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Alpha and Beta

Alpha and Beta

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Mathematics, Other

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11th Grade

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Hard

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KASSIA! LLTTF

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16 Slides • 0 Questions

1

Quadratics

Alpha and Beta

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2

Roots And Coefficients

 ax2+bx+c=0ax^2+bx+c=0  
The coefficients are a, b and c.

 ∝ β\propto\ \beta  - these are called roots
 ∝− alpha   β−beta\propto-\ alpha\ \ \ \beta-beta  
Note: the symbol for alpha can also be represented by  α\alpha  

There are relationships between coefficients of a quadratic equation and the roots alpha and beta. The relationship is based on the Sum of the Roots and the Product of the roots.

3

Representation of the relationships.

Find the values of x if  x2+4x−12=0x^2+4x-12=0  

 x2+4x−12=0x^2+4x-12=0  (factorize)
 (x+6)(x−2)=0\left(x+6\right)\left(x-2\right)=0  
 ∴x=−6 or x=2\therefore x=-6\ or\ x=2  

Let roots be  ∝ & β\propto\ \&\ \beta  
 ∝=−6  β=2\propto=-6\ \ \beta=2           Sum of Roots :   ∝+β=−6+2=−4\propto+\beta=-6+2=-4                                                   Product of Roots :  ∝β=−6(2)=−12\propto\beta=-6\left(2\right)=-12  
Therefore the relationship between the roots and coefficients is   ∝+β=−ba  &  ∝β=ca\propto+\beta=\frac{-b}{a}\ \ \&\ \ \propto\beta=\frac{c}{a}  

4

example 1.

If  x2−2x−8=0 has roots∝and βx^2-2x-8=0\ has\ roots\propto and\ \beta  find: a. ∝+β  b.∝β\propto+\beta\ \ b.\propto\beta  



 x2−3x−8=0x^2-3x-8=0  
 a=1  b=−2  c=−8a=1\ \ b=-2\ \ c=-8  

a)  ∝+β=−ba=−(−2)1=21=2\propto+\beta=\frac{-b}{a}=\frac{-\left(-2\right)}{1}=\frac{2}{1}=2  
b)  ∝β=ca=−81=−8\propto\beta=\frac{c}{a}=\frac{-8}{1}=-8  

5

Example 2.

If  ∝and \propto and\   are the roots of the equation  2x2−5x+3=02x^2-5x+3=0  find:

 1.  ∝+β\propto+\beta   2.  ∝β\propto\beta  


 2x2−5x+3=02x^2-5x+3=0  
 a=2  b=−5  c=3a=2\ \ b=-5\ \ c=3  

 ∝+β=−ba=−(−5)2=52\propto+\beta=\frac{-b}{a}=\frac{-\left(-5\right)}{2}=\frac{5}{2}  
 ∝β=ca=32\propto\beta=\frac{c}{a}=\frac{3}{2}  

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Operations with alpha and beta.

1.  ∝+β=−ba\propto+\beta=\frac{-b}{a}    
2.  ∝β=ca\propto\beta=\frac{c}{a}  
3.  ∝2β2=(∝β)2\propto^2\beta^2=\left(\propto\beta\right)^2  
4.  ∝2β+β2∝=∝β(∝+β)\propto^2\beta+\beta^2\propto=\propto\beta\left(\propto+\beta\right)  
5. x∝+xβ=x∝+xβ∝β=x(∝+β)∝β\frac{x}{\propto}+\frac{x}{\beta}=\frac{x\propto+x\beta}{\propto\beta}=\frac{x\left(\propto+\beta\right)}{\propto\beta}  where x is a constant (e.g 2,3,4,5)

6.  (∝+β)2=squared of ∝+β\left(\propto+\beta\right)^2=squared\ of\ \propto+\beta  
7.  ∝β+β∝=∝2+β2∝β= (∝+β)2−2∝β∝β\frac{\propto}{\beta}+\frac{\beta}{\propto}=\frac{\propto^2+\beta^2}{\propto\beta}=\ \frac{\left(\propto+\beta\right)^2-2\propto\beta}{\propto\beta}  

7

Examples of the different operations.

If  ∝and β\propto and\ \beta  are the roots of the equation  x2−3x−18=0x^2-3x-18=0  find:

 a.  ∝+β    b.∝β    c.∝2β2   d.∝2β+β2∝ \propto+\beta\ \ \ \ b.\propto\beta\ \ \ \ c.\propto^2\beta^2\ \ \ d.\propto^2\beta+\beta^2\propto\  

 a) ∝+β=−ba=−(−3)1=3\propto+\beta=\frac{-b}{a}=\frac{-\left(-3\right)}{1}=3  

b) ∝β=ca=−181=−18\propto\beta=\frac{c}{a}=\frac{-18}{1}=-18  

c)  ∝2β2⟹(∝β)2=(−18)2=324\propto^2\beta^2\Longrightarrow\left(\propto\beta\right)^2=\left(-18\right)^2=324  
d)  ∝2β+β2∝⟹∝β(∝+β)=−18(3)=−54\propto^2\beta+\beta^2\propto\Longrightarrow\propto\beta\left(\propto+\beta\right)=-18\left(3\right)=-54  

