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KS4 Mathematics

KS4 Mathematics. A6 Quadratic equations. A6 Quadratic equations. Contents. A. A6.2 Completing the square. A. A6.1 Solving quadratic equations by factorization. A6.3 Using the quadratic formula. A. A6.4 Equations involving algebraic fractions. A.

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KS4 Mathematics

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  1. KS4 Mathematics A6 Quadratic equations

  2. A6 Quadratic equations Contents • A A6.2 Completing the square • A A6.1 Solving quadratic equations by factorization A6.3 Using the quadratic formula • A A6.4 Equations involving algebraic fractions • A A6.5 Problems leading to quadratic equations • A

  3. Find the width of the rectangle x + 4 x The length of a rectangle is 4 cm more than its width. The area of the rectangle is 45 cm2. Find the width of the rectangle. If we call the width of the rectangle x we can draw the following diagram: Using the information about the area of the rectangle we can write an equation: x(x + 4) = 45

  4. Find the width of the rectangle The solution to x(x + 4) = 45 will give us the width of the rectangle. In this example, it should be quite easy to spot that x = 5 is a possible solution to this equation, because 5 × 9 = 45 The width of the rectangle is therefore 5 cm. However, there is another value of x that will also solve the equation x(x + 4) = 45. This is because x(x + 4) = 45 is an example of a quadratic equation.

  5. Quadratic equations This is easier to spot if we multiply out the bracket, x(x + 4) = 45 x2 + 4x = 45 We usually arrange quadratic equations so that all the terms are on the left-hand side of the equals sign, leaving a 0 on the right-hand side. In this example we would have x2 + 4x – 45 = 0 The general form of a quadratic equation is ax2 + bx + c = 0 Where a, b and c are constants and a≠ 0.

  6. Quadratic equations We can solve the quadratic equation x2 + 4x – 45 = 0 in full by factorizing the expression on the left-hand side. This means that we can write the equation in the form (x + ….)(x + ….) = 0 We need to find two integers that add together to make 4 and multiply together to make –45. Because –45 is negative, one of the numbers must be positive and one must be negative. By considering the factors of 45 we find that the two numbers must be 9 and –5. We can therefore write x2 + 4x – 45 = 0 as (x + 9)(x – 5) = 0

  7. Quadratic equations or x – 5 = 0 and x = 5 When two numbers multiply together to make 0, one of the numbers must be 0, so if (x + 9)(x – 5) = 0 we can conclude that either x + 9 = 0 This gives us two solutions that solve the quadratic equation: x = – 9 In the context of finding the width of a rectangle we cannot allow a negative length, and so x = 5 is the only valid solution. Many other problems that lead to quadratic equations however, would require both solutions.

  8. Solving quadratic equations by factorization So x = 0 or x – 3 = 0 Solve the equation x2 = 3x by factorization. Start by rearranging the equation so that the terms are on the left-hand side, x2 – 3x = 0 Factorizing the left-hand side gives us x(x – 3)= 0 x = 3

  9. Solving quadratic equations by factorization or x – 4 = 0 Solve the equation x2 – 5x = –4 by factorization. Start by rearranging the equation so that the terms are on the left-hand side. x2 – 5x + 4 = 0 We need to find two integers that add together to make –5 and multiply together to make 4. Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4. Factorizing the left-hand side gives us (x – 1)(x – 4)= 0 x – 1 = 0 x = 1 x = 4

  10. Solving quadratic equations by factorization

  11. Demonstrating solutions using graphs

  12. A6 Quadratic equations Contents A6.1 Solving quadratic equations by factorization • A • A A6.2 Completing the square A6.3 Using the quadratic formula • A A6.4 Equations involving algebraic fractions • A A6.5 Problems leading to quadratic equations • A

  13. Perfect squares In general, x2 + 2ax + a2 = (x + a)2 and x2 – 2ax + a2 = (x – a)2 Some quadratic expressions can be written as perfect squares. For example, x2 + 2x + 1 = (x + 1)2 x2 – 2x + 1 = (x – 1)2 x2 + 4x + 4 = (x + 2)2 x2 – 4x + 4 = (x – 2)2 x2 + 6x + 9 = (x + 3)2 x2 – 6x + 9 = (x – 3)2 How could the quadratic expression x2 + 6x be made into a perfect square? We could add 9 to it.

