Solving a second degree inequality with a quick sketch

By Marius, private math tutor in Montreal · 2026-10-05

To solve a second degree inequality, move everything to one side so that $a x^{2} + b x + c$ is compared with $0$ , find the zeroes, and sketch the parabola through them: upward if $a > 0$ , downward if $a < 0$ . The solution set is wherever the sketch is above the $x$ -axis for $>$ or $\geq$ , or below it for $<$ or $\leq$ , and the zeroes belong to it only when the sign includes "or equal to".

The method in four steps

Get zero on one side. The sketch is read against the $x$ -axis, which is the line $y = 0$ , so $x^{2} + 3 x \geq 10$ has to become $x^{2} + 3 x - 10 \geq 0$ first.
Find the zeroes of $a x^{2} + b x + c$ , by factoring or with the quadratic formula.
Sketch. Mark the zeroes on the $x$ -axis and draw a parabola through them, opening upward if $a > 0$ and downward if $a < 0$ . That is all the sketch needs: no vertex, no scale, no table of values.
Read and write. For $>$ or $\geq$ , keep the $x$ -values where the curve is above the axis; for $<$ or $\leq$ , where it is below. Include the zeroes only if the sign has "or equal to".

Worked example 1: a parabola that opens upward

Solve $x^{2} + 3 x \geq 10$ .

Bring everything to one side, then factor:

x^{2} + 3 x - 10 \geq 0

(x + 5) (x - 2) \geq 0

The zeroes are $- 5$ and $2$ , and $a = 1 > 0$ , so the parabola opens upward.

The sketch for

x^{2} + 3 x - 10 \geq 0

. The curve is above the axis to the left of

- 5

and to the right of

2

, and crosses it at the two zeroes, which are included because the sign is

\geq

Solution set:

] - \infty, - 5] \cup [2, + \infty [

A test value from each region confirms it. At $x = - 6$ : $36 - 18 = 18 \geq 10$ , true. At $x = 0$ : $0 \geq 10$ , false, and $0$ is indeed left out. At $x = 3$ : $9 + 9 = 18 \geq 10$ , true.

Worked example 2: a parabola that opens downward

Solve $- 2 x^{2} + 5 x + 3 < 0$ .

Zero is already on the right. Factor out $- 1$ to find the zeroes:

- 2 x^{2} + 5 x + 3 = - (2 x^{2} - 5 x - 3) = - (2 x + 1) (x - 3)

The zeroes are $- \frac{1}{2}$ and $3$ . This time $a = - 2 < 0$ , so the parabola opens downward: it is above the axis between the zeroes and below it outside them.

The sketch for

- 2 x^{2} + 5 x + 3 < 0

. The curve is below the axis outside the zeroes

- \frac{1}{2}

and

3

. The dots are open because the sign is strict.

Solution set:

] - \infty, - \frac{1}{2} [\cup] 3, + \infty [

Check: at $x = 0$ the expression is $3$ , which is not negative, so $0$ is correctly left out; at $x = 4$ it is $- 32 + 20 + 3 = - 9$ , which is.

Notice that nothing was divided by $- 2$ . The sketch copes with a negative $a$ directly. If you prefer a positive leading coefficient, multiply both sides by $- 1$ and reverse the sense of the inequality: $2 x^{2} - 5 x - 3 > 0$ has the same solution set.

When the parabola touches or misses the axis

Not every second degree inequality has two zeroes, and the sketch handles the other cases just as well. The discriminant $b^{2} - 4 a c$ tells you which case you are in before you solve anything.

One zero: the vertex sits on the axis. $x^{2} - 6 x + 9 = (x - 3)^{2}$ touches the axis at $3$ and is above it everywhere else. So $(x - 3)^{2} > 0$ has solution set $] - \infty, 3 [\cup] 3, + \infty [$ ; $(x - 3)^{2} \geq 0$ holds for all of $ℝ$ ; $(x - 3)^{2} \leq 0$ has the single solution ${3}$ ; and $(x - 3)^{2} < 0$ has none.
No zero: the parabola misses the axis. For $x^{2} + 2 x + 5$ , $b^{2} - 4 a c = 4 - 20 = - 16 < 0$ . The parabola opens upward and never reaches the axis, so it is above it everywhere: $x^{2} + 2 x + 5 > 0$ holds for all of $ℝ$ , and $x^{2} + 2 x + 5 < 0$ has solution set $\emptyset$ .

An answer of $ℝ$ , $\emptyset$ or a single number is not a sign that something went wrong. It is exactly what the picture says.

Writing the solution set

A Secondary 4 answer gives the solution set in interval notation, and sometimes in set-builder notation as well.

