Cambridge International AS and A Level Mathematics Pure Mathematics 1

Quadratic equations

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1

This equation, involving terms in x^2 and x as well as a constant term (i.e. a

number, in this case 6000), is an example of a quadratic equation. This is in

contrast to a linear equation. A linear equation in the variable x involves only

terms in x and constant terms.

It is usual to write a quadratic equation with the right-hand side equal to zero.

To solve it, you first factorise the left-hand side if possible, and this requires a

particular technique.

Quadratic factorisation

EXAMPLE 1.21 Factorise xa + xb + ya + yb.

SOLUTION

xa + xb + ya + yb = x (a + b) + y (a + b)

= (x + y)(a + b)

The expression is now in the form of two factors, (x + y) and (a + b), so this is the

answer.

You can see this result in terms of the area of the rectangle in figure 1.4. This can

be written as the product of its length (x + y) and its width (a + b), or as the

sum of the areas of the four smaller rectangles, xa, xb, ya and yb.

The same pattern is used for quadratic factorisation, but first you need to split

the middle term into two parts. This gives you four terms, which correspond to

the areas of the four regions in a diagram like figure 1.4.

Notice (a + b) is a

common factor.

x

a xa

b xb

ya

yb

y

Figure 1.4