  14. Completing the square We can write x2 + 6x = x2 + 6x + 9 – 9 b b x2 + bx = x + 2 – 2 In general, 2 2 Adding 9 to the expression x2 + 6x to make it into a perfect square is called completing the square. If we add 9 we then have to subtract 9 so that both sides are still equal. By writing x2 + 6x + 9 we have completed the square and so we can write this as x2 + 6x = (x + 3)2 – 9

  15. Completing the square Complete the square for x2 – 10x. Compare this expression to (x – 5)2 = x2 – 10x + 25 x2 – 10x = x2 – 10x + 25 – 25 = (x – 5)2 – 25 Complete the square for x2 – 3x. Compare this expression to (x – 1.5)2 = x2 – 3x + 2.25 x2 – 3x = x2 – 3x + 2.25 – 2.25 = (x + 1.5)2 – 2.25

  16. Expressions in the form x2 + bx

  17. Completing the square In general, b b x2 + bx + c = x + 2 – 2 + c 2 2 How can we complete the square for x2 + 8x + 9? Look at the coefficient of x. This is 8 so compare the expression to (x + 4)2 = x2 + 8x + 16. x2 + 8x + 9 = x2 + 8x + 16 – 16 + 9 = (x + 4)2 – 7

  18. Completing the square Complete the square for x2 + 12x – 5. Compare this expression to (x + 6)2 = x2 + 12x + 36 x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5 = (x2 + 6) – 41 Complete the square for x2 – 5x + 16 Compare this expression to (x – 2.5)2 = x2 – 5x + 6.25 x2 – 5x + 16 = x2 – 5x + 6.25 – 6.25 + 16 = (x2 – 2.5) + 9.75

  19. Expressions in the form x2 + bx + c

  20. Completing the square When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square. Complete the square for 2x2 + 8x + 3. Start by factorizing the first two terms by dividing by 2, 2x2 + 8x + 3 = 2(x2 + 4x) + 3 By completing the square, x2 + 4x = (x + 2)2 – 4 so, 2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3 = 2(x + 2)2 – 8 + 3 = 2(x + 2)2 – 5

  21. Completing the square Complete the square for 5 + 6x – 3x2. Start by factorizing the the terms containing x’s by –3. 5 + 6x – 3x2 = 5 – 3(–2x + x2) 5 + 6x – 3x2 = 5 – 3(x2 – 2x) By completing the square, x2 – 2x = (x – 1)2 – 1 so, 5 + 6x – 3x2= 5 – 3((x – 1)2 – 1) = 5 – 3(x – 1)2 + 3 = 8 – 3(x – 1)2

  22. Expressions in the form ax2 + bx + c

  23. Solving quadratics by completing the square (x – 2)2 – 7 = 0 simplify: (x – 2)2 = 7 add 7 to both sides: x – 2 = ±√7 square root both sides: or x = 2 – √7 Quadratic equations that cannot be solved by factorization can be solved by completing the square. For example, the quadratic equation, x2 – 4x – 3 = 0 can be solved by completing the square as follows, (x – 2)2 – 4 – 3 = 0 x = 2 + √7 x = 4.646 x = –0.646 (to 3 d.p.)

  24. Solving quadratics by completing the square (x + 4)2 – 11 = 0 simplify: (x + 4)2 = 11 add 11 to both sides: x + 4 = ±√11 square root both sides: or x = –4 – √11 Solve the equation x2 + 8x + 5 = 0 by completing the square. Write the answer to 3 decimal places. x2 + 8x + 5 = 0 Completing the square on the left-hand side, (x + 4)2 – 16 + 5 = 0 x = –4 + √11 x = –0.683 x = –7.317 (to 3 d.p.)

  25. Solving quadratics by completing the square Solve the equation 2x2– 4x + 1 = 0 by completing the square. Write the answer to 3 decimal places. We can complete the square for 2x2– 4x + 1 by first factorizing the terms containing x’s by the coefficient of x2, 2x2– 4x + 1 = 2(x2– 2x) + 1 Next complete the square for the expression in the bracket, = 2((x– 1)2 – 1) + 1 = 2(x– 1)2 – 2 + 1 = 2(x– 1)2– 1 We can now use this to solve the equation 2x2– 4x + 1 = 0.

  26. Solving quadratics by completing the square completing the square: 2(x – 1)2 – 1 = 0 2(x – 1)2 = 1 add 1 to both sides: 1 1 1 1 (x – 1)2 = divide both sides by 2: 2 2 2 2 square root both sides: x– 1 = ± or x = 1 + x = 1 – Solve the equation 2x2– 4x + 1 = 0 by completing the square. Write the answer to 3 decimal places. 2x2– 4x + 1 = 0 x = 1.707 x = 0.293 (to 3 d. p)

  27. A6 Quadratic equations Contents A6.1 Solving quadratic equations by factorization • A A6.2 Completing the square • A A6.3 Using the quadratic formula • A A6.4 Equations involving algebraic fractions • A A6.5 Problems leading to quadratic equations • A

  28. Using the quadratic formula –b± b2 – 4ac x = 2a Any quadratic equation of the form, ax2 + bx + c = 0 can be solved by substituting the values of a, b and c into the formula, This equation can be derived by completing the square on the general form of the quadratic equation.

  29. Using the quadratic formula –b± b2 – 4ac x = 2a 7± (–7)2 – (4 × 1 ×8) x = 2 × 1 7± 49 – 32 x = 2 7+ 17 7– 17 x = or x = 2 2 Use the quadratic formula to solve x2 – 7x + 8 = 0. 1x2– 7x+ 8 = 0 x = 5.562 x = 1.438 (to 3 d.p.)

  30. Using the quadratic formula –b± b2 – 4ac x = –5± 52 – (4 × 2 ×–1) 2a x = x = –5 ± 25 + 8 2 × 2 4 x = or x = –5 + 33 –5 – 33 4 4 Use the quadratic formula to solve 2x2 + 5x – 1 = 0. 2x2+ 5x– 1 = 0 x = 0.186 x = –2.686 (to 3 d.p.)

  31. Using the quadratic formula 12±  (–12)2 – (4 × 9 ×4) x = 2 × 9 –b± b2 – 4ac 12± 144 – 144 x = x = 2a 18 12± 0 2 x = 18 3 There is only one solution,x = Use the quadratic formula to solve 9x2 – 12x + 4 = 0. 9x2– 12x+ 4 = 0

  32. Using the quadratic formula –1± 12 – (4 × 1×3) x = 2 × 1 –b± b2 – 4ac –1± 1 – 12 x = x = 2a 2 –1± –11 x = 2 Use the quadratic formula to solve x2+ x + 3 = 0. 1x2+ 1x+ 3= 0 We cannot find –11 and so there are no solutions.

  33. Using b2 – 4ac –b± b2 – 4ac x = 2a From using the quadratic formula, we can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many solutions there are. • When b2 – 4ac is positive, there are two solutions. • When b2 – 4ac is equal to zero, there is one solution. • When b2 – 4ac is negative, there are no solutions.

  34. Using b2 – 4ac b2 – 4ac is positive b2 – 4ac is zero b2 – 4ac is negative y y y x x x We can demonstrate each of these possibilities using graphs. Remember, if we plot the graph of y = ax2 + bx + c the solutions to the equation ax2 + bx + c = 0 are given by the points where the graph crosses the x-axis. Two solutions One solution No solutions

  35. Using b2 – 4ac

  36. A6 Quadratic equations Contents A6.1 Solving quadratic equations by factorization • A A6.2 Completing the square • A A6.4 Equations involving algebraic fractions A6.3 Using the quadratic formula • A • A A6.5 Problems leading to quadratic equations • A

  37. Equations involving algebraic fractions 1 5 + = 2 x x + 4 x + 4 + 5x = 2x(x + 4) multiply byx(x+ 4): 6x + 4 = 2x2+ 8x expand brackets and simplify: 0 = 2x2+ 2x – 4 collect all terms on the r.h.s.: 0 = x2+ x – 2 divide by 2: 0 = (x + 2)(x – 1) factorize: x = –2 or x = 1 Some equations involving algebraic fractions rearrange to give quadratic equations. For example: The first step when solving equations involving fractions is to multiply through by the product of the denominators.

  38. Equations involving algebraic fractions 4 3 Solve – = 1 x + 2 x + 8 4x + 32 – 3x – 6 = x2 + 10x + 16 expand the brackets: x + 26 = x2 + 10x + 16 simplify: 0 = x2+ 9x – 10 collect all terms on the r.h.s.: 0 = (x + 10)(x – 1) factorize: x = –10 or x = 1 Start by multiplying through by x + 2 and x + 8 to remove the denominators, 4(x + 8) – 3(x + 2) = (x + 2)(x + 8)

  39. Equations involving algebraic fractions x 2 Solve, – = 3 4 – x x x2+ 2(4 – x) = 3x(4 – x) Multiply through by x(4 –x): x2+ 8 – 2x = 12x – 3x2 expand the brackets: 4x2– 14x + 8 = 0 collect terms on the l.h.s.: 2x2– 7x + 4 = 0 divide by 2: Equations involving algebraic fractions may also lead to quadratic equations that do not factorize. For example, This quadratic equation cannot be solved by factorization so we have to solve it using the quadratic formula.

  40. Equations involving algebraic fractions x = 7± 72 – (4 × 2 ×4) x = 2 × 2 7± 49 – 32 x = –b± b2 – 4ac 4 2a 7+ 17 7– 17 x = or x = 4 4 Using the quadratic formula to solve 2x2– 7x + 4 = 0 x = 2.781 x = 0.719 (to 3 d.p.)

  41. A6 Quadratic equations Contents A6.1 Solving quadratic equations by factorization • A A6.2 Completing the square • A A6.5 Problems leading to quadratic equations A6.3 Using the quadratic formula • A A6.4 Equations involving algebraic fractions • A • A

  42. Problems leading to quadratic equations distance Remember, time taken = average speed Some real-life problems can be solved using quadratic equations. For example,Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there. If her total journey time to work and back is 1 hour, what was her average speed on the way to work? Let Jenny’s average speed on the way to work be x.

  43. Problems leading to quadratic equations Jenny’s time taken to get to work = 24 24 x x 24 24 Jenny’s time taken to get home from work = x – 20 x – 20 Total time there and back = + = 1 Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour slower than on the way there. If her total journey time to work and back is 1 hour, what was her average speed on the way to work? Solving this equation will give us the value of x, Jenny’s average speed on the way to work.

  44. Problems leading to quadratic equations 24x – 480 + 24x = x2 – 20x expand the brackets: 24 48x – 480 = x2 – 20x x simplify: 24 0 = x2 – 68x + 480 collect terms on the r.h.s.: x – 20 0 = (x – 60)(x – 8) factorize: + = 1 Start by multiplying through by x(x – 20) to remove the fractions: 24(x – 20) + 24x = x(x – 20) We have two solutions x = 60 and x = 8. Which of these solutions is not possible in this situation?

  45. Problems leading to quadratic equations The only solution that makes sense is x = 60 miles per hour. If Jenny’s average speed on the way to work was 8 miles per hour her average speed on the way home would be –12 miles per hour, a negative number. We can therefore ignore the second solution. When practical problems lead to quadratic equations it is very often the case that only one of the solution will make sense in the context of the original problem. This is usually because many physical quantities, such as length, can only be positive.

  46. Problems leading to quadratic equations x + 1 x – 7 x The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle. Let’s start by drawing a diagram, We can use Pythagoras’ Theorem to write an equation in terms of x.

  47. Problems leading to quadratic equations x2 + x2 – 7x – 7x + 49 = x2 + x + x + 1 expand: 2x2 – 14x + 49 = x2 + 2x + 1 simplify: x2 – 16x + 48 = 0 collect on the l.h.s.: (x – 4)(x – 12)= 0 factorize: x = 4 or x = 12 The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle. x2 + (x – 7)2 = (x + 1)2 x2 + (x – 7)(x – 7) = (x + 1)(x + 1)

  48. Problems leading to quadratic equations 4 + 1 = 5 cm and 12 + 1 = 13 cm and The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle. If x = 4 then the lengths of the three sides are, 4 cm, 4 – 7 = –3 cm We cannot have a side of negative length and so x = 4 is not a valid solution. If x = 12 then the lengths of the three sides are, 12 cm, 12 – 7 = 5 cm So, the shorter sides are 12 cm and 5 cm and the hypotenuse is 13 cm.

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