06112020 Solving quadratics MATHSWATCH CLIP 191 209 GRADE

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 L. O. To be able to § Solve a quadratic by factorising § Solve a quadratic by using the formula § Solve a quadratic by completing the square Key Words quadratic, factor, coefficient, discriminant, significant figures, surd form, formula Starter

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 CONTENTS – Click to go to… BY FACTORISING BY USING THE FORMULA BY COMPLETING THE SQUARE PROBLEM SOLVING WITH QUADRATICS THE DISCRIMINANT

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 How to solve a quadratic Non calculator paper… Calculator paper… First try to FACTORISE Does it say give your answers to 3 s. f. ? If you cannot factorise, use the FORMULA or COMPLETE THE SQUARE and leave the answers in surd form NO See if you can FACTORISE first, as that is usually quickest. But remember you can always use your calculator to apply the FORMULA YES Use the FORMULA, put it in your calculator and round

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 BY FACTORISING

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – Spot the errors. How many errors can you find? If you think you found them all, then CORRECT THEM!

11/6/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

11/6/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Discussion - Below is an exam question. Discuss each bullet point in pairs and be ready to give feedback § What is the difference between the algebra given to you in parts a) and b)? § What type of answer should you get for part a)? What about part b)? How do you know? § What is the meaning of the word “hence” in part b)?

06/11/2020 Discussion Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics FACTORISE SOLVE BOTH equations MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics FACTORISE SOLVE BOTH equations MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics FACTORISE SOLVE BOTH equations MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 What special case of quadratic factorising is this…? FACTORISE SOLVE BOTH equations

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics Task – SOLVE by factorising (a) 5 x 2 = 10 x x = 0 or 2 (b) 4 x 2 - 6 x = 0 or 1½ (c) x 2 + 3 x + 2= 0 x = -1 or -2 (d) 4 x 2 - 9= 0 x = +/- 1½ (e) 2 t 2 - 9 t - 5= 0 x = -½ or 5 (f) 16 x 2 = 100 x = +/- 2½ (g) 5 x 2 = - 4 x 2 + 1 x = +/- 1/3 (h) 2(x 2 + 5 x) = -12 x = -2 or -3 (i) 12 x 2 - 13 x + 3= 0 x = ¾ or 1/3 MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics Task a) b) c) d) e) x² - 6 x + 8 x² + 7 x + 6 x² + x - 12 x² - 6 x + 9 x² - 5 x - 36 Hence solve the equations above MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – SOLVE by factorising x² + 7 x + 10 = 0 x² + 11 x + 30 = 0 x² + 4 x + 3 = 0 x² + 10 x + 21 = 0 x² + 7 x + 12 = 0 x² + 2 x + 1 = 0 x² + 5 x + 6 = 0 x² + x – 30 = 0 x² – 12 x + 35 = 0 x² – 4 = 0 x² + 5 x – 6 = 0 x² – 12 x + 35 = 0 x² – 2 x – 15 = 0 x² – 5 x – 36 = 0 x² – 64 = 0 4 x² - 121 = 0 x² – 2. 5 x + 2. 5 = 0 x² – x = 6 x² – 10 = 3 x x² – 36 = 2 x - 1

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – SOLUTIONS x = -5 or -2 x = -6 or -5 x = -3 or -1 x = -7 or -3 x = -4 or -3 x = -1 or -1 x = -3 or -2 x = -6 or 5 x = 7 or 5 x = 2 or -2 x = -6 or 1 x = 7 or 5 x = 5 or -3 x = 9 or -4 x = 8 or -8 x = 5. 5 or -5. 5 x = 1 or 1. 5 x = 3 or -2 x = 5 or -2 x = 7 or 5

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – SOLVE by factorising a) b) c) d) e) x² + 9 x + 18 = 0 x² + 9 x + 14 = 0 x² + 11 x + 24 = 0 x² + 10 x + 24 = 0 x² + 7 x + 12 = 0 a) b) c) d) e) x² - 8 x + 7 = 0 x² - 12 x + 11 = 0 x² - 5 x + 4 = 0 x² - 5 x + 6 = 0 x² - 9 x + 8 = 0 a) b) c) d) e) x² + 3 x – 18 = 0 x² - 3 x – 18 = 0 x² + 3 x – 28 = 0 x² - x – 12 = 0 x² + 2 x – 24 = 0

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 BY USING THE FORMULA

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 The quadratic formula MEMORISE THIS!!!!!

Solving quadratics 06/11/2020 Write out the formula SUBSTITUTE CALCULATE FULL answers ROUNDED answers MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

Solving quadratics 06/11/2020 Write out the formula SUBSTITUTE CALCULATE FULL answers ROUNDED answers MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Write out the formula SUBSTITUTE CALCULATE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Write out the formula SUBSTITUTE CALCULATE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Write out the formula SUBSTITUTE CALCULATE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics Task – On your mini whiteboard… SOLVE the following to 2 d. p. MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – SOLVE, giving your answers to 4 d. p. 1) 5) f=1. 8358 OR f=-0. 4358 V=-0. 2042 OR v=-9. 7958 2) 6) h=0. 1836 OR h=1. 8170 h=0. 1623 OR h=-6. 1623 7) 3) q=0. 3117 OR q=-4. 8117 x=0. 7320 OR x=-2. 7320 8) 4) f=-0. 5311 OR f=7. 5311 y=-1. 9361 OR y=0. 9361

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – SOLVE, giving your answers to 2 d. p. x = 6. 32 or – 0. 32 x = 3. 65 or – 1. 65 x = - 0. 27 or – 3. 73 x = - 0. 81 or – 6. 19 x = 1. 08 or – 11. 08 x = 2. 68 or – 1. 14 x = 3. 83 or – 1. 83 x = - 0. 23 or – 1. 43 x = -0. 68 or – 7. 32 x = 8. 27 or 0. 73 x = 2. 37 or – 3. 37 x = 1. 10 or – 9. 10 x = 0. 22 or 2. 28 x = 0. 65 or – 1. 15

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 BY COMPLETING THE SQUARE

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Perfect squares Some quadratic expressions can be written as perfect squares. For example, x 2 + 2 x + 1 = (x + 1)2 x 2 – 2 x + 1 = (x – 1)2 x 2 + 4 x + 4 = (x + 2)2 x 2 – 4 x + 4 = (x – 2)2 x 2 + 6 x + 9 = (x + 3)2 x 2 – 6 x + 9 = (x – 3)2 In general, x 2 + 2 ax + a 2 = (x + a)2 and x 2 – 2 ax + a 2 = (x – a)2 How could the quadratic expression x 2 + 6 x be made into a perfect square? We could add 9 to it.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Adding 9 to the expression x 2 + 6 x to make it into a perfect square is called completing the square. We can write x 2 + 6 x = x 2 + 6 x + 9 – 9 If we add 9 we then have to subtract 9 so that both sides are still equal. By writing x 2 + 6 x + 9 we have completed the square and so we can write this as x 2 + 6 x = (x + 3)2 – 9 In general, 2 2 2 b b x + bx = x + – 2 2

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Examples Complete the square for x 2 – 10 x. Compare this expression to (x – 5)2 = x 2 – 10 x + 25 x 2 – 10 x = x 2 – 10 x + 25 – 25 = (x – 5)2 – 25 Complete the square for x 2 – 3 x. Compare this expression to (x – 1. 5)2 = x 2 – 3 x + 2. 25 x 2 – 3 x = x 2 – 3 x + 2. 25 – 2. 25 = (x + 1. 5)2 – 2. 25

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – Write in the form (x + p)2 + q x² + 4 x x² + 10 x x² + 12 x x² + 24 x x² + 22 x x² – 14 x x² – 2 x x² – 16 x x² – 18 x x² – 3 x x² + 20 x + 96 x² + 16 x + 63 x² + 22 x + 132 x² + 36 x + 298 x² + 18 x + 76 y² – 14 y + 48 y² – 20 y + 98 y² – 2 y – 5 y² + 41 y + 420 y² + 56 y + 663

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – SOLUTIONS (x + 2)² – 4 (x + 5)² – 25 (x + 6)² – 36 (x + 12)² – 144 (x + 11)² – 121 (x – 1)² – 1 (x – 8)² – 64 (x – 9)² – 81 (x – 1. 5)² – 2. 25 (x + 10)² – 4 (x + 8)² – 1 (x + 11)² + 10 (x + 18)² – 26 (x + 9)² – 5 (y – 7)² – 1 (y – 10)² – 2 (y – 1)² – 6 (x + 20. 5)² +0. 25 (x + 28)² – 121

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Completing the square How can we complete the square for x 2 + 8 x + 9? Look at the coefficient of x. This is 8 so compare the expression to (x + 4)2 = x 2 + 8 x + 16 x 2 + 8 x + 9 = x 2 + 8 x + 16 – 16 + 9 In general, = (x + 4)2 – 7 2 2 2 b b x + bx + c = x + – + c 2 2

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Examples Complete the square for x 2 + 12 x – 5. Compare this expression to (x + 6)2 = x 2 + 12 x + 36 x 2 + 12 x – 5 = x 2 + 12 x + 36 – 5 = (x 2 + 6) – 41 Complete the square for x 2 – 5 x + 16 Compare this expression to (x – 2. 5)2 = x 2 – 5 x + 6. 25 x 2 – 5 x + 16 = x 2 – 5 x + 6. 25 – 6. 25 + 16 = (x 2 – 2. 5) + 9. 75

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – Write each of the following in “completed square” form 1. x 2 + 8 x + 10 2. x 2 - 6 x + 1 3. x 2 - 2 x + 2 4. x 2 + 10 x + 30 5. x 2 + 6 x - 5 6. x 2 - 12 x - 3 7. x 2 - x + 4 8. x 2 - 3 x + 7. 25 = (x + 4)2 - 6 = (x - 3)2 - 8 = (x - 1)2 + 1 = (x + 5)2 + 5 = (x + 3)2 - 14 = (x - 6)2 - 39 = (x - ½)2 + 3. 75 = (x - 1. 5)2 + 5

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 2 2 2 b b x + bx + c = x + – + c 2 2

06/11/2020 Solving quadratics Re-write using this method SUBSTITUTE SOLVE FULL answers ROUNDED answers MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Re-write using this method SUBSTITUTE SOLVE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Re-write using this method SUBSTITUTE SOLVE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Re-write using this method SUBSTITUTE SOLVE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Re-write using this method SUBSTITUTE SOLVE FULL answers ROUNDED answers Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Task – Solve by completing the square 1. x 2 + 6 x + 1 = 0 x = - 0. 17 and - 5. 83 2. x 2 - 8 x + 3 = 0 x = 7. 61 and 0. 39 3. x 2 - 4 x - 7 = 3 x = 5. 74 and - 1. 74 4. x 2 + 3 x + 1= 0 x = - 0. 38 and - 2. 62

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 When the coefficient of x 2 is not 1, quadratic equations in the form ax 2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square. Complete the square for 2 x 2 + 8 x + 3. Start by factorizing the first two terms by dividing by 2, 2 x 2 + 8 x + 3 = 2(x 2 + 4 x) + 3 By completing the square, x 2 + 4 x = (x + 2)2 – 4 so, 2 x 2 + 8 x + 3 = 2((x + 2)2 – 4) + 3 = 2(x + 2)2 – 8 + 3 = 2(x + 2)2 – 5

06/11/2020 Solving quadratics Complete the square for 5 + 6 x – 3 x 2. Start by factorising the terms containing x’s by – 3. 5 + 6 x – 3 x 2 = 5 – 3(– 2 x + x 2) 5 + 6 x – 3 x 2 = 5 – 3(x 2 – 2 x) By completing the square, x 2 – 2 x = (x – 1)2 – 1 so, 5 + 6 x – 3 x 2 = 5 – 3((x – 1)2 – 1) = 5 – 3(x – 1)2 + 3 = 8 – 3(x – 1)2 MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

11/6/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics Proof of the quadratic formula by completing the square… by a Complete the square MATHSWATCH CLIP 191, 209 GRADE 7, 8/9

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 PROBLEM SOLVING WITH QUADRATICS

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Example The distance of an object (d) after a certain amount of time (t), which is accelerating at a rate of ‘A’ metres per second (ms-2) and which is already travelling at a speed of u metres per second (ms-1) at time t=0 (i. e. when you start recording) is given by the equation: d = ut + ½ At 2 Calculate how long it takes a car which is already travelling at 80 ms-1 and is accelerating at 8 ms-2 to travel a further 500 metres. So d = 500, u = 80, A = 8

06/11/2020 Solving quadratics Example MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 So d = 500, u = 80, A = 8 Set up the equation: 500 = 80 t + ½ 8 t 2 From: d = ut + ½ At 2 Or: 500 = 80 t + 4 t 2 Rearrange this to give a quadratic equation with 0 on one side of the equals sign 0 = 4 t 2 + 80 t – 500 You can’t have – 25 seconds! or 4 t 2 + 80 t – 500 = 0 Factorising gives: (2 t + 50)(2 t – 10) = 0 So: 2 t + 50 = 0 … or: 2 t – 10 = 0 Therefore: 2 t = -50 Or: 2 t = +10 t = -25 seconds t = +5 seconds This is the answer!

