AS Mathematics Algebra – Quadratic equations

Objectives Be confident in the use of brackets Be able to factorise quadratic expressions Be able to solve quadratic equations by factorisation

Factorising quadratic expressions The general form of a quadratic expression is: where a, b, c are constants and a ≠ 0 ax 2 + bx +c

(i) x 2 + bx + c (a=1) You should be able to spot most of these by inspection

Example 1 Factorisex 2 + 5x + 6 All positive, so (x + )(x + ) coefficent of x 2 is 1, so (x )(x ) Need 2 numbers with a product of , 6 2, 3.. that sum to +5 ____ (x + 2)(x + 3) Check your answer by expanding the brackets.

Example 2 Factorise x 2 - 5x + 6 (x )(x ) (x - )(x - ) (x - 2)(x - 3) coefficent of x 2 is 1, so (x )(x ) need 2 numbers with a +ve product but a –ve sum… both numbers must be -ve. Need 2 numbers with a product of , -6 -2, -3.. that sum to -5 _____ Check your answer by expanding the brackets.

(ii) ax 2 + bx + c, a≠1,0 The inspection method is fine for quadratics where a=1, but for complex quadratics a more rigorous method is required.

Example 3 Factorise 3x 2 + 6x x x - 6x -24 3x(x + 4) - 6(x + 4) Find two numbers so that sum = b AND product = ac sum = +6 product = 3(-24) = -72 } split the middle term factorise by grouping (x + 4)(3x – 6) Check your answer by expanding the brackets.

Example 4 Factorise 6y y y y + 9y y(3y + 5) + 3(3y + 5) Find two numbers so that sum = b AND product = ac sum = +19 product = 6(15) = +90 } split the middle term factorise by grouping (3y + 5)(2y + 3) Check your answer by expanding the brackets.

Example 5 Factorise 8d d d d - 6d d(2d + 4) - 3(2d + 4) sum = +10 product = 8(-12) = -96 } (2d + 4)(4d - 3) Check your answer by expanding the brackets.

Solving quadratic equations 4 methods: (iv) graphically (iii) completing the square (ii) quadratic formula (i) factorisation We start with solving quadratic equations by factorisation

Solving quadratic equations by factorisation Example 1 Solve the equation 3x 2 – 1 = 2x 3x 2 - 3x + x – 1 = 0 3x(x - 1) + 1(x - 1) = 0 sum = -2 product = 3(-1) = -3 } (x - 1)(3x + 1) = 0 Have you checked your answer? Rearrange to get... = 0 3x 2 – 2x – 1 = 0 Either x - 1 = 0 or 3x + 1 = 0 x = 1 or x = -⅓ The roots of this equation are x = 1 and x = - 1 / 3 ___________

Example 2 Solve the equation (3x + 1)(2x-1) – (x + 2) 2 = 5 5x 2 - 5x – 10 = 0 5[x 2 – x – 2] = 0 sum = -1 product = 1(-2) = -2 } (x + 1)(x – 2) = 0 Have you checked your answer? Expand and rearrange 6x 2 – 3x + 2x – 1 – [x 2 + 4x + 4] = 5 Either x + 1 = 0 or x - 2 = 0 x = -1 or x = 2 ______ The roots of this equation are x = -1 and x = 2