Study notes

Equations, Inequalities & Quadratics

Quadratic graphs, solving linear inequalities, solving simultaneous linear equations, and solving quadratics by factorisation.

Learn it step by step

A quadratic graph is a U-shaped (or n-shaped) curve

The graph of $y = a x^{2} + b x + c$ is a smooth curve called a parabola. If $a > 0$ it opens upward, like a U, with a lowest point; if $a < 0$ it opens downward, like an n, with a highest point. This turning point is the minimum or maximum value of $y$ on the whole graph.

The roots of a quadratic are where its graph crosses the x-axis

A root of $a x^{2} + b x + c = 0$ is a value of $x$ that makes $y = 0$ , which is exactly where the curve touches or crosses the x-axis. A quadratic can have two roots (crosses twice), one root (touches once, at the turning point), or no real roots (never touches the axis at all).

The graph of

y = x^{2} - 4

crosses the x-axis at

x = - 2

and

x = 2

, the two roots of

x^{2} - 4 = 0

Solving a linear inequality is almost like solving an equation, with one key rule

Add, subtract, multiply, or divide both sides exactly as you would for an equation, EXCEPT: multiplying or dividing by a NEGATIVE number flips the inequality sign. For example, $- 2 x > 8$ becomes $x < - 4$ after dividing by $- 2$ , because you divided by a negative number.

Simultaneous equations by substitution: make one letter the subject, then swap it in

Rearrange one equation to write one letter in terms of the other, then substitute that expression into the second equation so it has only one unknown left. For $x + y = 7$ and $x - y = 3$ : from the second equation $x = y + 3$ , substituting into the first gives $(y + 3) + y = 7$ , so $y = 2$ and then $x = 5$ .

Simultaneous equations by elimination: add or subtract the equations to remove a letter

If the coefficients of one letter are the same size (possibly after multiplying an equation through), adding or subtracting the two equations removes that letter entirely. For $2 x + y = 11$ and $x - y = 1$ : the $y$ terms are $+ y$ and $- y$ , so ADDING the equations eliminates $y$ : $3 x = 12$ , giving $x = 4$ , then $y = 3$ .

Simultaneous equations by graph: the solution is where the two lines meet

Plot both equations as straight lines on the same axes. The single point where the two lines cross is the solution to both equations at once, since it's the only $(x, y)$ pair that lies on both lines. This is slower than substitution or elimination for exact answers, but shows clearly why there's usually exactly one solution.

Solving a quadratic equation by factorisation uses the zero product rule

If two factors multiply to give zero, at least one of them must be zero: if $P Q = 0$ then $P = 0$ or $Q = 0$ . To solve $x^{2} + 5 x + 6 = 0$ , factorise first to $(x + 2) (x + 3) = 0$ , then set each bracket to zero separately: $x = - 2$ or $x = - 3$ .

Worked examples

Solve the inequality:

4 x + 3 \leq 2 x + 15

Collect the $x$ terms on one side: subtract $2 x$ from both sides, giving $2 x + 3 \leq 15$ .
Subtract 3 from both sides: $2 x \leq 12$ .
Divide both sides by 2 (positive, so the sign stays the same): $x \leq 6$ .

Solve the simultaneous equations by substitution:

y = 2 x - 1

and

3 x + y = 14

The first equation already gives $y$ in terms of $x$ : $y = 2 x - 1$ .
Substitute this into the second equation: $3 x + (2 x - 1) = 14$ , so $5 x - 1 = 14$ .
Solve for $x$ : $5 x = 15$ , so $x = 3$ . Then $y = 2 (3) - 1 = 5$ .

Solve the simultaneous equations by elimination:

x + 2 y = 8

and

x - y = 2

Both equations have the same coefficient of $x$ (both $1 x$ ), so subtract the second equation from the first to eliminate $x$ : $(x + 2 y) - (x - y) = 8 - 2$ .
This simplifies to $3 y = 6$ , so $y = 2$ .
Substitute $y = 2$ into $x - y = 2$ : $x - 2 = 2$ , so $x = 4$ . Check in the first equation: $4 + 2 (2) = 8$ , which is correct.

Solve graphically:

y = x + 1

and

y = - x + 5

. At what point do the two lines cross?

The lines cross where both equations give the same $y$ for the same $x$ : $x + 1 = - x + 5$ .
Solve for $x$ : $2 x = 4$ , so $x = 2$ .
Substitute back into either equation: $y = 2 + 1 = 3$ . The lines cross at $(2, 3)$ , which is the solution.

Solve the quadratic equation by factorisation:

3 x^{2} - 5 x - 2 = 0

Look for factors of the form $(3 x + q) (x + s)$ with $q s = - 2$ and cross terms giving $- 5 x$ .
Try $(3 x + 1) (x - 2)$ : expanding gives $3 x^{2} - 6 x + x - 2 = 3 x^{2} - 5 x - 2$ , which matches.
Set each factor to zero: $3 x + 1 = 0$ gives $x = - \frac{1}{3}$ ; $x - 2 = 0$ gives $x = 2$ .

Mind map

Equations, inequalities, quadratics

Quadratic graphs

U-shape (a>0) or n-shape (a<0) · roots = x-axis crossings · turning point = min or max

Linear inequalities

solve like an equation · flip sign when x negative

Simultaneous equations

substitution: swap one letter in · elimination: add/subtract to remove a letter · graphical: solution is where the lines cross

Quadratic equations

factorise first · zero product rule

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