A quadratic inequality is solved by finding where the parabola crosses the axis, then reading off which side of the axis you need. The sketch does the work.
Today's hook, You cannot solve a quadratic inequality by the same steps as a linear one. Dividing by the variable is illegal and the answer often comes in two separate pieces, so a different method is needed.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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You know $x^2 - 4 = 0$ has solutions $x = \pm 2$. Now think about $x^2 - 4 > 0$. Test $x = 0$, $x = 3$ and $x = -3$. What pattern do you notice about which values work?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Find the roots, then read the parabola
Work through the core explanation before applying it.
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Find the roots, then read the parabola
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Solve the matching equation first to find where the parabola cuts the $x$-axis. Those roots split the number line into regions, and the parabola is entirely above or entirely below the axis within each region.
For $x^2 - x - 6 > 0$, the roots are $-2$ and 3. Since the parabola opens upward, it is above the axis outside the roots and below between them. So the solution is $x < -2$ or $x > 3$.
$a > 0$: positive OUTSIDE the roots, negative BETWEEN them $a < 0$: the other way round
Never divide by the variable
Dividing $x^2 > 4x$ by $x$ is invalid, because $x$ could be negative, which would flip the sign, or zero.
Outside or between
For an upward parabola, $> 0$ gives two outside pieces and $< 0$ gives one piece between the roots.
Check the boundary
Strict $>$ or $<$ excludes the roots. Non-strict $\geq$ or $\leq$ includes them.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
Solve the matching equation first to find the roots, which are the boundaries of the solution.
For $a > 0$ the expression is positive outside the roots and negative between them.
For $a < 0$ the expression is negative outside the roots and positive between them.
If there are no real roots, the expression has the same sign everywhere.
Understand
Concepts
Why the roots split the number line into regions of constant sign.
Why dividing an inequality by the variable is never valid.
Why the answer to a quadratic inequality is often two separate intervals.
Can do
Skills
Solve a quadratic inequality by factorising and using a sketch or sign analysis.
Handle a negative leading coefficient correctly.
Express the solution in inequality form and in interval notation.
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Key terms
Quadratic inequalityAn inequality where the highest power of the variable is 2. Like this: $x^2 - x - 6 > 0$ is a quadratic inequality.
Critical valueA root of the matching equation, marking where the expression changes sign. Like this: for $x^2 - x - 6$ the critical values are $-2$ and 3.
Sign diagramA number line marked with the critical values and the sign of the expression in each region. Like this: for $x^2-x-6$ it reads plus, minus, plus from left to right.
Leading coefficientThe number multiplying $x^2$, which decides whether the parabola opens up or down. Like this: in $-x^2 + 2x + 8$ the leading coefficient is $-1$, so the parabola opens downward.
Concave upA parabola that opens upward, which happens when the leading coefficient is positive. Like this: $y = x^2 - 4$ is concave up, so it is above the axis outside its roots.
Union of intervalsA solution made of two separate pieces rather than one. Like this: $x < -2$ or $x > 3$, written $(-\infty, -2) \cup (3, \infty)$.
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Why linear methods fail here
Work through the core explanation before applying it.
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Why linear methods fail here
core concept
With a linear inequality you isolate the variable. With a quadratic you cannot, because the variable appears to two different powers.
Dividing by the variable is not allowed. In $x^2 > 4x$, dividing by $x$ assumes $x$ is positive; if $x$ is negative the sign must flip, and if $x = 0$ the step is undefined.
Instead, bring everything to one side, factorise, and analyse the sign of the product. $x^2 > 4x$ becomes $x^2 - 4x > 0$, so $x(x-4) > 0$.
Zero on one side, always. Sign analysis compares a product against zero. $x(x-4) > 0$ can be reasoned about; $x^2 > 4x$ cannot.
Quick check: what is the correct first step for $x^2 \leq 3x$?
Never divide a quadratic inequality by the variable, because the variable may be negative or zero. Move everything to one side so the inequality compares an expression with zero, then factorise.
Pause, copy the rule that you must never divide by the variable, the reason (sign may flip, or the step is undefined at zero), and the correct first step of moving everything to one side, into your book.
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Solving with the roots and the parabola
core concept
We just saw that a quadratic inequality must be rearranged so one side is zero and then factorised. That raises a question: once it is factorised, how do you decide which values of $x$ satisfy it? This card answers it → the roots split the number line into regions, and the parabola keeps one sign throughout each region.
Solve the matching equation to find the roots. For $x^2 - x - 6 > 0$, factorising gives $(x-3)(x+2) = 0$, so the roots are $x = 3$ and $x = -2$.
These roots are the only places the expression can change sign, so they split the number line into three regions: $x < -2$, $-2 < x < 3$, and $x > 3$.
Because the leading coefficient is positive, the parabola opens upward, so it sits above the axis outside the roots and below between them. For $> 0$ the answer is $x < -2$ or $x > 3$.
Test a point if unsure. Pick $x = 0$, which is between the roots: $0 - 0 - 6 = -6$, which is negative. That confirms the middle region fails a $> 0$ test.
For an upward parabola with roots at 1 and 5, which statement is FALSE?
Find the roots, which split the number line into regions of constant sign. For a positive leading coefficient the expression is positive outside the roots and negative between them. Test a point in a region if you are unsure.