8

more examples/continued

If  ∝and β\propto and\ \beta  are the roots of the equation  3x2−x−5=03x^2-x-5=0  . Find the following a. ∝+β\propto+\beta   b. ∝β\propto\beta    c. ∝2+β2 \propto^2+\beta^2\    d. 6∝+6β\frac{6}{\propto}+\frac{6}{\beta}   e. (∝β)2\left(\propto\beta\right)^2  


a. ∝+β=−ba=−(−1)3=13\propto+\beta=\frac{-b}{a}=\frac{-\left(-1\right)}{3}=\frac{1}{3}  
b.  ∝β=ca=−53\propto\beta=\frac{c}{a}=\frac{-5}{3}  
c.  ∝2+β2=(∝+β)2−2∝β=(13)2−2(−53)\propto^2+\beta^2=\left(\propto+\beta\right)^2-2\propto\beta=\left(\frac{1}{3}\right)^2-2\left(-\frac{5}{3}\right)          
=   19+103=319\frac{1}{9}+\frac{10}{3}=\frac{31}{9}  

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d.  6∝+6β=6β+6∝∝β=6(β+∝)∝β\frac{6}{\propto}+\frac{6}{\beta}=\frac{6\beta+6\propto}{\propto\beta}=\frac{6\left(\beta+\propto\right)}{\propto\beta}  


 =6(13)−53=2−53=−65=\frac{6\left(\frac{1}{3}\right)}{-\frac{5}{3}}=\frac{2}{-\frac{5}{3}}=-\frac{6}{5}  

Note :
 21÷−53=21×−35=−65\frac{2}{1}\div-\frac{5}{3}=\frac{2}{1}\times-\frac{3}{5}=-\frac{6}{5}     and  β+∝≡∝+β\beta+\propto\equiv\propto+\beta  

e. (∝+β)2=(13)2=19\left(\propto+\beta\right)^2=\left(\frac{1}{3}\right)^2=\frac{1}{9}  

10

New Roots

Steps:

1. State coefficients a,b,c .
2. Find  ∝+β, ∝β\propto+\beta,\ \propto\beta  
3. For new roots find: sum of roots and product of roots.
4. Substitute  ∝+β, ∝β \propto+\beta,\ \propto\beta\   from step 2 into  x2−(sum)x+ product =0x^2-\left(sum\right)x+\ product\ =0  
N.B your answer will be a quadratic equation.

11

Common new roots

 ⋅∝+1 , β+1\cdot\propto+1\ ,\ \beta+1  
 ⋅x∝, xβ\cdot\frac{x}{\propto},\ \frac{x}{\beta}  where x is a constant 
 ⋅∝−1 , β−1\cdot\propto-1\ ,\ \beta-1  
 ⋅1∝, 1β\cdot\frac{1}{\propto},\ \frac{1}{\beta}  
 ⋅∝β, β∝\cdot\frac{\propto}{\beta},\ \frac{\beta}{\propto}  

12

Eg 1

If  ∝ and β\propto\ and\ \beta  are the roots of  x2+4x−12=0x^2+4x-12=0  find an equation who's roots are  ∝+1 and  β+1\propto+1\ and\ \ \beta+1  .

 x2+4x−12=0     a=1, b=4, c=−12x^2+4x-12=0\ \ \ \ \ a=1,\ b=4,\ c=-12   ∝+β=−ba=−41=−4\propto+\beta=-\frac{b}{a}=\frac{-4}{1}=-4  
 ∝β=ca=−121=−12\propto\beta=\frac{c}{a}=\frac{-12}{1}=-12  
 Sum: ∝+1 +β+1=∝+β+2=−4+2=−2Sum:\ \propto+1\ +\beta+1=\propto+\beta+2=-4+2=-2   Pr⁡oduct : (∝+1)(β+1)=∝β+∝+β+1\Pr oduct\ :\ \left(\propto+1\right)\left(\beta+1\right)=\propto\beta+\propto+\beta+1                                                                                        =−12−4+1=−15=-12-4+1=-15  
Equation  ⟹x2−(−2)x+(−15)=0\Longrightarrow x^2-\left(-2\right)x+\left(-15\right)=0                                                                     =x2+2x−15=0=x^2+2x-15=0  

13

Eg 2

If  ∝ and β\propto\ and\ \beta  are the roots of the equation  x2 −2x−8=0x^{2\ }-2x-8=0  find the equation with roots  3∝ and 3β\frac{3}{\propto}\ and\ \frac{3}{\beta}  .

