GCSE Completing The Square Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com @DrFrostMaths Last modified: 5th March 2018

Starter Expand the following brackets. Do you notice any relationship between the original expression and coefficient of 𝑥 in the expanded expression in each case? Key Term: The coefficient of a term is the number on front of it. So the coefficient of 3𝑥 is 3 and the coefficient of 5 𝑥 2 is 5. 𝑥+1 2 = 𝑥 2 +𝟐𝑥+1 𝑥+3 2 = 𝑥 2 +𝟔𝑥+9 𝑥−4 2 = 𝑥 2 −𝟖𝑥+16 𝑥+𝑎 2 = 𝑥 2 +𝟐𝒂𝑥+ 𝑎 2 ? ? ? ? ? Relationship The coefficient of 𝒙 is double the number after the 𝒙 in the bracket.

Therefore… ? ? ? ? ? ? ? 𝑥 2 +10𝑥 → 𝑥+5 2 −25 𝑥 2 +8𝑥 → 𝑥+4 2 −16 Therefore, it seems as if we can halve the coefficient of 𝒙 to get the missing number in 𝑥+__ 2 Put the following in the form 𝑥+𝑎 2 +𝑏 Consider the expansion of 𝑥+5 2 : 𝑥+5 2 = 𝑥 2 +10𝑥+25 We only want the “ 𝑥 2 +10𝑥” so we ‘throw away’ the 25 by subtracting it. 𝑥 2 +10𝑥 → 𝑥+5 2 −25 ? 𝑥 2 +8𝑥 → 𝑥+4 2 −16 ? Because the form required was “ 𝑥+𝑎 2 +𝑏”, we might expect to have to write our answer as: 𝑥+5 2 +[−25] However, since this is effectively the same as 𝑥+5 2 −25, this form is preferred; it is cleaner! 𝑥 2 +2𝑥 → 𝑥+1 2 −1 ? 𝑥 2 +𝑥 → 𝑥+ 1 2 2 − 1 4 ? 𝑥 2 −6𝑥 → 𝑥−3 2 −9 ? Because of the − sign, you might be tempted to think that the answer is 𝑥−3 2 +9. But note that: 𝑥−3 2 = 𝑥 2 −6𝑥+9 We don’t want the +9, so we subtract just as before. 𝑥 2 −20𝑥 → 𝑥−10 2 −100 ? 𝑥 2 −12𝑥 → 𝑥−6 2 −36 ?

Further Examples ? ? ? ? 𝑥 2 +8𝑥+3 → 𝑥+4 2 −16+3 = 𝑥+4 2 −13 ! Putting a quadratic expression in the form 𝑥+𝑎 2 +𝑏 is known as ‘completing the square’. The key motivation is that while 𝑥 2 +4𝑥+9 contains an 𝑥 2 term and a 𝑥 term, its ‘completed square’ 𝑥+2 2 +5 only contains a single 𝒙. We will see later why this has a number of useful applications. Put the following in the form 𝑥+𝑎 2 +𝑏 𝑥 2 +8𝑥+3 → 𝑥+4 2 −16+3 = 𝑥+4 2 −13 ? The +3 is still there! 𝑥 2 −2𝑥+10 → 𝑥−1 2 −1+10 = 𝑥+4 2 +9 ? 𝑥 2 +12𝑥−1 → 𝑥+6 2 −36−1 = 𝑥+4 2 −37 ? 𝑥 2 −𝑥+1 → 𝑥− 1 2 2 − 1 4 +1 = 𝑥− 1 2 2 + 3 4 ?

Test Your Understanding Put the following in the form 𝑥+𝑎 2 +𝑏 ? 𝑥 2 +16𝑥 → 𝑥+8 2 −16 𝑥 2 +10𝑥+31 → 𝑥+5 2 −25+31 = 𝑥+5 2 +6 ? 𝑥 2 −6𝑥−5 → 𝑥−3 2 −9−5 = 𝑥−3 2 −14 ?

