Absolute value and irrational equations in Year 3

End of September 2026, Year 3 in Italian technical institutes (istituti tecnici): the first weeks go on revising equations and inequalities, then absolute value and irrational equations (those with the unknown under a root) are added, before moving on to the parabola and the circle between October and November. With 3 hours of maths a week and a test of 4-6 exercises (3-4 tests per term), the first test of the year falls between four and seven weeks from now, so between mid-October and early November, and the revision counts as a topic in every respect, not as a formality before the real programme.

Absolute value and irrational equations are not a chapter on their own: they are the tools you will use in a few weeks for the parabola and the circle (where distance appears, which is always a root), and later in trigonometry, where both come back in the same form. The time spent now fixing the cases well pays off already in the next chapter.

Absolute value: what it is, before the rules

The absolute value of a number is its distance from zero on the number line, always positive or zero. That is why $|5|=5$ and $|-5|=5$: both are 5 away from zero. The formal definition, the one you need for solving equations and inequalities, is written in cases:

That is: if what is inside the absolute value is already positive or zero, the absolute value does nothing; if it is negative, the absolute value makes it positive by changing its sign. All the rules that follow come from this definition; they are not independent rules to be memorised.

Equations with one absolute value

Case $|A|=k$, with $k$ a number. If $k<0$, the equation has no solutions: an absolute value can never be negative. If $k\ge 0$, it splits into two equations:

Case $|A|=|B|$, two absolute values. Two quantities have the same absolute value if they are equal or opposite:

This case is more convenient than the next one: you need no additional condition, because an absolute value is automatically $\ge 0$ on both sides.

Case $|A|=B$, with $B$ an expression containing $x$. Here you need care: $|A|$ is always $\ge 0$, so for the equality to be possible $B$ must be too. The condition $B\ge 0$ must be written before solving, not checked only at the end on the result:

Whoever skips the condition on $B$ and just solves the two equations risks accepting a solution that makes no sense: if $B$ turns out negative for that value of $x$, the equality $|A|=B$ can’t hold; a positive number can’t equal a negative one.

Inequalities with one absolute value: the two quick rules

For an inequality of the form $|A|<k$ or $|A|>k$, with $k>0$, two direct rules hold, which come from thinking of the absolute value as distance from zero:

The first reads: “the distance of $A$ from zero is less than $k$”, so $A$ lies in an interval centred on zero. The second reads: “the distance is greater than $k$”, so $A$ lies outside that interval, on one side or the other. The same rules hold with $\le$ and $\ge$, including the endpoints.

A two-absolute-value case: when you need the sign table

When two absolute values with different contents appear in an expression, like $|x-2|+|x+1|$, there is no direct shortcut: you have to remove both absolute values together, and the way to do it without getting a case wrong is a sign table, exactly as for a rational inequality.

You find the zeros of each content ($x-2=0\Rightarrow x=2$; $x+1=0\Rightarrow x=-1$), you divide the real numbers into three zones with these two points, and in each zone you write explicitly how each absolute value behaves:

Once this table is written, the equation or inequality with two absolute values becomes three separate equations or inequalities, one per zone, each without absolute values any more, and at the end you check that every solution found really falls inside the zone it came from. The fourth exercise below shows it in full.

Irrational equations: $\sqrt{A}=B$

An irrational equation has the unknown under a root. The most common case, a square root equal to an expression, is solved by respecting two conditions, not one:

1. Condition of existence (C.E.): $A\ge 0$. Under a square root there can’t be a negative number, so it must be imposed before any other step.
2. Condition on the right-hand side: $B\ge 0$. A square root, when it exists, is by definition a positive number or zero. If $B$ were negative, the equality $\sqrt{A}=B$ could never hold, however the calculation goes.

Only after writing both conditions do you square:

The reason the condition on $B$ is needed, and it isn’t enough to square and then check: squaring is an operation that is not reversible with the same sign: it turns both $\sqrt{A}=B$ and the “twin” equation $\sqrt{A}=-B$ into the same equation $A=B^2$. Solving $A=B^2$ you find the solutions of both, mixed together, and only the condition $B\ge 0$ written before tells you which to keep. Waiting until the end and “checking by eye” works almost always, but it is an after-the-fact check, not a condition: in an exercise with more unknowns or parameters it becomes easy to get it wrong.

Irrational inequalities: why they become systems

With $\sqrt{A}<B$ or $\sqrt{A}>B$ there is no direct rule as there is for absolute values: you have to reason about the sign of $B$, because a root is always $\ge 0$ and the comparison with a negative or positive number changes everything.

$\sqrt{A}<B$. For the inequality to hold, the right-hand side must be positive (a quantity $\ge 0$ can’t be less than a negative number or zero). So you need, all together: $A\ge 0$ (C.E.), $B>0$, and, only under these conditions, you can square keeping the direction: $A<B^2$. It is a system of three conditions in “and”, not a single inequality:

$\sqrt{A}>B$. Here instead you have to distinguish two cases, because the inequality can hold even when $B$ is negative (a root, which is $\ge 0$, is automatically greater than any negative number). The case $B<0$ therefore gives the whole C.E. as the solution, with no need to square; the case $B\ge 0$ instead requires squaring to compare the two quantities, both non-negative:

The formulas must be understood as systems of conditions to be checked in order, not as schemes to apply without knowing why, because elsewhere in the year, with the circle, the same logical structure returns when you compare a distance (always positive) with a radius.

