had: it is the first real jump in technical difficulty of the three-year course, and it is perceived as abrupt. Up to here the calculation was largely mechanical; here you must first recognise what kind of equation it is, and only then choose how to solve it. The vocational approach to this topic is applied: no proof of the quadratic formula, the time goes on the procedure and on checking the result.

The starting point: not all quadratic equations are the same

A quadratic equation in normal form is written

a x^{2} + b x + c = 0 with a \neq = 0

The reason this chapter seems harder than expected is that most mistakes don’t come from the calculation, but from skipping a step before the calculation: recognising the type of equation. Pure and spurious equations are solved in three lines, with no formula. Using the general formula there too isn’t wrong in the result, but wastes time, and in a timed test that is time missing elsewhere.

Pure (missing the first-degree term, $b = 0$ ): $a x^{2} + c = 0$ . You isolate $x^{2}$ and take the root:

3 x^{2} - 27 = 0 ⟹ x^{2} = 9 ⟹ x = \pm 3

If $- \frac{c}{a}$ comes out negative, the equation has no real solutions: there is no real number whose square is negative.

Spurious (missing the constant term, $c = 0$ ): you take $x$ out as a common factor.

2 x^{2} - 10 x = 0 ⟹ x (2 x - 10) = 0 ⟹ x = 0 \lor x = 5

The commonest mistake on this form is stopping at the second solution and losing $x = 0$ , which is always a valid solution when the constant term is missing.

Complete (no coefficient is zero): you need the quadratic formula, seen in the next section.

Quadratic formula and discriminant

For a complete equation, the solutions are

x_{1, 2} = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The quantity under the root, $Δ = b^{2} - 4 a c$ , is called the discriminant and its sign determines how many real solutions exist:

$Δ > 0$ : two distinct real solutions;
$Δ = 0$ : one real solution (called a double one);
$Δ < 0$ : no real solution.

Sum and product of the roots, as a quick check. Once you have found $x_{1}$ and $x_{2}$ , it always holds that:

x_{1} + x_{2} = - \frac{b}{a} x_{1} \cdot x_{2} = \frac{c}{a}

It is a check that costs ten seconds and exposes a calculation error before you hand in: if the sum or product you find doesn’t match what the formula gives, there is a wrong sign somewhere.

A typical mistake in the formula

The most frequent concerns the sign of $b$ when $b$ is negative. In $x^{2} - 6 x + 8 = 0$ , $b = - 6$ , so $- b = 6$ , not $- 6$ : whoever substitutes $- b$ without first rewriting the sign ends up with $x_{1, 2} = \frac{- 6 \pm \dots}{2}$ instead of $\frac{6 \pm \dots}{2}$ , and gets both solutions wrong even though $Δ$ was calculated correctly. The practical rule: write the value of $b$ in brackets with its sign, then substitute it into the formula; don’t trust yourself to do it in your head.

Quadratic inequalities: the parabola method

An inequality like $a x^{2} + b x + c > 0$ is solved by looking at where the graph of the parabola $y = a x^{2} + b x + c$ lies above or below the $x$ -axis. The procedure:

Bring the inequality to normal form, with zero on one side.
Solve the associated equation $a x^{2} + b x + c = 0$ , finding the solutions (if they exist).
If $a > 0$ , the parabola is concave up: it lies above the $x$ -axis (so the expression is positive) outside the roots, and below (negative) between the roots.
If $a < 0$ , the opposite holds: the parabola is concave down, positive between the roots, negative outside.
If $Δ < 0$ , the parabola never touches the $x$ -axis: it always has the same sign as $a$ , for every value of $x$ .

The typical mistake here is choosing the wrong interval because you don’t check the direction of the concavity: you find the two roots correctly, but then take “between the roots” when you needed “outside”, or vice versa. One way never to go wrong: after choosing the interval, try a random value inside that interval in the starting expression, and check that the sign you get is really the one the inequality asks for.

Why it pays to arrive with method, not with more hours of exercises

The previous chapter, equations, inequalities and systems in Year 2 at a vocational school, closed with first-degree equations and systems: here the jump is that you have to handle two decisions in one: recognising the type of equation, and only then applying the right method, instead of always applying the same procedure. Whoever arrives at this chapter used to always applying the general formula to everything (the habit picked up in the first month, when complete equations are the majority of the examples seen in class) struggles more precisely in recognising the incomplete ones: it isn’t a calculation problem, it is a method problem, and it is solved by training first the recognition of the type, then the solving.

First worked exercise: pure, spurious and complete compared

Pure. Solve $4 x^{2} - 100 = 0$ .

4 x^{2} = 100 ⟹ x^{2} = 25 ⟹ x = \pm 5

Three lines, no formula.