Exercise 1 Put the following in the form 𝑥+𝑎 2 +𝑏 𝑥 2 +7𝑥+2 = 𝒙+ 𝟕 𝟐 𝟐 − 𝟒𝟏 𝟒 𝑥 2 −𝑥+3 = 𝒙− 𝟏 𝟐 𝟐 + 𝟏𝟏 𝟒 𝑥 2 − 1 3 𝑥+ 1 4 = 𝒙− 𝟏 𝟔 𝟐 + 𝟐 𝟗 ? 𝑥 2 +20𝑥 = 𝒙+𝟏𝟎 𝟐 −𝟏𝟎𝟎 𝑥 2 −2𝑥 = 𝒙−𝟏 𝟐 −𝟏 𝑥 2 +4𝑥−1= 𝒙+𝟐 𝟐 −𝟓 𝑥 2 +6𝑥+3= 𝒙+𝟑 𝟐 −𝟔 𝑥 2 −2𝑥+5= 𝒙−𝟏 𝟐 +𝟒 𝑥 2 −10𝑥−2= 𝒙−𝟓 𝟐 −𝟐𝟕 𝑥 2 +20𝑥+100= 𝒙+𝟏𝟎 𝟐 𝑥 2 −8𝑥+1= 𝒙−𝟒 𝟐 −𝟏𝟓 𝑥 2 +3𝑥 = 𝒙+ 𝟑 𝟐 𝟐 − 𝟗 𝟒 𝑥 2 −5𝑥+1 = 𝒙− 𝟓 𝟐 𝟐 − 𝟐𝟏 𝟒 ? 3 a 1 a ? ? b b ? c ? c ? d ? e 𝑥 2 +𝑎𝑥 = 𝒙+ 𝒂 𝟐 𝟐 − 𝒂 𝟐 𝟒 ? N a ? f ? ? 𝑥 2 +2𝑎𝑥+ 𝑎 2 = 𝒙+𝒂 𝟐 b g ? h ? 2 a ? b

What if the coefficient of 𝑥 2 is not 1? Put 2 𝑥 2 +8𝑥+7 in the form 𝑎 𝑥+𝑏 2 +𝑐 2 𝑥 2 +4𝑥 +7 =2 𝑥+2 2 −4 +7 =2 𝑥+2 2 −8+7 =2 𝑥+2 2 −1 ? Factorise the coefficient of 𝑥 2 out of the first two terms (leave last term outside brackets) ? Complete the square within brackets (you should have a bracket within a bracket) ? Expand outer bracket ? Final simplification. Put 3 𝑥 2 −6𝑥+11 in the form 𝑎 𝑥+𝑏 2 +𝑐 Put 4 𝑥 2 +40𝑥−5 in the form 𝑎 𝑥+𝑏 2 +𝑐 3 𝑥 2 −2𝑥 +11 =3 𝑥−1 2 −1 +11 =3 𝑥−1 2 −3+11 =3 𝑥−1 2 +8 ? 4 𝑥 2 −2𝑥 +11 =4 𝑥−1 2 −1 +11 =4 𝑥−1 2 −4+11 =4 𝑥−1 2 +7 ?

Harder Examples ? ? Put 3−12𝑥− 𝑥 2 in the form 𝑎 𝑥+𝑏 2 +𝑐 − 1𝑥 2 −12𝑥+3 =−1 𝑥 2 +12𝑥 +3 =−1 𝑥+6 2 −36 +3 =− 𝑥+6 2 +36+3 =− 𝑥+6 2 +39 𝑜𝑟 39− 𝑥+6 2 Put 1 10 𝑥 2 +4𝑥−3 in the form 𝑎 𝑥+𝑏 2 +𝑐 = 1 10 𝑥 2 +40𝑥 −3 = 1 10 𝑥+20 2 −400 −3 = 1 10 𝑥+20 2 −40−3 = 1 10 𝑥+20 2 −43 ?

Test Your Understanding Put 5 𝑥 2 +20𝑥+7 in the form 𝑎 𝑥+𝑏 2 +𝑐 =5 𝑥 2 +4𝑥 +7 =5 𝑥+2 2 −4 +7 =5 𝑥+2 2 −20+7 =5 𝑥+2 2 −13 ? Put 1 4 𝑥 2 −3𝑥+7 in the form 𝑎 𝑥+𝑏 2 +𝑐 = 1 4 𝑥 2 −12𝑥 +7 = 1 4 𝑥−6 2 −36 +7 = 1 4 𝑥−6 2 −9+7 = 1 4 𝑥−6 2 −2 ?

Exercise 2 Put the following in the form 𝑎 𝑥+𝑏 2 +𝑐 3 𝑥 2 +6𝑥−2 =𝟑 𝒙+𝟏 𝟐 −𝟓 2 𝑥 2 −12𝑥+1 =𝟐 𝒙−𝟑 𝟐 −𝟏𝟕 4 𝑥 2 −40𝑥+11 =𝟒 𝒙−𝟓 𝟐 −𝟖𝟗 2 𝑥 2 +24𝑥+70 =𝟐 𝒙+𝟔 𝟐 −𝟐 5 𝑥 2 −60𝑥+10 =𝟓 𝒙−𝟔 𝟐 −𝟏𝟕𝟎 3 𝑥 2 +12𝑥−1 =𝟑 𝒙+𝟐 𝟐 −𝟏𝟑 ? 1 a b ? c ? d ? e ? f ? − 𝑥 2 +8𝑥+6 =− 𝒙−𝟒 𝟐 +𝟐𝟐 − 𝑥 2 −10𝑥+5 =− 𝒙+𝟓 𝟐 +𝟑𝟎 − 𝑥 2 +6𝑥−2 =− 𝒙−𝟑 𝟐 +𝟕 1−2𝑥− 𝑥 2 =− 𝒙+𝟏 𝟐 +𝟐 −2 𝑥 2 +12𝑥−20 =−𝟐 𝒙−𝟑 𝟐 −𝟐 5+10𝑥−5 𝑥 2 =−𝟓 𝒙−𝟏 𝟐 +𝟏𝟎 2 a ? b ? c ? d ? e ? f ? 1 3 𝑥 2 +2𝑥+1 = 𝟏 𝟑 𝒙+𝟑 𝟐 −𝟐 1 5 𝑥 2 −4𝑥+13= 𝟏 𝟓 𝒙−𝟏𝟎 𝟐 −𝟕 1 6 𝑥 2 +𝑥−1 = 𝟏 𝟔 𝒙+𝟑 𝟐 − 𝟓 𝟐 ? 3 a ? b ? c

Why complete the square? There are 3 major applications of completing the square, two of which we will look at: 1 :: Solving Equations “(a) Write 𝑥 2 +4𝑥−7 in the form 𝑥+𝑎 2 +𝑏. (b) Hence determine the exact solutions of: 𝑥 2 +4𝑥−7=0 Solving quadratic equations in this way also allows us to derive the quadratic formula! (The proof is later in these slides) 2 :: Finding the Turning Point of a Parabola “Determine the turning point of the line with equation 𝑦= 𝑥 2 −4𝑥+5” Key Term: A parabola is the name of a line which has a quadratic equation. 3 :: (Further Maths A Level) Integrating reciprocals of quadratics “Determine 1 𝑥 2 −4𝑥+5 𝑑𝑥 ” ∫ means “integrate”. This allows us to find the area under a graph. You won’t need to worry about this for some time!

Application #1 :: Solving by Completing the Square a) Write 𝑥 2 +4𝑥−7 in the form 𝑥+𝑎 2 +𝑏 b) Hence, determine the exact solutions of 𝑥 2 +4𝑥−7=0 a ? 𝑥+2 2 −4−7 = 𝑥+2 2 −11 𝑥+2 2 −11=0 𝑥+2 2 =11 𝑥+2 =± 11 𝑥=−2± 11 b ? Because 𝑥 now only appears once in the equation, we can use our ‘changing the subject’ skills to make 𝑥 the subject. ? ? Don’t forget the ±. Suppose for example we were solving 𝑥 2 =4. 𝑥=±2 because 2 2 =4 and −2 2 =4 We tend to write 𝑎± 𝑏 rather than ± 𝑏 +𝑎 to avoid ambiguity of what is included under the √.