Five worked exercises

1. $|A|=k$, the basic case

Since $7>0$, the equation splits in two:

Check: $|2\cdot 5-3|=|7|=7$, correct; $|2\cdot (-2)-3|=|-7|=7$, correct. Both solutions are valid, there is no extra condition to check: $k=7$ is already a number, not an expression with $x$.

2. $|A|=|B|$, two absolute values

No additional condition: it splits directly into two equations.

Check: for $x=-5$, $|-5-1|=6$ and $|2\cdot (-5)+4|=|-6|=6$, they coincide; for $x=-1$, $|-1-1|=2$ and $|2\cdot (-1)+4|=|2|=2$, they coincide. Both valid.

3. $|A|=B$, with a condition to write first

Condition: $3x-6\ge 0⟹x\ge 2$. Only inside this condition does it make sense to look for solutions. The two equations:

Now the check that makes the difference: $x=4$ respects $x\ge 2$, it is kept. $x=1$ does not respect $x\ge 2$: it must be discarded, and you don’t even need to substitute it to notice: the condition written at the start excludes it directly. (Substituting anyway as a check: $|1+2|=3$, while $3\cdot 1-6=-3$; they don’t coincide, consistent with the discarding.) Only $x=4$ remains.

4. Two absolute values, with a sign table

Zeros of the contents: $x=2$ and $x=-1$. Three zones: $x<-1$, $-1\le x<2$, $x\ge 2$.

Zone $x<-1$: both contents are negative, so $|x-2|=-(x-2)$ and $|x+1|=-(x+1)$:

$x=-2$ respects $x<-1$: it is kept.

Zone $-1\le x<2$: $|x-2|=-(x-2)$, but $|x+1|=x+1$ (here the content is already $\ge 0$):

A false equality, independent of $x$: in this zone there is no solution.

Zone $x\ge 2$: both contents are positive, $|x-2|=x-2$ and $|x+1|=x+1$:

$x=3$ respects $x\ge 2$: it is kept.

Final solutions: $x=-2$ and $x=3$. Check on $x=3$: $|3-2|+|3+1|=1+4=5$, correct.

5. An irrational equation, checking the conditions

C.E.: $2x+3\ge 0⟹x\ge -\frac{3}{2}$. Condition on the right-hand side: $x\ge 0$ (the right-hand side must be $\ge 0$ because a root is never negative). Of the two, the tighter is $x\ge 0$: it is the one we will use to check the solutions.

I square both sides:

I factorise, looking for two numbers with product $-3$ and sum $-2$: they are $-3$ and $1$.

Check with the condition $x\ge 0$: $x=3$ respects it, it is kept; $x=-1$ doesn’t respect it, it must be discarded: it is the extraneous solution introduced by squaring, exactly the case explained above: on squaring, $\sqrt{2x+3}=x$ and $\sqrt{2x+3}=-x$ become the same equation, and $x=-1$ is the solution of the “twin” equation we weren’t interested in.

Check on $x=3$: $\sqrt{2\cdot 3+3}=\sqrt{9}=3$, and the right-hand side is $3$: they coincide. Only $x=3$ remains.

Typical mistakes

• Squaring without conditions. Writing straight away $A=B^2$ without having imposed $A\ge 0$ and $B\ge 0$ produces equations that seem correctly solved but return solutions that don’t solve the original equation.
• Extraneous solutions not discarded. Every solution of an irrational equation must be rechecked against the conditions written at the start (or substituted directly into the original equation): the algebra can be perfect and the solution still invalid.
• Forgetting a case in the absolute value or in the zeros of a double absolute value. Both $|A|=k$ and the two-absolute-value case require considering all the zones or both equations: stopping at the first solution found, ignoring the other case, is the most frequent mistake on this topic.

Why it comes straight back in the next few weeks

With the parabola, the vertex and the sign of the discriminant you will use again the reasoning by cases you have just practised; with the circle, the radius is always a distance, so always a value under a root, never negative, and the comparison between distance and radius has the same logical structure as the irrational inequalities seen above. Later, in trigonometry, sine and cosine appear again inside absolute values in the same way. The time spent now on cases, not on formulas learned by heart, is what makes those chapters faster.

For the chapter that comes before this one, the revision of quadratic equations and inequalities, you will find quadratic equations in Year 3 at a vocational school: same programme level, different track. The chapter from a year earlier in the same technical cycle, line and factorisation, is told in factorisation and the line: the first test of Year 2. Rational inequalities, in which the sign is studied factor by factor, are in rational inequalities in Year 3.

On absolute value and irrational equations we don’t yet have a dedicated practice test in the catalogue: in the meantime you will find the complete index of tests by topic at /simulazioni/verifiche-liceo/, as the chapter grows in the class programme.

The full maths picture for the technical track is on the maths page. If you want a hand preparing for the mid-October test, start from the free 30-minute introductory call: half an hour to work out where to concentrate the revision.