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Solving by factorization 1. Solving by factorization 2. Solving by completing the square 3. Solving by using the quadratic equation formula 1 of 48 © Boardworks Ltd 2005 Solving by factorization 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 2 of 48 © Boardworks Ltd 2005 Solving by factorization 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 or x–5=0 This gives us two solutions that solve the quadratic equation: x=–9 3 of 48 and x=5 © Boardworks Ltd 2005 Solving by factorization – Example 1 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 So x=0 or x–3=0 x=3 4 of 48 © Boardworks Ltd 2005 Solving by factorization – Example 2 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 or x–4=0 x=1 x=4 5 of 48 © Boardworks Ltd 2005 Solving by factorization – Practice 1 6 of 48 © Boardworks Ltd 2005 Completing the square Adding 9 to the expression x2 + 6x to make it into a perfect square is called completing the square. We can write x2 + 6x = x2 + 6x + 9 – 9 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 In general, 7 of 48 x2 + bx = x + b 2 2 – b 2 2 © Boardworks Ltd 2005 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 8 of 48 © Boardworks Ltd 2005 Expressions in the form x2 + bx 9 of 48 © Boardworks Ltd 2005 Completing the square 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 In general, x2 10 of 48 + bx + c = x + b 2 2 – b 2 2 +c © Boardworks Ltd 2005 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 11 of 48 © Boardworks Ltd 2005 Expressions in the form x2 + bx + c 12 of 48 © Boardworks Ltd 2005 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 13 of 48 © Boardworks Ltd 2005 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 14 of 48 © Boardworks Ltd 2005 Expressions in the form ax2 + bx + c 15 of 48 © Boardworks Ltd 2005 Solving by completing the square – Example 1 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)2 – 7 = 0 simplify: add 7 to both sides: (x – 2)2 = 7 square root both sides: x = 2 + √7 x = 4.646 16 of 48 x – 2 = ±√7 or x = 2 – √7 x = –0.646 (to 3 d.p.) © Boardworks Ltd 2005 Solving by completing the square – Example 2 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)2 – 11 = 0 simplify: add 11 to both sides: (x + 4)2 = 11 square root both sides: x + 4 = ±√11 x = –4 + √11 x = –0.683 17 of 48 or x = –4 – √11 x = –7.317 (to 3 d.p.) © Boardworks Ltd 2005 Solving by completing the square – Example 3 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. 18 of 48 © Boardworks Ltd 2005 Solving by completing the square – Example 4 Solve the equation 2x2 – 4x + 1 = 0 by completing the square. Write the answer to 3 decimal places. 2x2 – 4x + 1 = 0 completing the square: 2(x – 1)2 – 1 = 0 2(x – 1)2 = 1 add 1 to both sides: (x – 1)2 = divide both sides by 2: x–1=± square root both sides: x=1+ x = 1.707 19 of 48 1 2 1 2 or 1 2 x=1– 1 2 x = 0.293 (to 3 d. p) © Boardworks Ltd 2005 Using the quadratic formula 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, –b ± b2 – 4ac x= 2a This equation can be derived by completing the square on the general form of the quadratic equation. 20 of 48 © Boardworks Ltd 2005 Using the quadratic formula – Example 1 Use the quadratic formula to solve x2 – 7x + 8 = 0. 1x2 – 7x + 8 = 0 –b ± b2 – 4ac x= 2a 7 ± (–7)2 – (4 × 1 × 8) x= 2×1 7 ± 49 – 32 x= 2 7 + 17 x= 2 x = 5.562 21 of 48 or 7 – 17 x= 2 x = 1.438 (to 3 d.p.) © Boardworks Ltd 2005 Using the quadratic formula – Example 2 Use the quadratic formula to solve 2x2 + 5x – 1 = 0. 2x2 + 5x – 1 = 0 –b ± b2 – 4ac x= 2a –5 ± 52 – (4 × 2 × –1) x= 2×2 –5 ± 25 + 8 x= 4 –5 + 33 x= 4 x = 0.186 22 of 48 or –5 – 33 x= 4 x = –2.686 (to 3 d.p.) © Boardworks Ltd 2005 Using the quadratic formula – Example 3 Use the quadratic formula to solve 9x2 – 12x + 4 = 0. 9x2 – 12x + 4 = 0 –b ± b2 – 4ac x= 2a 12 ± (–12)2 – (4 × 9 × 4) x= 2×9 12 ± 144 – 144 x= 18 12 ± 0 x= 18 2 There is only one solution, x = 3 23 of 48 © Boardworks Ltd 2005 Using the quadratic formula – Example 4 Use the quadratic formula to solve x2 + x + 3 = 0. 1x2 + 1x + 3 = 0 –b ± b2 – 4ac x= 2a –1 ± 12 – (4 × 1 × 3) x= 2×1 –1 ± 1 – 12 x= 2 –1 ± –11 x= 2 We cannot find –11 and so there are no solutions. 24 of 48 © Boardworks Ltd 2005