Further Examples ? ? a) Write 𝑥 2 −6𝑥+3 in the form 𝑥+𝑎 2 +𝑏 b) Hence, determine the exact solutions of 𝑥 2 −6𝑥+3=0 ? 𝑥−3 2 −9+3=0 𝑥−3 2 −6=0 𝑥−3 2 =6 𝑥−3=± 6 𝑥=3± 6 a) Write 2 𝑥 2 +8𝑥−1 in the form 𝑎 𝑥+𝑏 2 +𝑐 b) Hence, determine the exact solutions of 2 𝑥 2 +8𝑥+1=0 ? 2 𝑥 2 +4𝑥 −1 =2 𝑥+2 2 −4 −1 =2 𝑥+2 2 −8−1 =2 𝑥+2 2 −9 2 𝑥+2 2 −9=0 2 𝑥+2 2 =9 𝑥+2 2 = 9 2 𝑥+2=± 3 2 𝑥=−2± 3 2 2

Test Your Understanding a) Write 𝑥 2 +10𝑥−4 in the form 𝑥+𝑎 2 +𝑏 b) Hence, determine the exact solutions of 𝑥 2 +10𝑥−4=0 ? 𝑥+5 2 −29=0 𝑥+5=± 29 𝑥=−5± 29 a) Write 3 𝑥 2 −30𝑥+71 in the form 𝑎 𝑥+𝑏 2 +𝑐 b) Hence, determine the exact solutions of 3 𝑥 2 −30𝑥+11=0 ? 3 𝑥 2 −10𝑥 +71 =3 𝑥−5 2 −25 +71 =3 𝑥−5 2 −75+71 =3 𝑥−5 2 −4 3 𝑥−5 2 −4=0 𝑥−5 2 = 4 3 𝑥−5=± 2 3 𝑥=5± 2 3 3

Application #2 :: Minimum/Maximum Values Completing the square also allows us to find the minimum or maximum value of a quadratic. For the quadratic expression 𝑥 2 −4𝑥+11, Determine its minimum value. Determine the value of 𝑥 for which this minimum occurs. ? 𝑥−2 2 −4+11 = 𝑥−2 2 +7 First complete the square. Let’s experiment with different values of 𝑥 to see what gives us the smallest value: 𝑥=0 → 0−2 2 +7=11 𝑥=1 → 1−2 2 +7=8 𝑥=2 → 2−2 2 +7=7 𝑥=3 → 32 2 +7=8 So the minimum value of 𝑥 2 −4𝑥+11 appears to be 7, which occurs when 𝑥=2. But why is this? Anything squared is at least 0. So we choose 𝒙 such that the squared term is 0 in order to minimise it. ? ? ? ? ? ? ?

Further Examples ? ? For the quadratic expression 𝑥 2 +6𝑥+5, Determine its minimum value. Determine the value of 𝑥 for which this minimum occurs. ? = 𝑥+3 2 −9+5 = 𝑥+3 2 −4 We want the squared term to be 0. This occurs when 𝑥=−3. Then the minimum value will be 0 2 −4=−4 ! The minimum value of 𝑥+𝑎 2 +𝑏 is 𝑏, which occurs when 𝑥=−𝑎. For the quadratic expression − 𝑥 2 +8𝑥+3, Determine its maximum value. Determine the value of 𝑥 for which this maximum occurs. ? =− 𝑥 2 −8𝑥 +3 =− 𝑥−4 2 −16 +3 =− 𝑥−4 2 +19 =19− 𝑥−4 2 We are subtracting a number which is at least 0. Therefore to maximise the result, we should subtract 0. Max value: 19 𝑥 at which this occurs: 4

Turning Points of Quadratics A curve with equation 𝑦= 𝑥 2 −8𝑥+17 has a turning point at 𝑃. Determine the coordinates of 𝑃. 𝑃 At the turning point, the value of 𝑦, i.e. 𝑥 2 −8𝑥+17, is minimised. We know how to do this! ? 𝑦= 𝑥−4 2 −16+17 = 𝑥−4 2 +1 ∴𝑃(4,1) Quickfire Questions Completed Square Turning Point 𝑦= 𝑥 2 +2𝑥+8 = 𝑥+1 2 +7 −1,7 𝑦= 𝑥 2 −6𝑥+3 = 𝑥−3 2 −6 3,−6 𝑦= 𝑥 2 +10𝑥+4 = 𝑥+5 2 −21 −5,−21 𝑦=2 𝑥 2 +8𝑥+1 =2 𝑥+2 2 −7 (−2,−7) ? ? ? ? ? ? ? ?