Spurious. Solve $5 x^{2} - 15 x = 0$ .

5 x (x - 3) = 0 ⟹ x = 0 \lor x = 3

Here too, no formula: just taking out a common factor.

Complete. Solve $x^{2} - 6 x + 8 = 0$ . Here $a = 1$ , $b = - 6$ , $c = 8$ :

Δ = (- 6)^{2} - 4 \cdot 1 \cdot 8 = 36 - 32 = 4

x_{1, 2} = \frac{6 \pm 4}{2} = \frac{6 \pm 2}{2} ⟹ x_{1} = 4, x_{2} = 2

Check with sum and product: $x_{1} + x_{2} = 6 = - \frac{- 6}{1}$ , correct; $x_{1} \cdot x_{2} = 8 = \frac{8}{1}$ , correct.

Applying the general formula to the pure or spurious one as well would have given the same final result, but with three times the steps: the time saved by recognising the type before starting is exactly the time needed for the last exercise of the test.

Second worked exercise: negative delta and zero delta

Negative delta. Solve $x^{2} + 4 x + 9 = 0$ .

Δ = 4^{2} - 4 \cdot 1 \cdot 9 = 16 - 36 = - 20

$Δ$ negative: the equation has no real solutions. It isn’t a calculation to be redone: it is the correct result, and it must be written as such, “no real solution”, not left blank as if it were an incomplete exercise.

Zero delta. Solve $x^{2} - 8 x + 16 = 0$ .

Δ = (- 8)^{2} - 4 \cdot 1 \cdot 16 = 64 - 64 = 0

$Δ = 0$ means one real solution, a double one:

x_{1, 2} = \frac{8 \pm 0}{2} = 4

Check by factorising: $x^{2} - 8 x + 16 = (x - 4)^{2}$ , which vanishes only for $x = 4$ : consistent with the double solution just found.

On these two cases you can practise with immediate correction in the maths test: quadratic equations: an honest warning, that practice test is catalogued under “Year 2” because at the science high school quadratic equations are studied a year earlier; at vocational schools they arrive in Year 3, but the level of the exercises is exactly right for this chapter, not simplified.

Third worked exercise: an inequality with the parabola method

Solve $x^{2} - x - 6 > 0$ .

The associated equation: $x^{2} - x - 6 = 0$ . I look for two numbers with product $- 6$ and sum $- 1$ : they are $2$ and $- 3$ .

x^{2} - x - 6 = (x + 2) (x - 3) ⟹ x = - 2 \lor x = 3

Here $a = 1 > 0$ : the parabola is concave up, so it is positive (greater than zero) outside the two roots:

x < - 2 \lor x > 3

Check with a test value inside the interval found, for example $x = 4$ : $4^{2} - 4 - 6 = 16 - 4 - 6 = 6 > 0$ , consistent. Also check with an excluded value, $x = 0$ : $0 - 0 - 6 = - 6$ , which is not greater than zero, consistent with the exclusion of the interval between the roots.

Fourth worked exercise: an applied problem

A craft workshop produces rectangular trays whose long side exceeds the short one by 4 cm. If the area of a tray is $96 cm^{2}$ , what are the dimensions?

I call the short side $x$ : the long side is $x + 4$ , and the area is their product.

x (x + 4) = 96 ⟹ x^{2} + 4 x - 96 = 0

Δ = 4^{2} - 4 \cdot 1 \cdot (- 96) = 16 + 384 = 400 = 2 0^{2}

x_{1, 2} = \frac{- 4 \pm 20}{2} ⟹ x_{1} = 8, x_{2} = - 12

The solution $x = - 12$ is discarded: not because “negative numbers are no good in general”, but because here $x$ represents the length of a side, and a negative length makes no sense in the real problem. That leaves short side $8 cm$ and long side $12 cm$ : check, $8 \cdot 12 = 96$ .

How we work in lessons

With a term-based schedule and a report arriving at the end of January, the first test on this chapter counts more than elsewhere: a low grade here doesn’t have much time to be recovered before the grading meeting. In lessons I always start from a recognition mini-test (given a list of equations, say for each one whether it is pure, spurious or complete, without solving it), because that is where speed in a test is decided, not in calculating the discriminant. Only afterwards do we move to the formula, inequalities and applied problems.

On the same topic, but in another track and with a different rhythm (2 hours a week, as against the applied approach of the vocational school), I wrote quadratic equations in Year 3 at the art high school. The chapter leading up to this one, from a year earlier in the same track, is told in equations, inequalities and systems in Year 2 at a vocational school.

The full maths programme for the track is on the maths page. If you want a hand on this block before the November test, write to me via contacts.

Quadratic equations in Year 3 of a vocational school