Pause, copy the method (rearrange, factorise, find roots, sketch or test a point), and the rule that a concave-up parabola is positive outside its roots and negative between them, into your book.
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Negative leading coefficients, and no real roots
core concept
We just saw the standard case where the parabola opens upward. That raises a question: what changes when the coefficient of $x^2$ is negative, or when the quadratic has no roots at all? This card answers it → a negative leading coefficient reverses which region is which, and no real roots means the sign never changes.
If $a < 0$ the parabola opens downward, so it is above the axis between the roots and below outside them. That is the exact reverse of the standard case.
For $-x^2 + 2x + 8 \geq 0$, multiply through by $-1$ and flip the inequality to get $x^2 - 2x - 8 \leq 0$, which is now a standard upward case with roots $-2$ and 4, giving $-2 \leq x \leq 4$.
If the discriminant is negative there are no real roots, so the parabola never crosses the axis and the expression keeps one sign for every value of $x$. Then the answer is either all real numbers or no solution.
Multiplying by $-1$ flips the inequality. This is the same flip rule you met with linear inequalities, and it is the safest way to handle a negative leading coefficient.
Fill the blank: since $x^2 + 4 > 0$ has no real roots and opens upward, the solution is all real .
A negative leading coefficient reverses the regions, so multiply through by $-1$ and flip the inequality to return to the standard case. With no real roots the expression never changes sign, so the answer is either every real number or no solution at all.
Pause, copy the multiply-by-negative-one-and-flip strategy with the worked $-x^2+2x+8 \geq 0$ giving $-2 \leq x \leq 4$, and the no-real-roots rule, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · STANDARD UPWARD CASE
Solve $x^2 - x - 6 > 0$, giving your answer in interval notation.
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Factorise: $(x-3)(x+2) > 0$, so the roots are $3$ and $-2$
One side is already zero.
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Leading coefficient $1 > 0$, so the parabola opens upward
Above the axis outside the roots, below between them.
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$x < -2$ or $x > 3$, that is $(-\infty, -2) \cup (3, \infty)$
Strict inequality, so the roots are excluded and both brackets are round.
PROBLEM 2 · BETWEEN THE ROOTS
Solve $x^2 - 4x + 3 \leq 0$, giving your answer in interval notation.
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Factorise: $(x-1)(x-3) \leq 0$, so the roots are 1 and 3
Rearranged already, one side zero.
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Upward parabola, and we want where it is $\leq 0$
That is the region between the roots.
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$1 \leq x \leq 3$, that is $[1, 3]$
Non-strict, so the roots are included and both brackets are square.
PROBLEM 3 · NEGATIVE LEADING COEFFICIENT
Solve $-x^2 + 2x + 8 \geq 0$.
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Multiply through by $-1$ and flip: $x^2 - 2x - 8 \leq 0$
Multiplying an inequality by a negative reverses the sign.
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Factorise: $(x-4)(x+2) \leq 0$, roots $4$ and $-2$
Now a standard upward case.
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$-2 \leq x \leq 4$, that is $[-2, 4]$
At or below the axis means between the roots, endpoints included.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Solve $x^2 - 9 < 0$.
Solve $(x-1)(x+5) \geq 0$.
Solve $x^2 \leq 3x$.
Solve $x^2 + 1 > 0$.
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Revisit your test values
Run the quick drill and copy the summary into your book.
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Revisit your test values
At the start you tested $x = 0$, $x = 3$ and $x = -3$ in $x^2 - 4 > 0$. Explain how those three results match the shape of the parabola, and write the full solution in interval notation.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Solve $x^2 + 2x - 15 \geq 0$, giving your answer in interval notation. (3 marks)
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ApplyBand 54 marks
Q2. Solve $-x^2 + 5x - 4 > 0$, showing clearly how you handle the negative leading coefficient. (4 marks)
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UnderstandBand 42 marks
Q3. Explain why $x^2 > 4x$ cannot be solved by dividing both sides by $x$, and state the correct first step. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $(x-3)(x+3) < 0$, so $-3 < x < 3$, that is $(-3, 3)$. Practice 2: $x \leq -5$ or $x \geq 1$, that is $(-\infty, -5] \cup [1, \infty)$. Practice 3: $x^2 - 3x \leq 0$, so $x(x-3) \leq 0$, giving $0 \leq x \leq 3$. Practice 4: no real roots and always positive, so all real $x$.
Q1 (3 marks): $(x+5)(x-3) \geq 0$, roots $-5$ and 3 [1]. Upward parabola, at or above the axis outside the roots [1]. $x \leq -5$ or $x \geq 3$, that is $(-\infty, -5] \cup [3, \infty)$ [1].
Q2 (4 marks): Multiply by $-1$ and flip: $x^2 - 5x + 4 < 0$ [1]. Factorise: $(x-1)(x-4) < 0$, roots 1 and 4 [1]. Upward parabola below the axis between the roots [1]. $1 < x < 4$, that is $(1, 4)$ [1].
Q3 (2 marks): Dividing by $x$ is invalid because $x$ may be negative, which would require flipping the inequality, and may be zero, which makes the division undefined [1]. Correct first step: rewrite as $x^2 - 4x > 0$ and factorise to $x(x-4) > 0$ [1].
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Review and finish
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Boss battle · Parabola Patrol
earn bronze · silver · gold
Solve quadratic inequalities at speed, reading the answer off the shape of the parabola. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.