STROUD Worked examples and exercises are in the text PROGRAMME F6 POLYNOMIAL EQUATIONS

STROUD Worked examples and exercises are in the text Polynomial equations Quadratic equations Solution of cubic equations having at least one linear factor Solution of quartic equations having at least two linear factors Programme F6: Polynomial equations

STROUD Worked examples and exercises are in the text Quadratic equations, ax 2 + bx + c = 0 Solution by factors Solution by completing the square Solution by formula Programme F6: Polynomial equations

STROUD Worked examples and exercises are in the text Quadratic equations, ax 2 + bx + c = 0 Solution by factors Programme F6: Polynomial equations Where simple factors exist the solution can be derived from those. For example: x 2 + 5x – 14 can be factorized as (x + 7)(x – 2) so if: x 2 + 5x – 14 = 0 then (x + 7)(x – 2) = 0 and so x = −7 or x = 2

STROUD Worked examples and exercises are in the text Quadratic equations, ax 2 + bx + c = 0 Solution by completing the square Programme F6: Polynomial equations Where simple factors do not exist the solution can be derived from completing the square. For example to solve x 2 – 6x – 4 = 0 it is noted that x 2 – 6x – 4 does not have simple factors so add 4 to both sides to give: x 2 – 6x = 4 Now, add the square of half the x-coefficient to both sides to give: x 2 – 6x + ( – 3) 2 = 4 + (–3) 2 that is x 2 – 6x + 9 = (x – 3) 2 = 13 Therefore x – 3 = ±√13 so x = or x = −0.606 to 3 dp

STROUD Worked examples and exercises are in the text Quadratic equations, ax 2 + bx + c = 0 Solution by formula Programme F6: Polynomial equations To solve ax 2 + bx + c = 0 use can be made of the formula: