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IGCSE Completing the Square

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Presentation on theme: "IGCSE Completing the Square"β€” Presentation transcript:

1 IGCSE Completing the Square
Dr J Frost Objectives: (from the specification) Last modified: 22nd August 2015

2 RECAP π‘₯ 2 βˆ’4π‘₯=𝒙 π’™βˆ’πŸ’ π‘₯ 2 βˆ’3π‘₯βˆ’40= 𝒙+πŸ“ π’™βˆ’πŸ– π‘₯ 2 βˆ’9= 𝒙+πŸ‘ π’™βˆ’πŸ‘ 2 π‘₯ 2 βˆ’π‘₯βˆ’6=(πŸπ’™+πŸ‘)(π’™βˆ’πŸ) ? ? ? ?

3 What makes this topic Further Maths-ey?
You’re used to expressing for example π‘₯ 2 +4π‘₯βˆ’3 in the form π‘₯+2 2 βˆ’7 But you’ve (probably) never had to deal with the coefficient of π‘₯ 2 not being 1!

4 Reminder ? π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐 π‘Ž π‘₯+__ 2 +__ ?
What the devil is β€˜completing the square’? ? π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐 π‘Ž π‘₯+__ 2 +__ It means putting a quadratic expressions in the form on the right, i.e. where π‘₯ only appears once. What’s the point? ? It has four uses, the first two of which we will explore: Solving quadratic equations (including deriving the quadratic formula!). Sketching quadratic equations. Helps us to β€˜integrate’ certain expressions (an A Level topic!) Helps us do something called β€˜Laplace Transforms’ (a university topic!)

5 π‘₯ 2 βˆ’2π‘₯= π‘₯βˆ’1 2 βˆ’1 π‘₯ 2 βˆ’6π‘₯+4= π‘₯βˆ’3 2 βˆ’5 π‘₯ 2 +8π‘₯+1= π‘₯+4 2 βˆ’15
Recap of π‘₯+𝑏 2 +𝑐 π‘₯ 2 βˆ’2π‘₯= π‘₯βˆ’1 2 βˆ’1 π‘₯ 2 βˆ’6π‘₯+4= π‘₯βˆ’3 2 βˆ’5 π‘₯ 2 +8π‘₯+1= π‘₯+4 2 βˆ’15 π‘₯ 2 +10π‘₯βˆ’3= π‘₯+5 2 βˆ’28 π‘₯ 2 +4π‘₯+3= π‘₯+2 2 βˆ’1 π‘₯ 2 βˆ’20π‘₯+150= π‘₯βˆ’ ? ? ? ? ? ? Reminder of method: π‘₯ 2 βˆ’6π‘₯+4 = π‘₯βˆ’3 2 βˆ’9+4 = π‘₯βˆ’3 2 βˆ’5 π‘₯ 2 +8π‘₯+1= π‘₯+4 2 βˆ’16+1 = π‘₯+4 2 βˆ’15 Remember we halve the coefficient of π‘₯, then square it and β€˜throw it away’.

6 π‘Ž π‘₯ 2 +… So far the coefficient of the π‘₯ 2 term has been 1. What if it isn’t? Express 3 π‘₯ 2 +12π‘₯βˆ’6 in the form π‘Ž π‘₯+𝑏 2 +𝑐 3 π‘₯ 2 +12π‘₯βˆ’6 =3 π‘₯ 2 +4π‘₯βˆ’2 =3 π‘₯+2 2 βˆ’4βˆ’2 =3 π‘₯+2 2 βˆ’6 =3 π‘₯+2 2 βˆ’18 Just factorise out the coefficient of the π‘₯ 2 term. Now we have an expression just like before for which we can complete the square! ? ? Now expand out the outer brackets. To be sure about your answer you could always expand and check you get the original expr. ? Express 2βˆ’4π‘₯βˆ’2 π‘₯ 2 in the form π‘Žβˆ’π‘ π‘₯+𝑐 2 βˆ’2 π‘₯ 2 βˆ’4π‘₯+2 =βˆ’2 π‘₯ 2 +2π‘₯βˆ’1 =βˆ’2 π‘₯+1 2 βˆ’1βˆ’1 =βˆ’2 π‘₯+1 2 βˆ’2 =βˆ’2 π‘₯ =4βˆ’2 π‘₯+1 2 ? Bro Tip: Reorder the terms so you always start with something in the form π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐 ? ? Bro Tip: Be jolly careful with your signs! Bro Tip: You were technically done on the previous line, but it’s nice to reorder the terms so it’s more explicitly in the requested form. ? ?