Test Your Understanding ? 𝒚= 𝒙+𝟓 𝟐 −𝟕 −𝟓,−𝟕 = 𝒙−𝒎 𝟐 − 𝒎 𝟐 Minimum value: − 𝒎 𝟐 (which occurs when 𝒙=𝒎) ?

Exercise 3 Find the turning point of the curves with the following equations: 𝑦= 𝑥 2 +6𝑥+1 −𝟑,−𝟖 𝑦= 𝑥 2 −4𝑥+13 𝟐,𝟗 𝑦= 𝑥 2 +10𝑥+31 −𝟓,𝟔 𝑦= 𝑥 2 −12𝑥+30 𝟔,−𝟔 𝑦=3 𝑥 2 +12𝑥+1 −𝟐,−𝟏𝟏 𝑦=2 𝑥 2 −12𝑥+21 𝟑,𝟑 𝑦=8−4𝑥− 𝑥 2 −𝟐,𝟏𝟐 𝑦=1−8𝑥−2 𝑥 2 −𝟐,𝟗 Find the minimum value of the following expressions: 𝑥 2 +2𝑥+2 𝟏 𝑥 2 +10𝑥+4 −𝟐𝟏 𝑥 2 −4𝑥+7 𝟑 𝑥 2 +𝑥+2 𝟕 𝟒 𝑥 2 +𝑎𝑥 − 𝒂 𝟐 1 3 [Edexcel GCSE(9-1) Mock Set 1 Autumn 2016 - 1H Q15] Here is a sketch of a vertical cross section through the centre of a bowl. The cross section is the shaded region between the curve and the 𝑥 -axis. The curve has equation 𝑦= 𝑥 2 10 −3𝑥  where 𝑥  and 𝑦 are both measured in centimetres. Find the depth of the bowl. 𝒚= 𝟏 𝟏𝟎 𝒙 𝟐 −𝟑𝟎𝒙 = 𝟏 𝟏𝟎 𝒙−𝟏𝟓 𝟐 −𝟐𝟐𝟓 = 𝟏 𝟏𝟎 𝒙−𝟏𝟓 𝟐 −𝟐𝟐.𝟓 Minimum point is 𝟏𝟓,−𝟐𝟐.𝟓 Therefore depth is 22.5 cm ? a ? b c ? ? d ? e ? f ? g ? h 2 a ? ? b ? c ? ? d e ?

Spot the Student Error ? ? ? “Complete the square using 𝑥 2 +4𝑥−12.” Student Answer: “(𝑥+6)(𝑥−2)” What’s wrong: The student factorised rather than completed the square. We factorise when we want to find the roots of a quadratic. Meanwhile, we complete the square when we want to find the turning point of a quadratic. ? 𝑥 2 −6𝑥+11 → 𝑥−3 2 +9+11=… What’s wrong: The 9 should have been subtracted (and it should always be a subtraction regardless of whether the coefficient of 𝑥 is positive or negative. ? “Write 2 𝑥 2 +8𝑥+11” in the form 𝑎 𝑥+𝑏 2 +𝑐” Student Answer: =2 𝑥 2 +4𝑥 +11 =2 𝑥+2 2 −4 +11 =2 𝑥+2 2 +7 What’s wrong: In the last line of working, they did −4+11 → +7 They forgot to multiply the −4 by 2. ?

Proof of the Quadratic Formula By completing the square, show that the solution of 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 is 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 ? 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 𝑥 2 + 𝑏 𝑎 𝑥+ 𝑐 𝑎 =0 𝑥+ 𝑏 2𝑎 2 − 𝑏 2 4 𝑎 2 + 𝑐 𝑎 =0 𝑥+ 𝑏 2𝑎 2 = 𝑏 2 4 𝑎 2 − 𝑐 𝑎 = 𝑏 2 −4𝑎𝑐 4 𝑎 2 𝑥+ 𝑏 2𝑎 =± 𝑏 2 −4𝑎𝑐 4 𝑎 2 =± 𝑏 2 −4𝑎𝑐 2𝑎 𝑥=− 𝑏 2𝑎 ± 𝑏 2 −4𝑎𝑐 2𝑎 = −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