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Slideshow 22, Mathematics Mr Richard Sasaki

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1 Slideshow 22, Mathematics Mr Richard Sasaki
Vieta’s Formulae Slideshow 22, Mathematics Mr Richard Sasaki

2 Objectives Understand how two solutions for a quadratic equation can be written as two simultaneous equations Understand how to use these equations Solve such equations to find solutions for π‘₯

3 Vieta Vieta was a French man. He knew how to factorise an equation π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐=0 and considered this as π‘₯ 2 + 𝑏 π‘Ž π‘₯+ 𝑐 π‘Ž =0. When we factorise this we get… π‘₯βˆ’ π‘₯ 1 π‘₯βˆ’ π‘₯ 2 =0 for some two solutions π‘₯ 1 and π‘₯ 2 . How do they relate to 𝑏 π‘Ž and 𝑐 π‘Ž ? 𝑏 π‘Ž We know that βˆ’ π‘₯ 1 + βˆ’ π‘₯ 2 = and βˆ’ π‘₯ 1 βˆ™ βˆ’ π‘₯ 2 = 𝑐 π‘Ž π‘₯ 1 + π‘₯ 2 =βˆ’ 𝑏 π‘Ž So… and π‘₯ 1 βˆ™ π‘₯ 2 = 𝑐 π‘Ž

4 Vieta’s Formulae π‘₯ 1 + π‘₯ 2 =βˆ’ 𝑏 π‘Ž and . π‘₯ 1 βˆ™ π‘₯ 2 = 𝑐 π‘Ž
In the simple case π‘₯ 2 +𝑏π‘₯+𝑐=0, we get… π‘₯ 1 + π‘₯ 2 =βˆ’π‘ π‘₯ 1 βˆ™ π‘₯ 2 =𝑐 Example Solve π‘₯ 2 βˆ’3π‘₯βˆ’10=0 using Vieta’s Formulae. We have 𝑏=βˆ’3 and 𝑐=βˆ’10 so… π‘₯ 1 + π‘₯ 2 =3 π‘₯ 1 π‘₯ 2 =βˆ’10 Now let’s try to find the solutions for π‘₯ 1 and π‘₯ 2 by solving the simultaneous equations above!

5 Vieta’s Formulae β‘  π‘₯ 1 + π‘₯ 2 =3 β‘‘ π‘₯ 1 π‘₯ 2 =βˆ’10
π‘₯ 1 π‘₯ 2 =βˆ’10 It doesn’t matter which I’ll substitute into which. I will substitute β‘‘ into β‘ . βˆ’ 10 π‘₯ 2 β‘‘ π‘₯ 1 π‘₯ 2 =βˆ’10 β‡’ π‘₯ 1 = βˆ’ 10 π‘₯ 2 + π‘₯ 2 =3 β‘  π‘₯ 1 + π‘₯ 2 =3 β‡’ β‡’ βˆ’10+ π‘₯ 2 2 =3 π‘₯ 2 β‡’ π‘₯ 2 2 βˆ’3 π‘₯ 2 βˆ’10=0 I’m back where I started!!

6 Vieta’s Formulae Because the formulae are derived from the quadratic anyway, we can’t solve them like that. Vieta’s formulae can only be used to solve quadratics visually! π‘₯ 1 + π‘₯ 2 =3 π‘₯ 1 π‘₯ 2 =βˆ’10 What numbers add together to make 3 and multiply together to make βˆ’10? βˆ’2 and 5! (It’s very similar to factorisation.) So the solutions are π‘₯=βˆ’2 and 5.

7 Vieta’s Formulae Let’s try another! Example We have 𝑏= , 𝑐= , so… 13
We have 𝑏= , 𝑐= , so… 13 42 π‘₯ 1 + π‘₯ 2 =βˆ’π‘ π‘₯ 1 + π‘₯ 2 =βˆ’13 β‡’ π‘₯ 1 π‘₯ 2 =𝑐 π‘₯ 1 π‘₯ 2 =42 What numbers add together to make βˆ’13 and multiply together to make 42? βˆ’6 and βˆ’7! So the solutions are π‘₯=βˆ’6 and βˆ’7.

8 Answers - Easy 3 2 βˆ’3 2 βˆ’1, βˆ’2 7 10 βˆ’7 10 βˆ’2, βˆ’5 2 βˆ’3 βˆ’2 βˆ’3 1, βˆ’3 βˆ’5 6
2, 3 6 9 βˆ’6 9 βˆ’3 6 βˆ’7 βˆ’6 βˆ’7 βˆ’7, 1 10 24 βˆ’10 24 βˆ’4, βˆ’6 βˆ’10 25 10 25 5 βˆ’2 βˆ’35 2 βˆ’35 7, βˆ’5

9 Answers - Hard 14 49 βˆ’14 49 βˆ’7 βˆ’11 βˆ’26 11 βˆ’26 βˆ’2, 13 βˆ’13 βˆ’14 13 βˆ’14
βˆ’1, 14 βˆ’18 81 18 81 9 βˆ’9 βˆ’9 Β±3 βˆ’17 72 17 72 8, 9 19 70 βˆ’19 70 βˆ’14,βˆ’5 2 βˆ’2 βˆ’2, 0 βˆ’169 βˆ’169 Β±13

10 Vieta’s Formulae If the numbers are easy, it is a similar process solving equations where π‘Žβ‰ 1. π‘₯ 1 + π‘₯ 2 =βˆ’ 𝑏 π‘Ž and π‘₯ 1 βˆ™ π‘₯ 2 = 𝑐 π‘Ž Example Solve 2π‘₯ 2 βˆ’2π‘₯βˆ’4=0 using Vieta’s Formulae. π‘Ž=2, 𝑏=βˆ’2, 𝑐=βˆ’4, π‘ π‘œβ€¦ π‘₯ 1 + π‘₯ 2 =βˆ’ 𝑏 π‘Ž π‘₯ 1 + π‘₯ 2 =1 β‡’ π‘₯ 1 π‘₯ 2 = 𝑐 π‘Ž π‘₯ 1 π‘₯ 2 =βˆ’2 What numbers add together to make 1 and multiply together to make βˆ’2? βˆ’1 and 2! So the solutions are π‘₯=βˆ’1 and 2.

11 π‘₯=4 π‘œπ‘Ÿ βˆ’1 π‘₯=3 π‘œπ‘Ÿ βˆ’1 π‘₯=3 π‘œπ‘Ÿ βˆ’4 π‘₯=βˆ’1 π‘₯=βˆ’8 π‘œπ‘Ÿ 7 π‘₯=5 π‘œπ‘Ÿ βˆ’5


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