 x2−2x−8=0 x^2-2x-8=0\    a=1, b=-2, c=-8
 ∝+β=−ba=−(−2)1=2\propto+\beta=\frac{-b}{a}=\frac{-\left(-2\right)}{1}=2  
 ∝β=ca =−81=−8\propto\beta=\frac{c}{a\ }=\frac{-8}{1}=-8  
 Sum: 3∝+3β=3β+3∝∝β=3(∝+β)∝β=3(2)−8=−68Sum:\ \frac{3}{\propto}+\frac{3}{\beta}=\frac{3\beta+3\propto}{\propto\beta}=\frac{3\left(\propto+\beta\right)}{\propto\beta}=\frac{3\left(2\right)}{-8}=-\frac{6}{8}   Pr⁡oduct: 3∝×3β=9∝β=9−8=−98\Pr oduct:\ \frac{3}{\propto}\times\frac{3}{\beta}=\frac{9}{\propto\beta}=\frac{9}{-8}=-\frac{9}{8}  
 Equation ⟹ x2−(−68)x+(−98)=0 (×8)Equation\ \Longrightarrow\ x^2-\left(-\frac{6}{8}\right)x+\left(-\frac{9}{8}\right)=0\ \left(\times8\right)   =8x2+6x−9=0=8x^2+6x-9=0  

14

Eg 3

If  ∝and β\propto and\ \beta  are the roots of the equation  2x2+5x−3=02x^2+5x-3=0  .  Find the equation who's roots are  ∝−1 and β−1\propto-1\ and\ \beta-1  .

 2x2+5x−3=02x^2+5x-3=0     ∝+β=−ba=−52\propto+\beta=\frac{-b}{a}=-\frac{5}{2}  
a=2, b=5 , c=-3               ∝β=ca=−32\propto\beta=\frac{c}{a}=-\frac{3}{2}  
 Sum: ∝−1+β−1=∝+β−2=−52−2=−92Sum:\ \propto-1+\beta-1=\propto+\beta-2=-\frac{5}{2}-2=-\frac{9}{2}   Pr⁡oduct: (∝−1)(β−1)=∝β−∝−β+1\Pr oduct:\ \left(\propto-1\right)\left(\beta-1\right)=\propto\beta-\propto-\beta+1   =∝β−1(∝+β)+1=−32−1(−52)+1=2=\propto\beta-1\left(\propto+\beta\right)+1=-\frac{3}{2}-1\left(-\frac{5}{2}\right)+1=2 
Equation
 ⟹x2−(−92)x+(2)=0(×2)=x2+9x+4=0\Longrightarrow x^2-\left(-\frac{9}{2}\right)x+\left(2\right)=0\left(\times2\right)=x^2+9x+4=0 

15

Eg4

If  ∝and β\propto and\ \beta  are the roots of the equation  x2+2x−1=0x^2+2x-1=0  find a quadratic equation who's roots are  1∝and 1β\frac{1}{\propto}and\ \frac{1}{\beta}  .
 x2+2x−1=0x^2+2x-1=0                 ∝+β=−ba=−21=−2\propto+\beta=\frac{-b}{a}=\frac{-2}{1}=-2  
a=1 , b=2 , c=-1                       ∝β=ca=−11=−1\propto\beta=\frac{c}{a}=\frac{-1}{1}=-1  
 Sum: 1∝+1β=β+∝∝β=(∝+β)∝β=−21=−2Sum:\ \frac{1}{\propto}+\frac{1}{\beta}=\frac{\beta+\propto}{\propto\beta}=\frac{\left(\propto+\beta\right)}{\propto\beta}=\frac{-2}{1}=-2   Pr⁡oduct:1∝×1β=1−1=−1\Pr oduct:\frac{1}{\propto}\times\frac{1}{\beta}=\frac{1}{-1}=-1  
Equation  ⟹x2−(−2)x+(−1)=0\Longrightarrow x^2-\left(-2\right)x+\left(-1\right)=0  
 x2+2x−1=0x^2+2x-1=0  

16

Eg 5

If  ∝ and β\propto\ and\ \beta  are the roots of the equation  2x2+3x−1=02x^2+3x-1=0  .Find the equation who's roots are  ∝βandβ∝\frac{\propto}{\beta}and\frac{\beta}{\propto}  .
 2x2+3x−1=02x^2+3x-1=0         ∝+β=−ba=−32\propto+\beta=\frac{-b}{a}=-\frac{3}{2}  
a=2, b=3 c=-1                         ∝β=ca=−12=−12\propto\beta=\frac{c}{a}=\frac{-1}{2}=-\frac{1}{2}  
 Sum:∝β+β∝=∝2+β2∝β=(∝+β)2−2∝β∝βSum:\frac{\propto}{\beta}+\frac{\beta}{\propto}=\frac{\propto^2+\beta^2}{\propto\beta}=\frac{\left(\propto+\beta\right)^2-2\propto\beta}{\propto\beta}   =(−32)2−2(−12)−12=−132=\frac{\left(-\frac{3}{2}\right)^2-2\left(-\frac{1}{2}\right)}{-\frac{1}{2}}=-\frac{13}{2}     E: x2−(−132)x+(1)=0 (×2)x^2-\left(-\frac{13}{2}\right)x+\left(1\right)=0\ \left(\times2\right)  
 Pr⁡oduct:(∝β)(β∝)=1 \Pr oduct:\left(\frac{\propto}{\beta}\right)\left(\frac{\beta}{\propto}\right)=1\                   2x2+13x+2=02x^2+13x+2=0  

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Quadratics

Alpha and Beta

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