Factorising Quadratic Equations (extract)

Solving Quadratics by Factorisation. (This has been pasted from Word, so there could be a loss of formatting and special symbols.)

Example (2). Solve the equation x^2 = 7x.

We might be tempted to divide both sides by x to give x = 7, but that does not give the complete result. True, 7^2 = 49, but x = 0 is also a solution, since 0^2 = 0. By dividing by x, we have `lost` a possible solution of the equation.

The correct procedure is to re-express the equation as x^2 - 7x = 0.

Example (2a): Factorise x^2 - 7x and hence solve x^2 - 7x = 0.

x^2 - 7x = 0 can be rewritten as x(x - 7) = 0.

x^2 - 7x is equal to 0 if and only if x = 0, or x - 7 = 0.

The solutions of the quadratic are thus x = 0 and x = 7.

They are also known as the roots of the equation.

Example (3): Factorise x^2 + 7x + 10 and hence find the roots of x^2 + 7x + 10 = 0.

The factorised expression would be of the form (x+A) (x+B) where A and B are numbers to be determined. On expanding the brackets, we have the expression x^2 + (A+B)x + AB.

The coefficient of x is A + B, given here as 7. The constant term is AB, given here as 10.

We must therefore find two numbers A and B that add to give 7 and multiply to give 10. Such a pair of numbers is A = 5, B = 2.

x^2 + 7x + 10 = (x+5) (x+2), and thus the roots of x2 + 7x + 10 = 0 are the solutions of x + 5 = 0, namely x = -5, and of x + 2 = 0, or x = -2.

In general, if the quadratic takes the form x^2 + bx + c = 0 where b and c are positive, then the factorised form (if it exists) will be (x + p) (x + q) where p and q are to be determined.

The above method can be used to try and factorise any expression of the form x^2 + bx + c = 0, namely where the coefficient of x^2 is unity. Although the coefficients in the last example were all positive, the method also works when b and/or c are negative.

Example (4): Factorise x^2 - 13x + 40 and hence solve x^2 - 13x + 40 = 0.

Find two numbers A and B whose sum is -13 and whose product is 40. Such a number pair is -5 and -8. So, x^2 - 13x + 40 = (x-5) (x-8), and hence the roots of x^2 - 13x + 40 = 0 are x = 5 and x = 8.

In general, if the quadratic takes the form x^2 - bx + c = 0 where b is negative but c is positive, then the factorised form (if it exists) will be (x - p) (x - q) where p and q are to be determined. This is because two negative numbers have a positive product but a negative sum.

Interestingly, another valid factorisation is x^2 - 13x + 40 = (5 - x) (8 - x) leading to the same solutions.

This is because, for any numbers a and b, ab = (-a) (-b).

Example (5): Factorise x^2 - 2x - 15 and hence solve x^2 - 2x - 8 = 7.

Note how the constant term in the expression x^2 - 2x - 15 is now negative. The factorised version will therefore be of the form (x + a) (x - b) with one positive solution and one negative one.

Two numbers whose sum is -2 and whose product is -15 are -5 and 3.

x^2 - 2x - 15 = (x-5) (x+3).

The equation x2 - 2x - 8 = 7 might look unrelated to the earlier part of the question at first.

What we must not do is to try and factorise the LHS as (x + 2) (x - 4) and solve x + 2 = 7 or x - 4 = 7. This is completely wrong ! The correct thing to do is to subtract 7 from both sides, to obtain the equation x^2 - 2x - 15 = 0, with the important zero on the RHS.

From the earlier factorisation, the solutions of x%2 - 2x - 15 = 0 are x = 5 and x = -3.