SOLVING QUADRATIC EQUATIONS BY FACTORISING

Transcription

1 SOLVING QUADRATIC EQUATIONS BY FACTORISING SOLVING EQUATIONS Solving any equation means finding the values of x which satisfy the equation. Example Solve x 6 = 0 x = 3 satisfies this equation. SOLVING QUADRATIC EQUATIONS The solution to a quadratic equation means finding the value or values of x which satisfies the equation EXAMPLE Solve x² 3x + = 0 SOLUTION x = 1 satisfies the equation since 1² 3(1) + = 0 x = also satisfies the equation since ² 3() + = 0 METHOD1 - GUESSING In some simple cases, the solutions can be found by trial and error guessing EXAMPLE Solve x² + x 6 = 0 by guessing! METHOD - FACTORISING In the first example, the solutions to x² 3x + = 0 were x = 1 and x = x² 3x + can be factorised as (x - 1)(x - ) Notice the connection between the solutions and the factors. The equation x² 3x + = 0 can be written as (x - 1)(x - ) = 0 Since zero multiplied by anything is zero, we just need to find a value of x which makes the first bracket zero or makes the second bracket zero In this case x - 1 = 0 gives x = 1 and quadraticequations RSH 6-Mar-10 Page 1 of 14

2 x - = 0 gives x = The solutions to the equation x² 3x + = 0 are x = 1 and x = EXAMPLE Solve x + x - 6 = 0 Factorise x + x - 6 = (x + 3)(x - ) x + x - 6 = 0 is the same as (x + 3)(x - ) = 0 x + 3 = 0 gives x = -3 x - = 0 gives x = The solution is x = -3 and x = EXAMPLE Solve x - 5x - 3 = 0 Factorise x - 5x - 3 = (x + 1)(x - 3) x - 5x - 3 = 0 is the same as (x + 1)(x - 3) = 0 x + 1 = 0 gives x = x - 3 = 0 gives x = 3 The solution is x = -0.5 and x =3 quadraticequations RSH 6-Mar-10 Page of 14

3 EXERCISE 1 Solve by factorising 1. x² + 3x + = 0. x² + 6x + 5 = 0 3. x² + 7x + 1 = 0 4. x² + 8x + 15 = 0 5. x² + 1x + 0 = 0 6. x² + 8x + 7 = 0 7. x² + 8x + 1 = 0 8. x² + 13x + 1 = 0 9. x² + x + 40 = x² + 11x + 30 = x² 9x + 8 = 0 1. x² 17x + 30 = x² 5x + 6 = x² 18x + 3 = x² 16x + 63 = x² x 6 = x² + x 0 = x² x 1 = x² +x 15 = 0 0. x² 8x 0 = 0 1. x² + 9x + 14 = 0. x² 10x + 1 = 0 3. x² 10x + 5 = 0 4. x² + x 80 = 0 5. x² 11x + 30 = 0 6. x² 1x 8 = 0 7. x² + 11x + 4 = 0 8. x² 11x 4 = 0 9. x² 18x + 3 = x² + 7x 60 = 0 quadraticequations RSH 6-Mar-10 Page 3 of 14

4 EXERCISE Solve by factorising 1. x² + 3x + 1 = 0. 3x² 5x + = x² + 7x + 3 = 0 4. x² 7x +3 = x² + 13x + 4 = x² 8x + 4 = 0 7. x² + 9x + 4 = x² 17x + 6 = 0 9. x² + 11x + 1 = x² 9x + 4 = x² 3x = x² + x 4 = 0 quadraticequations RSH 6-Mar-10 Page 4 of 14

5 USING THE FORMULA When a quadratic equation cannot easily be solved by factorising, can be solved by using the formula Examples 1. Solve x² +10x + 1 = 0 Compare with ax² + bx + c = 0 a = 1, b = 10, c = 1 b b 4ac x a x 0.1 or x 9.9 x = 01 or x = 99 ( d.p) quadraticequations RSH 6-Mar-10 Page 5 of 14

6 . Solve x² + 10x 3 = 0 a =, b = 10, c = 3 b b 4ac x a ( 3) x or x x = 0.8 or 5.8 ( d.p.) 3. Solve 3x² = 9x First re-arrange this equation into the form ax² + bx + c = 0 3x² 9x + = 0 a = 3, b = 9, c = b b 4ac x a ( 9) ( 9) x or x x =.76 or 0.4 ( d.p.) quadraticequations RSH 6-Mar-10 Page 6 of 14

7 EXERCISE 3 Solve the following equations. Give your answers correct to 1 decimal place. 1. x² + 4x 6 = 0. x² + x 4 = x² + x 4 = 0 4. x² + x + 4 = 0 5. x² + 3x 1 = 0 6. x² x 7 = x² = 5x 8. x² = 1 4x 9. x² 5x + =0 10. x² + 3x 1 = 0 Solve the following equations. Give your answers correct to decimal places x² 4x = x² + x 1 = x² x = 14. x² = 4x x x² = x² + 7x = x² 5 = x(x + ) = (x )(x + 1) = 3 0. x² + 9x + 1 = 0 quadraticequations RSH 6-Mar-10 Page 7 of 14

8 GRAPHICAL SOLUTION One graph can be used to solve many equations Example Draw the graph of y = x² for values of x between 4 and 4. Use this graph to solve the following equations a) x² = 4 b) x² = 0 c) x² = 6 d) x² 9 = 0 e) x² 4 = 0 Step 1: Table of Values for y = x² X Y Step : Draw the axes Plot the points Join with a smooth curve quadraticequations RSH 6-Mar-10 Page 8 of 14

9 a) x² = 4 Draw the horizontal line y = 4 Read off the x - values x² = 4 when x = and x = b) x² = 0 The horizontal line y = 0 is the x- axis. x² = 0 when x = 0 c) x² = 6 Draw the horizontal line y = 6 Read off the x - values x² = 6 when x =.45 and x =.45 quadraticequations RSH 6-Mar-10 Page 9 of 14

10 d) x² 9 = 0 Rearrange to x² = 9 Draw the horizontal line y = 9 Read off the x values x² 9 = 0 when x = 3 and x = 3 e) x² 4 = 0 Rearrange to x² = 4 And again to x² = Draw the horizontal line y = Read off the x values x² 4 = 0 when x = 1.4 and x = 1.4 quadraticequations RSH 6-Mar-10 Page 10 of 14

11 Example Draw the graph of y = x² x 15 in the range 5 x 6 Use your graph to solve the equations a) x² x 15 = 0 b) x² x 15 = 10 c) x² x = 14 d) x² x 6 = 0 Step 1: Table of Values x y Step : Draw the axes Plot the points Join with a smooth curve a) x² x 15 = 0 The line y = 0 is the x axis y = x² x 15 cuts the x-axis when x = 3 and 5 x² x 15 = 0 when x = 3 and 5 b) x² x 15 = 10 Draw the line y = 10 Read off the x-values x² x 15 = 10 when x = 4.1 and 6.1 quadraticequations RSH 6-Mar-10 Page 11 of 14

12 *c) x² x = 14 Rearrange x² x = 14 so that the left hand side is x² x 15 x² x 15 = x² x 15 = 1 Draw the horizontal line y = 1 Read off the x values x² x 15 = 1 when x = -.9 and 4.9 *d) x² x 6 = 0 Rearrange x² x 6 = 0 so that the left hand side is x² x 15 x² x 6 = 0 x² x = 6 x² x 15 = 6 15 x² x 15 = 9 Draw the horizontal line y = 9 Read off the x values x² x 6 = 0 when x = -1.6 and 3.6 quadraticequations RSH 6-Mar-10 Page 1 of 14

13 PROBLEM SOLVING Quadratic equations are often used to solve practical problems. The method is o read the question carefully o form a quadratic equation o solve this quadratic equation either by factorising or by using the quadratic formula. Example 1 The sum of the squares of two consecutive whole numbers is 1. Find the whole numbers. Let one of the numbers be n The next number is (n + 1) The equation is Expanding Simplify The square of n is n² The square of (n + 1) is (n + 1)² (n + 1)² + n² = 1 n² + n n² = 1 n² + n + 1 = 1 n² + n 0 = 0 n² + n 110 = 0 (n + 11)(n 10) = 0 Solve n = 11 or n = 10 If n = 11, the next number is 10 If n = 10, the next number is 11 There are two answers 11 and 10 or 10 and 11 In some problems, one of the solutions may not be an acceptable answer. quadraticequations RSH 6-Mar-10 Page 13 of 14

14 Example A rectangle is 1m longer than it is wide. If the area of the rectangle is 0m², find its dimensions. Let the width be x, then the length will be (x + 1) Area of rectangle = x(x + 1) = x² + x The equation is x² + x = 0 x² + x 0 = 0 (x + 5)(x 4) = 0 x = 5 or x = 4 Clearly lengths cannot be negative, so x 5 Width = 4 m and the length = 5 m GCSE Question ABC is a triangle with a right angle at B. The length of its three sides are: x cm, (x + ) cm and (x + 4) cm. a) Use Pythagoras Theorem to show that x satisfies the equation x² 4x 1 = 0 [3] (x + 4)² = (x + )² + x² x² + 8x + 16 = x² + 4x x² x² 4x 1 = 0 b) Solve the equation x² 4x 1 = 0 [] (x 6)(x + ) = 0 x = 6 or x = c) Use your solutions in (b) to write down the lengths of the sides of the triangle. [1] Lengths cannot be negative so x x = 6 cm AB = 6 cm, BC = 8 cm, AC = 10 cm quadraticequations RSH 6-Mar-10 Page 14 of 14