P1 1.7, P1 1.9 - Solving inequalities | Edexcel A-Level International Mathematics

Find complete ranges of values that satisfy linear and quadratic inequalities. P1 1.9 supplies the algebraic solutions; P1 1.7 connects them to graph positions, including a curve below a line and fractions with excluded values.

Linear inequalities with brackets and fractions

If $x^2<4$ , is every number less than $2$ allowed? The answer is no: $-3$ fails. Inequalities ask for a whole set of values, so a correct boundary is only the beginning.

First consider linear expressions. Adding or subtracting the same quantity preserves order. Multiplying by a positive number also preserves order: $2<5$ gives $6<15$ . Multiplying by $-1$ reverses their positions on the number line: $-2>-5$ .

Order rules

For real $a,b$ , if $a<b$ , then

a+c<b+c.

For $c>0$ , $ac<bc$ ; for $c<0$ , $ac>bc$ . Division follows the same sign rule.

Not in the formula booklet: learn it.

Expand brackets carefully. Clear fixed positive denominators by multiplying every term by their lowest common multiple. If you later divide by a negative coefficient, reverse the inequality.

Worked example 1: brackets and a negative coefficient

Question. Solve $3(2-x)<2x+1$ .

Plan. Expand, collect $x$ -terms on the left, then divide.

\begin{aligned} 6-3x&<2x+1 &&\text{expand the bracket}\\ -5x&<-5 &&\text{subtract }2x+6\\ x&>1 &&\text{divide by }-5\text{ and reverse} \end{aligned}

Answer. $x>1$ .

Check. $x=2$ gives $0<5$ , true. At the excluded boundary $x=1$ , both expressions equal $3$ .

Change only the left expression by dividing it by $2$ . Clearing this positive denominator preserves the inequality, although the boundary changes.

Worked example 2: a fixed denominator

Question. Solve $\frac{3(2-x)}2<2x+1$ .

Plan. Multiply both sides by $2$ before expanding and collecting.

\begin{aligned} 3(2-x)&<2(2x+1) &&\text{multiply by positive }2\\ 6-3x&<4x+2 &&\text{expand both sides}\\ -7x&<-4 &&\text{collect terms}\\ x&>\frac47 &&\text{divide by }-7\text{ and reverse} \end{aligned}

Answer. $x>\frac47$ .

Check. At $x=1$ , $3/2<3$ , true. At $x=0$ , $3<1$ , false, matching the two sides of the boundary.

A strict inequality excludes its endpoint. Replacing $<$ with $\le$ throughout the same calculation includes the endpoint instead.

Watch out

From $-7x<-4$ , the answer is $x>4/7$ , not $x<4/7$ . Division by a negative reverses order; adding a negative does not.

Use a positive common denominator first.

Quadratic sign diagrams

A quadratic's roots split the number line into intervals where its value is positive or negative.

For $f(x)=(x+1)(x-3)$ , take one value in each interval. At $x=-2$ , both factors are negative, so their product is positive. At $x=0$ , the factors have opposite signs, giving a negative product. At $x=4$ , both are positive. The quadratic is continuous and has no other zeros, so its sign cannot change inside one of these intervals.

Sign diagram

A sign diagram records where an expression is positive, zero or negative. Put its zeros in order, then label the sign on each intervening interval. For a fraction, also mark any excluded values.

The sign diagram describes the same information as a graph's position above or below the horizontal axis.

[DIAGRAM: asset_name: Roots divide a quadratic into sign intervals; asset_slug: quadratic-sign-intervals; caption: A positive-leading quadratic is negative between its two distinct roots.; description: The graph y equals x plus one times x minus three crosses at minus one and three. A number-line sign strip below shows positive, negative, positive on the three intervals.]

Worked example 3: positive leading coefficient

Question. State where $(x+1)(x-3)$ is positive, negative and zero.

Plan. Locate the roots and examine the factors in each interval.

\begin{array}{c|ccc} x&x<-1&-1<x<3&x>3\\\hline x+1&-&+&+\\ x-3&-&-&+\\\hline (x+1)(x-3)&+&-&+ \end{array}

Answer. Positive for $x<-1$ or $x>3$ ; negative for $-1<x<3$ ; zero at $x=-1,3$ .

Check. The test values $-2,0,4$ give $5,-3,5$ , respectively.

Change only the sign of the entire expression. The roots stay fixed while every nonzero output changes sign.

Worked example 4: negative leading coefficient

Question. State where $-(x+1)(x-3)$ is positive, negative and zero.

Plan. Use the same boundaries and reverse the product's signs.

\begin{array}{c|ccc} x&x<-1&-1<x<3&x>3\\\hline -(x+1)(x-3)&-&+&- \end{array}

Answer. Positive for $-1<x<3$ ; negative for $x<-1$ or $x>3$ ; zero at $x=-1,3$ .

Check. Its expanded form is $-x^2+2x+3$ , a downward-opening parabola. At $x=0$ , its value is $3>0$ , between the roots.

A repeated root behaves differently. Since $(x-2)^2\ge0$ , its sign is positive on both sides of $2$ and zero at $2$ . A repeated factor need not change sign at its root.

Watch out

The pattern positive, negative, positive requires a positive-leading quadratic with two distinct real roots. Check the leading coefficient and repeated roots before using a remembered picture.

Build the signs rather than assuming the outer intervals are positive.

Solving quadratic inequalities

To solve a quadratic inequality, first find the boundary roots, then select the intervals with the required sign.

For $x^2-4<0$ , factorising gives boundaries $-2$ and $2$ . The parabola opens upwards, so it is below the axis between them. This answers our opening question: $x^2<4$ means $-2<x<2$ , rather than just $x<2$ .

Key idea

Roots locate the boundaries. The sign of the quadratic selects the intervals. The symbols $\le$ and $\ge$ include zeros; $<$ and $>$ exclude them.

Move all terms to one side before analysing signs, so the comparison is with zero.

Worked example 5: values between roots

Question. Solve $2x^2-5x-3<0$ .

Plan. Factorise, identify the graph's shape and select the negative interval.

\begin{aligned} 2x^2-5x-3&=(2x+1)(x-3)\\ (2x+1)(x-3)&=0\implies x=-\frac12,3 \end{aligned}

The leading coefficient is positive, so the sign pattern is positive, negative, positive.

Answer. $-\frac12<x<3$ .

Check. $x=0$ gives $-3<0$ , while $x=4$ gives $9<0$ , false. The strict sign excludes both roots.

Change only the requested sign, from $<0$ to $\ge0$ . Use the same roots and signs, but choose the opposite intervals and include the endpoints.

Worked example 6: two outer intervals

Question. Solve $2x^2-5x-3\ge0$ .

Plan. Reuse the roots $-1/2$ and $3$ , selecting positive values and zeros.

\begin{array}{c|ccc} x&x<-1/2&-1/2<x<3&x>3\\\hline 2x^2-5x-3&+&-&+ \end{array}

Answer. $x\le-\frac12$ or $x\ge3$ .

Check. $x=-1$ gives $4\ge0$ , and $x=4$ gives $9\ge0$ . At either boundary the value is zero, which is allowed.

Some quadratics have no distinct sign intervals. For example, $(x-2)^2<0$ has no real solution, whereas $(x-2)^2\le0$ has just $x=2$ . Also, $(x-2)^2+1>0$ holds for every real $x$ .

Watch out

The two outer intervals are joined by or. Writing $x\le-1/2$ and $x\ge3$ would require one number to meet both conditions, which is impossible.

Identify the sign and endpoint conditions separately.

Curve below a line and fractional inequalities with exclusions

Comparing a curve's height with a line's height gives an inequality in $x$ , and the same sign reasoning can handle simple fractional expressions.

For the curve $y=x^2$ to lie below $y=x+2$ , we need $x^2<x+2$ . Their intersections mark where the vertical difference is zero.

Worked example 7: where a curve is below a line

Question. Find the values of $x$ for which $y=x^2$ is below $y=x+2$ .

Plan. Subtract the line's height and select where the difference is negative.

x^2<x+2\implies x^2-x-2<0\implies(x-2)(x+1)<0.

The positive-leading quadratic is negative between its roots.

Answer. $-1<x<2$ .

Check. At $x=0$ , the curve's height $0$ is below the line's height $2$ . At $x=3$ , $9$ is above $5$ , as the interval predicts.

Change only the line's constant term, from $2$ to $6$ . The algebra still compares heights; only the intersection boundaries move.

Worked example 8: new intersection boundaries

Question. Find where $y=x^2$ is below $y=x+6$ .

Plan. Form and solve the quadratic inequality for the height difference.

x^2<x+6\implies(x-3)(x+2)<0.

Answer. $-2<x<3$ .

Check. At $x=2$ , $4<8$ , true; at $x=4$ , $16<10$ , false. Both endpoints give equal heights and are excluded.

A variable denominator introduces a different issue: its sign is unknown. In $a/x<b$ , multiplying by $x$ without a case split is unsafe. Instead, record $x\ne0$ and multiply by $x^2$ , which is strictly positive for every allowed $x$ :

\frac ax<b\iff ax<bx^2,\qquad x\ne0.

This gives a quadratic inequality while preserving its direction. The excluded value must survive the transformation.

Worked example 9: a positive numerator

Question. Solve $\frac3x<1$ .

Plan. Exclude zero, multiply by positive $x^2$ , then use signs.

\frac3x<1\iff3x<x^2\iff x(x-3)>0,\qquad x\ne0.

The product is positive outside its roots $0$ and $3$ .

Answer. $x<0$ or $x>3$ .

Check. $x=-1$ gives $-3<1$ , and $x=4$ gives $3/4<1$ . In the rejected interval, $x=1$ gives $3<1$ , false.

Change only the numerator's sign. The denominator restriction stays the same, but the other boundary moves to the left of zero.

Worked example 10: a negative numerator

Question. Solve $-\frac3x<1$ .

Plan. Preserve $x\ne0$ and multiply by $x^2$ again.

-\frac3x<1\iff-3x<x^2\iff x(x+3)>0,\qquad x\ne0.

Answer. $x<-3$ or $x>0$ .

Check. At $x=-4$ , $3/4<1$ ; at $x=1$ , $-3<1$ . At $x=-1$ , $3<1$ is false.

In a non-strict fractional inequality, a transformed boundary may look admissible even though the original expression is undefined there.

Watch out

For $3/x\le1$ , the quadratic gives $x\le0$ or $x\ge3$ , but $x=0$ is forbidden. The final answer is $x<0$ or $x\ge3$ .

Carry the exclusion through your own solution.

Quick Recap

Reverse an inequality when multiplying or dividing by a negative number.
Find quadratic roots, then select whole sign intervals with the correct endpoints.
A curve is below a line where curve height minus line height is negative.
Preserve denominator exclusions; multiplying by $x^2>0$ avoids guessing the sign of $x$ .

Our opening inequality $x^2<4$ has the complete solution $-2<x<2$ .

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P1 1.7, P1 1.9 - Solving inequalities

Linear inequalities with brackets and fractions

Quadratic sign diagrams

Solving quadratic inequalities

Curve below a line and fractional inequalities with exclusions

Quick Recap

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