Quadratic Expressions and Equations (I)

Syllabus tag: KCSE | Mathematics | Form 2 | Topic 15 Quadratic Expressions and Equations (I)

Lesson objectives

By the end of this topic, you should be able to:

• Expand algebraic expressions that form quadratic equations.
• Derive and use the three quadratic identities.
• Factorise quadratic expressions.
• Solve quadratic equations by factorisation and form them from word problems.

Quadratic Expressions and Equations (I)

A quadratic expression contains a squared term as its highest power, such as x² + 7x + 12.

a) Expanding

Every term in the first bracket multiplies every term in the second, giving four products.

Some remember this as FOIL: First, Outer, Inner, Last.

Collect the like terms in the middle.

b) The three identities

These follow directly from expanding, and recognising them saves time.

(a + b)² is a² + 2ab + b². The middle term is twice the product, not just ab. Writing (a + b)² as a² + b² is the classic error.

(a − b)² gives the same with a negative middle term. The b² stays positive, since a negative squared is positive.

(a + b)(a − b) gives a² − b², the difference of two squares. The middle terms cancel exactly.

That last identity makes some arithmetic easy. 51 × 49 is (50+1)(50−1), which is 2500 − 1 = 2499.

c) Factorising a quadratic

To factorise x² + bx + c, find two numbers that multiply to c and add to b.

Take x² + 7x + 12. The numbers are 3 and 4, since 3 × 4 = 12 and 3 + 4 = 7.

So it factorises to (x + 3)(x + 4).

Watch the signs. If c is positive, both numbers share the sign of b. If c is negative, the numbers have opposite signs.

d) When a is not 1

For ax² + bx + c, find two numbers multiplying to ac and adding to b.

Split the middle term using them, then factorise in pairs.

Take 2x² + 7x + 3. Here ac is 6, and 6 and 1 work. Split it as 2x² + 6x + x + 3. That becomes 2x(x + 3) + 1(x + 3), which is (x + 3)(2x + 1).

e) Solving by factorisation

First get the equation into the form = 0. This step is essential, not cosmetic.

Factorise the left side.

Then apply the null factor law: if a product equals zero, at least one factor must be zero.

That law only works against zero. Given (x+3)(x+4) = 2, you cannot say x + 3 = 2. Many pairs of numbers multiply to 2.

Set each factor to zero and solve. A quadratic usually has two solutions.

f) Forming from word problems

Name the unknown and state what it represents.

Translate the information into an equation, expand, and rearrange into = 0 form.

Solve, then check both answers against the original situation. Negative lengths and fractional people must be rejected, even though they satisfy the equation.

g) Where this is used

Areas where a dimension is unknown. Projectile paths. Profit models. Any relationship where a quantity multiplies by itself.

Words to know

• Quadratic expression -- one in which the highest power of the unknown is 2.
• Identity -- an equation true for all values of the letters involved.
• Difference of two squares -- the expression a² − b², factorising as (a + b)(a − b).
• Root -- a solution of an equation; where the graph crosses the x-axis.
• Parabola -- the curved graph of a quadratic function.

:::checkpoint Check yourself

1. Expand (x + 5)(x + 2).
2. Expand (x − 3)² using the identity.
3. Factorise x² + 9x + 20.
4. Solve x² − 5x + 6 = 0. :::

Bridge to practice

The exercises begin with expanding and the three identities, move through factorising with a = 1 and a ≠ 1, and finish with solving and word problems. Before factorising to solve, always check that one side of the equation is zero.