LearnYMA01WMA11 · Pure Mathematics 1Algebra and functionsQuadratics, simultaneous equations and inequalities

Algebra and functions · Subtopic 2 of 3

Quadratics, Simultaneous Equations and Inequalities

Completing the square finds the turning point of a quadratic; the discriminant $b^2-4ac$ tells you how many roots it has. Solve a line and a curve together by substitution, and solve quadratic inequalities by sketching the parabola and reading off where it is above or below the axis.

Edexcel YMA01 WMA11≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Complete the square and find the turning point.
Use $b^2-4ac$ to decide the number of real roots, and find unknown constants.
Solve simultaneous equations where one is quadratic.
Solve linear and quadratic inequalities, including with $x$ in a denominator.

Pause and think

The line $y=kx-1$ is slid around by changing $k$ . How many times can it meet the parabola $y=x^2+3$ ?

Section 2 of 14

Core idea

Start hereHow many times does the ball cross the line?

Every quadratic graph is a parabola. Slide it up or down and it may cut the $x$ -axis twice, touch it once, or miss completely. One number, the discriminant, tells you which, before you draw anything.

The big idea

$ax^2+bx+c=0$ is solved by factorising, completing the square or the formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . $b^2-4ac$ decides the number of roots. Inequalities: sketch and read off.

1. Completing the square and the formula

From completing the square to the formula

$ax^2+bx+c=0\Rightarrow x^2+\tfrac bax=-\tfrac ca$
Divide by $a\neq0$ .
$\left(x+\tfrac b{2a}\right)^2-\tfrac{b^2}{4a^2}=-\tfrac ca$
$(x+k)^2=x^2+2kx+k^2$ with $k=\frac b{2a}$ ; subtract the extra $k^2$ .
$\left(x+\tfrac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}$
Common denominator.
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Square root both sides ( $\pm$ ), subtract $\frac b{2a}$ .

Worked example $x^2-6x+5=0$ three ways

Solve.

Goal: Both roots.
Strategy: Factorise, complete the square, formula: all must agree.

$(x-1)(x-5)=0\Rightarrow x=1,5$
Numbers multiplying to 5, adding to −6.
$(x-3)^2-4=0\Rightarrow x-3=\pm2$
Half of −6 is −3; $9-4=5$ .
$x=\frac{6\pm\sqrt{36-20}}2=\frac{6\pm4}2$
Formula.

Answer

x=1

x=5

✓ Check

1-6+5=0

✓;

25-30+5=0

✓.

2. The discriminant

y = x² − 4x + c: sliding the curve up changes the number of roots

c = 0.0

b^2-4ac=16-4c

. For

c<4

two roots,

c=4

one (touching),

c>4

none.

b^2-4ac\ \begin{cases}>0&\text{two distinct real roots}\\=0&\text{one repeated root}\\<0&\text{no real roots}\end{cases}

$b^2-4ac$: the quantity under the square root in the formula.

Worked exampleEqual roots

Find $k$ so $x^2+kx+9=0$ has equal roots.

Goal: $k$ .
Strategy: Set the discriminant to zero.

$k^2-4(1)(9)=0$
$a=1$ , $b=k$ , $c=9$ .
$k=\pm6$
$k^2=36$ .

Answer

k=6

k=-6

✓ Check

x^2+6x+9=(x+3)^2

✓.

3. Inequalities and simultaneous equations

Line meets curve at two points; the curve is below the line between them

The shaded strip is where

x^2-x-2<x+1

, i.e.

x^2-2x-3<0

: the solution of the inequality is

-1<x<3

Worked example $x^2-x-2<x+1$

Solve.

Goal: Interval of $x$ .
Strategy: Collect on one side, find roots, sketch.

$x^2-2x-3<0$
Subtract $x+1$ .
$(x-3)(x+1)<0$
Critical values $-1$ , $3$ .
$-1<x<3$
A U-shape is below the axis between its roots.

Answer

-1<x<3

✓ Check

x=0

-2<1

✓;

x=4

10<5

✗ ✓.

Worked exampleLine and circle

Solve $y=x+1$ , $x^2+y^2=13$ .

Goal: Intersection points.
Strategy: Substitute the linear one into the quadratic.

$x^2+(x+1)^2=13\Rightarrow2x^2+2x-12=0$
Expand.
$x^2+x-6=(x+3)(x-2)=0$
Divide by 2.
$x=2\Rightarrow y=3;\quad x=-3\Rightarrow y=-2$
Back into the linear equation.

Answer

(2,3)

and

(-3,-2)

✓ Check

4+9=13

✓;

9+4=13

✓.

Common misconceptionDividing by x Never divide an inequality or equation by

x

: you lose

x=0

or flip the sign wrongly.

Pause and think

For which $k$ does $x^2+4x+k=0$ have no real roots?

Summary

Three solving methods.
Discriminant counts roots.
Sketch for inequalities; substitute for simultaneous.

Discriminant: $b^2-4ac$ .
Completing the square: Writing $(x+p)^2+q$ .

HistoryCompleting a real square. Around

820

AD the Baghdad scholar Muhammad ibn Musa al-Khwarizmi solved quadratics by literally completing a square: drawing

x^2

and

10x

as areas and adding the missing corner. His book's title gave us the word algebra, and his name gave us algorithm.

Fun factWhy "quad"? "Quadratic" comes from the Latin quadratus, meaning "square", because

x^2

is the area of a square of side

x

. That is why an equation with a squared term is called quadratic even though "quad" usually means four.

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