A bracket that faces the number includes it, and a bracket turned away excludes it: $[2, + \infty [$ contains $2$ , and $] 3, + \infty [$ does not contain $3$ .
$- \infty$ and $+ \infty$ always get a bracket turned away.
Two separate pieces are joined with $\cup$ , the piece on the left first.
The answer to Worked example 1 in set-builder notation is ${x \in ℝ ∣ x \leq - 5 or x \geq 2}$ .

The most common mistake: solving each factor on its own. The zero product property says when a product equals zero; it says nothing about when a product is positive. $(x + 5) (x - 2)$ is positive where the two factors have the same sign, and that happens in two separate places, which is why the answer has two pieces. Treating each bracket as an equation and keeping only " $x$ is at least $2$ " throws away the entire left-hand piece. A sketch cannot make that mistake, because both arms of the parabola are on the page.

Check your understanding

No calculator needed. Give each answer as a solution set, in interval notation wherever the answer is an interval.

Solve $x^{2} \leq 7 x$ .
Solve $x^{2} - 2 x - 24 \geq 0$ .
Solve $- x^{2} + 2 x + 15 > 0$ .
Solve $2 x^{2} + x - 6 \leq 0$ .
Solve $x^{2} - 8 x + 16 \leq 0$ .
Without solving, explain why $x^{2} - x + 3 < 0$ has no solution. What is the solution set of $x^{2} - x + 3 > 0$ ?
The solution set of an inequality $a x^{2} + b x + c < 0$ is $] - \infty, 1 [\cup] 5, + \infty [$ . Is $a$ positive or negative? Explain.
Find the error. To solve $- x^{2} + 3 x + 10 \geq 0$ , Sam multiplies both sides by $- 1$ to get $x^{2} - 3 x - 10 \geq 0$ , factors it as $(x - 5) (x + 2) \geq 0$ , and answers $] - \infty, - 2] \cup [5, + \infty [$ .

Show the answers

$[0, 7]$ . First get zero on one side: $x^{2} - 7 x \leq 0$ , so $x (x - 7) \leq 0$ . The parabola opens upward through $0$ and $7$ and is below the axis between them; the zeroes are included.
$] - \infty, - 4] \cup [6, + \infty [$ . $(x + 4) (x - 6) \geq 0$ . The parabola opens upward, so it is on or above the axis outside the zeroes $- 4$ and $6$ .
$] - 3, 5 [$ . $- x^{2} + 2 x + 15 = - (x - 5) (x + 3)$ , with zeroes $- 3$ and $5$ . Since $a = - 1 < 0$ the parabola opens downward, so it is above the axis between the zeroes, which are excluded because the sign is strict.
$[- 2, \frac{3}{2}]$ . $2 x^{2} + x - 6 = (2 x - 3) (x + 2)$ , with zeroes $- 2$ and $\frac{3}{2}$ . The parabola opens upward, so it is below the axis between them, and $\leq$ includes them.
${4}$ . $x^{2} - 8 x + 16 = (x - 4)^{2}$ , which is never negative and equals zero only at $x = 4$ . The parabola touches the axis there and is above it everywhere else.
$b^{2} - 4 a c = 1 - 12 = - 11 < 0$ , so there are no zeroes, and $a = 1 > 0$ , so the whole parabola is above the axis. The expression is never negative: $x^{2} - x + 3 < 0$ has solution set $\emptyset$ , and $x^{2} - x + 3 > 0$ has solution set $ℝ$ , that is $] - \infty, + \infty [$ .
Negative. The expression is below the axis outside its zeroes $1$ and $5$ , and therefore above it between them: the parabola opens downward. With $a > 0$ the solution set would have been $] 1, 5 [$ .
Multiplying by $- 1$ reverses the sense of the inequality, so the correct one is $x^{2} - 3 x - 10 \leq 0$ , and the answer is $[- 2, 5]$ . A quick test exposes the error: $x = 0$ gives $10 \geq 0$ in the original, so $0$ must be in the solution set, and Sam's answer leaves it out.

If question 5 or 6 surprised you, reread the section on parabolas that touch or miss the axis. If question 8 did, the sketch is the safer habit: it never needs the sense of the inequality reversed.

Practise it

The free Secondary 4 Equations and Inequalities worksheet has a sheet on second degree equations and inequalities, next to the square root, exponential and greatest integer ones, plus a sheet on writing solution sets. For the parabola itself (zeroes, the discriminant, the vertex and the direction of opening), the Secondary 4 Quadratic Functions worksheet is the companion.

Want one-on-one help with this? I offer Secondary 4 and 5 math tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.