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This page sets out the method for quadratic inequalities in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Quadratic inequalities topic page.
When you see any inequality with an term, do this.
When you want to check a sketch, or the curve is hard to picture, do this.
Solve the matching equation. The roots are the boundary of the region.
If a is positive, the curve is negative between the roots and positive outside them.
A quick U-shape with two crossings shows the region without a table of ten values.
A strict inequality leaves the roots out. An inclusive one keeps them.
2x^2-8x+6<0 has the same roots and the same region as x^2-4x+3<0. Dividing by a positive 2 does not flip the sign.
Write a double inequality between the roots, or two inequalities joined with or.
Treat x^2-5x+6≤ 0 as the equation x^2-5x+6=0 for one line only. It factorises as (x-2)(x-3)=0, so the curve meets the axis at x=2 and x=3. Those two numbers split the number line into three regions: less than 2, between 2 and 3, and greater than 3. You have not solved the inequality until you say which of those regions you want.
Check a factorisation by expanding it back. (x-2)(x-3)=x^2-5x+6. If the quadratic does not factorise with integers, the quadratic formula still gives the critical values, and you sketch from those. A repeated root, such as (x-4)^2≥ 0, is a different picture: the curve touches the axis and the expression is never negative, so the inequality holds for every x.
Method: 1. Replace the inequality sign by = for one line. 2. Factorise or use the formula to find the critical values. 3. Mark them on a sketch of the parabola. 4. You have not finished until you choose a region.
With a positive coefficient of x^2, shade below the axis for ≤ or <, and above the axis for ≥ or >. For x^2-5x+6≤ 0 the curve is below or on the axis from 2 to 3, so the solution is 2≤ x≤ 3. A test point confirms it. At x=2.5 the value is 6.25-12.5+6=-0.25, which is negative, so it belongs. At x=1 the value is 1-5+6=2, which is positive, so it does not belong to ≤ 0.
The same sketch read the other way answers the opposite inequality. x^2-5x+6>0 is the outside: x<2 or x>3. Joining those two parts with "and" is impossible, because no number is both less than 2 and greater than 3. Write "or", or draw the number line with two separate rays. Include the roots only when the sign allows equality.
Method: 1. A positive coefficient of x^2 gives a U-shaped curve. 2. The curve is below the axis between the roots and above it outside them. 3. Include the roots for ≤ and ≥. Leave them out for < and >. 4. Test one point inside your region and one point outside it.
For 2x^2-3x-2>0, factorise first. (2x+1)(x-2)=2x^2-4x+x-2=2x^2-3x-2. The critical values are x=-1/2 and x=2. The coefficient of x^2 is still positive, so the expression is positive outside the roots: x<-1/2 or x>2. At x=3 the value is 18-9-2=7, which is positive. At x=0 the value is -2, which is negative and correctly left out.
Do not divide the inequality by 2 and then forget the other factor. Dividing by a positive number keeps the sign, so x^2-3/2x-1>0 is a fair rewrite, but the roots are easier to see from the integer factorisation. If you ever multiply or divide by a negative number, the inequality sign turns over. The sketch method avoids that trap, because you never move terms across an inequality sign after the quadratic is arranged with a positive x^2 term.
Method: 1. Factorise, including when a is not 1. 2. The same U-shape rule applies if a is positive. 3. Outside the roots, join the two parts with "or". 4. Do not write "and" between two rays that do not overlap.
The inequality x^2-7x+10≤ 0 factorises as (x-2)(x-5)≤ 0, so 2≤ x≤ 5. The integers in that closed interval are 2, 3, 4 and 5. Check the ends: both give zero, and ≤ keeps zero. Check just outside: x=1 gives 1-7+10=4, and x=6 gives 36-42+10=4. Both are positive, so both stay out. The answer to "find the integers" is the list, not the inequality 2≤ x≤ 5 on its own.
A strict inequality drops the ends. If the same quadratic were <0, the integers would be 3 and 4 only. Read the sign before you write the list. If the solution is two rays, the integer list can be infinite, and the honest answer is then the inequality itself, with a note that every integer on those rays works. A finite list is what you get when the region is a closed or open interval between two roots.
Method: 1. Solve the inequality first, as a region. 2. Then list the integers inside it. 3. A strict inequality drops the endpoints. 4. If the question asked for integers, the list is the answer, not the inequality alone.
Exam questions rarely hand you the inequality ready to factorise. You may be given x^2>3x+10 or 3x^2-2x<2x^2+3x+36 with terms on both sides. Move everything to one side so the other side is 0, and keep the x^2 term positive if you can. The two examples become x^2-3x-10>0 and x^2-5x-36<0. Each of these moves keeps the sign of the inequality the same, because only multiplying or dividing by a negative number reverses it.
Only then do you factorise and find the critical values. Never divide both sides by x or take a square root, because this loses a region of the solution. Moving terms across changes their sign, just as it does in a normal equation. Check that you have collected every term before you factorise.
Method: 1. Expand any brackets and gather all terms on one side. 2. Leave 0 on the other side and keep the sign of the inequality. 3. Factorise the quadratic to find the critical values. 4. Sketch the curve and choose the region that matches the inequality.
Tip: If you multiply or divide both sides by a negative number, reverse the inequality sign.
A quick sketch removes guessing. Draw a U shape that meets the x-axis at the critical values. If the question asks for values where the expression is less than zero, shade between the roots, where the curve is below the axis. For greater than zero, take the two outside regions, where the curve is above the axis.
The answer is written as an inequality: -3<x<8 for the region between, and x<-5 or x>4 for the outside. Use strict signs for < and > and include the end values only for ≤ and ≥. The word or joins two separate regions. A number line with open and closed circles is accepted when the question asks you to show the solution on one.
Method: 1. Mark the critical values on a quick sketch of a U shaped curve. 2. Decide whether the question wants above the axis or below it. 3. Write the region as one inequality or as two joined by or. 4. Test one value from your region in the original inequality.
If the x^2 term is negative, as in -x^2+4x+12≥ 0, the curve is an upside down U. The easiest approach is to multiply both sides by -1 and reverse the sign, giving x^2-4x-12≤ 0. Then solve as normal. Check the answer by testing a point, because reversing the sign is where marks are lost.
Special cases appear too. A perfect square such as x^2-6x+9≥ 0 is true for every value of x, because a square is never negative. The inequality x^2-6x+9≤ 0 has only one solution, x=3. When the quadratic does not factorise, you can use the quadratic formula or completing the square to find the critical values as surds.
Method: 1. If x^2 is negative, multiply through by -1 and reverse the inequality sign. 2. Look for a perfect square before you do anything else. 3. For surd roots, use completing the square or the formula to find the critical values. 4. Test a value in each region of your sketch.
Some questions ask for the integer values that satisfy two conditions together, such as a quadratic and a linear inequality. Solve each separately, then find where the regions overlap, and list the whole numbers in the overlap. A number line helps you see the overlap. Take care at the end values: whether they are included depends on the sign.
Worded contexts use the same method. A rectangle of width x and length x+7 having an area less than 60 gives x(x+7)<60, which rearranges to x^2+7x-60<0. After solving, apply the context: a width cannot be negative, so keep only the positive part. Always return to the words and give the allowed range of x.
Method: 1. Write the condition from the words as an inequality in x. 2. Solve each inequality separately and mark the regions on a number line. 3. Take the overlap of the regions. 4. List integers or state the range, and discard values that do not make sense in the context.
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