7 One more example ? 2 π‘₯ 2 +6π‘₯+7=2 π‘₯ 2 +3π‘₯ =2 π‘₯ βˆ’ =2 π‘₯ =2 π‘₯ ? ? ? This was the actual example on the specification!

8 Test Your Understanding
Put the expression 3 π‘₯ 2 βˆ’12π‘₯+5 in the form π‘Ž π‘₯+𝑏 2 +𝑐. ? =3 π‘₯ 2 βˆ’4π‘₯ =3 π‘₯βˆ’2 2 βˆ’ =3 π‘₯βˆ’2 2 βˆ’ =3 π‘₯βˆ’2 2 βˆ’7

9 Proof of the Quadratic Formula!
by completing the square… π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐=0 π‘₯ 2 + 𝑏 π‘Ž π‘₯+ 𝑐 π‘Ž =0 π‘₯+ 𝑏 2π‘Ž 2 βˆ’ 𝑏 2 4 π‘Ž 2 + 𝑐 π‘Ž =0 π‘₯+ 𝑏 2π‘Ž π‘Žπ‘βˆ’ 𝑏 2 4 π‘Ž 2 =0 π‘₯+ 𝑏 2π‘Ž 2 = 𝑏 2 βˆ’4π‘Žπ‘ 4 π‘Ž 2 π‘₯+ 𝑏 2π‘Ž =Β± 𝑏 2 βˆ’4π‘Žπ‘ 2π‘Ž π‘₯= βˆ’π‘Β± 𝑏 2 βˆ’4π‘Žπ‘ 2π‘Ž ? ? ? ? ? ?

10 Exercises Express π‘₯ 2 βˆ’4π‘₯+5 in the form π‘₯βˆ’π‘Ž 2 +𝑏: π’™βˆ’πŸ 𝟐 +𝟏 Work out the values of π‘Ž and 𝑏 such that π‘₯ 2 βˆ’6π‘₯+5≑ π‘₯+π‘Ž 2 +𝑏 𝒂=βˆ’πŸ‘, 𝒃=βˆ’πŸ’ [June 2013 Paper 1] Express 2 π‘₯ 2 βˆ’12π‘₯βˆ’7 in the form π‘Ž π‘₯+𝑏 2 +𝑐. 𝟐 π’™βˆ’πŸ‘ 𝟐 βˆ’πŸπŸ“ 2 π‘₯ 2 βˆ’4π‘₯+5β‰‘π‘Ž π‘₯+𝑏 2 +𝑐 Work out the values of π‘Ž, 𝑏, 𝑐 𝒂=𝟐, 𝒃=βˆ’πŸ, 𝒄=πŸ‘ Express the following in the form π‘Ž π‘₯+𝑏 2 +𝑐 2 π‘₯ 2 +16π‘₯=𝟐 𝒙+πŸ’ 𝟐 βˆ’πŸ‘πŸ 5 π‘₯ 2 +20π‘₯βˆ’10=πŸ“ 𝒙+𝟐 𝟐 βˆ’πŸ‘πŸŽ 9 π‘₯ 2 βˆ’18π‘₯+27=πŸ— π’™βˆ’πŸ 𝟐 +πŸπŸ– 3 π‘₯ 2 βˆ’6π‘₯+4=πŸ‘ π’™βˆ’πŸ 𝟐 +𝟏 4 π‘₯ 2 +16π‘₯βˆ’1=πŸ’ 𝒙+𝟐 𝟐 βˆ’πŸπŸ• 1 6 Express the following in the form π‘Ž π‘₯+𝑏 2 +𝑐: 3 π‘₯ 2 βˆ’π‘₯=πŸ‘ π’™βˆ’ 𝟏 πŸ” 𝟐 βˆ’ 𝟏 𝟏𝟐 4 π‘₯ 2 +π‘₯βˆ’1=πŸ’ 𝒙+ 𝟏 πŸ– 𝟐 Express the following in the form π‘Žβˆ’π‘ π‘₯+𝑐 2 : 3+6π‘₯βˆ’ π‘₯ 2 =πŸπŸβˆ’ π’™βˆ’πŸ‘ 𝟐 10βˆ’8π‘₯βˆ’ π‘₯ 2 =πŸπŸ”βˆ’ 𝒙+πŸ’ 𝟐 10π‘₯βˆ’8βˆ’5 π‘₯ 2 =βˆ’πŸ‘βˆ’πŸ“ π’™βˆ’πŸ 𝟐 1βˆ’36π‘₯βˆ’6 π‘₯ 2 =πŸ“πŸ“βˆ’πŸ” 𝒙+πŸ‘ 𝟐 ? ? a 2 ? ? b 3 7 ? ? a 4 b ? c ? ? d ? 5 a ? b ? c ? d ? e ?


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