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Example The unit cost (c) of production of a very exclusive and expensive product is given by the following formula: c = 75 q + 25 q 2 …where q is the quantity produced. The company has had a restriction placed (by the Office of Fair Trading) on the selling price of their product. This is going to restrict the cost per unit to £ 250, in order to maintain profits. So c = 250. Calculate how many units the firm can produce.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Example Set up the equation: From: c = 75 q + 25 q 2 250 = 75 q + 25 q 2 Rearrange this to give a quadratic equation with 0 on one side of the equals sign You can’t have – 5 units! 0 = 25 q 2 + 75 q – 250 or 25 q 2 + 75 q – 250 = 0 Factorising gives: (5 q + 25)(5 q - 10) = 0 So: 5 q + 25 = 0 … or: 5 q - 10 = 0 Therefore: 5 q = - 25 Or: 5 q = +10 q = - 5 units q = + 2 units This is the answer!

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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? Remember, time taken = distance average speed Let Jenny’s average speed on the way to work be x.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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? Jenny’s time taken to get to work = 24 x Jenny’s time taken to get home from work = 24 x – 20 24 24 Total time there and back = + = 1 x – 20 x Solving this equation will give us the value of x, Jenny’s average speed on the way to work.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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? 24 24 + = 1 x – 20 x Start by multiplying through by x(x – 20) to remove the fractions: 24(x – 20) + 24 x = x(x – 20) 24 x – 480 + 24 x = x 2 – 20 x expand the brackets: simplify: 48 x – 480 = x 2 – 20 x 0 = x 2 – 68 x + 480 We have two solutions x = 60 and x = 8. 0 = (x – 60)(x – 8) factorise: collect terms on the r. h. s. : Which of these solutions is not possible in this situation?

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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, x + 1 x – 7 x We can use Pythagoras’ Theorem to write an equation in terms of x.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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. x 2 + (x – 7)2 = (x + 1)2 x 2 + (x – 7) = (x + 1) expand: x 2 + x 2 – 7 x + 49 = x 2 + x + 1 simplify: collect on the l. h. s. : factorize: 2 x 2 – 14 x + 49 = x 2 + 2 x + 1 x 2 – 16 x + 48 = 0 (x – 4)(x – 12) = 0 x = 4 or x = 12

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 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 and 4 + 1 = 5 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 and 12 + 1 = 13 cm So, the shorter sides are 12 cm and 5 cm and the hypotenuse is 13 cm.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 THE DISCRIMINANT

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Use the quadratic formula to solve 9 x 2 – 12 x + 4 = 0 x = –b ± b 2 – 4 ac 2 a 12 ± (– 12)2 – (4 × 9 × 4) x = 2 × 9 12 ± 144 – 144 x = 18 12 ± 0 x = 18 There is only one solution, x = 2 3

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 Use the quadratic formula to solve x 2 + x + 3 = 0. 1 x 2 + 1 x + 3 = 0 x = –b ± b 2 – 4 ac 2 a – 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.

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 The discriminant From using the quadratic formula, x = –b ± b 2 – 4 ac 2 a we can see that we can use the expression under the square root sign, b 2 – 4 ac, to decide how many solutions there are. § When b 2 – 4 ac is positive, there are two solutions. § When b 2 – 4 ac is equal to zero, there is one solution. § When b 2 – 4 ac is negative, there are no solutions. b 2 – 4 ac is called the DISCRIMINANT

06/11/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9 We can demonstrate each of these possibilities using graphs. Remember, if we plot the graph of y = ax 2 + bx + c the solutions to the equation ax 2 + bx + c = 0 are given by the points where the graph crosses the x-axis. b 2 – 4 ac is positive b 2 – 4 ac is zero y y x Two solutions b 2 – 4 ac is negative y x One solution x No solutions

11/6/2020 Solving quadratics MATHSWATCH CLIP 191, 209 GRADE 7, 8/9