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Education for Sustainable Development and Science Teaching
Submitted:
28 November 2023
Posted:
30 November 2023
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Where AB = c = , |
βββBC = a = |
βββAC = b |
Also, β BAC = β ACB = . |
AB= BC |
ββββ B = - ( β BAC + β BAC ) |
ββββ= |
= |
ββ= 2 |
ββ |
β |
β |
ββ |
b = |
Current Method: |
Let the square root of the be - for which x, y βR. Therefore |
βββ |
= x + y β 2 |
Comparing the both sides, |
Weβve , x + y = 12 |
also - 2 |
Divide 2sides by β 2, |
Square 2 sides , xy = 27 β¦β¦β¦β¦β¦..β¦..(ii) |
From (i), y = 12 β x β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(iii) |
Substitute equation (iii) in equation (ii), |
ββββx ( 12 -x ) = 27 |
ββββ12 x - = 27. |
βββ - 12 x + 27 = 0 |
βββ( x β 9 ) ( x β 3 ) = 0 |
Therefore, either x β 9 = 0, x β 3 = 0 |
x = 9, 3 |
We now need to find the values for y, therefore |
When x = 9, from (iii), y = 12 -9 = 3 . |
And when x = 3 , from (iii) , y = 12 β 3 = 9 |
Hence , the square root of the 12 β 6 are |
ββ - = |
= |
and + = |
= |
βββ΄ = ( ). |
Since AC cannot be negative, thens AC = () cm . [ is negative ]. |
Peter Chew . |
= |
Cause - 12x + 27 = 0, then x = 9, 3 |
βββ΄ = - |
= - |
β΄ Length AC = - |
Example: |
sum of 2 real number in surd form. |
Current Method, |
i) Solution 1: |
= |
βββββ= |
βββββ= |
βββββ |
ii) Solution 2: If be + |
βββββ = ( + |
= x + y + 2 |
Comparing the both sides, |
Weβve βx + y = 12 |
ββββy = 12- x β¦.. i) |
βalsoββxy =35 β¦.β¦.ii) |
From i) and ii), x (12 - x) = 35 |
βββββ - 12x + 35 = 0 |
ββββ( x β 7 ) ( x β 5 ) = 0 |
βββ Therefore,ββx = 7, 5 |
From i), If x = 7, y = 12- 7 = 5 |
βββIfβx = 5, y = 12- 5 = 7 |
ββββ΄ = + |
iii) Peter Chew Theorem, |
βCauseββ- 12x + 35 = 0, then x = 7, 5 |
βββ΄ = + |
Example 1 : Find the solution to the equation: . |
Solution : |
The auxiliary equation is: m 2 + 3 m + = 0 |
Current Method, |
If = - |
Then = ( - |
ββ΄ββ = x + y - 2 |
Comparing the both sides , |
Weβveββx + y = 9 |
ββββy = 9 - x β¦.. i) |
alsoββx y = 8ββ¦.β¦.ii) |
From i) and ii),βx (9- x) = 8 |
βββββ - 9x +8 = 0 |
ββββ(x - 8)(x - 1) = 0 |
x = 8, 1 |
From i), If x = 8, y = 9- 8 = 1 |
βββIf x = 1, y = 9- 1 = 8 |
βββ΄ = - |
βββββ |
iii) Peter Chew Theorem, |
Causeβ- 9x + 8 = 0, then x = 8, 1 |
βββ΄ = - |
βββββ |
Therefore |
βββ |
ββββ , |
ββββ , |
For , Solution To The Differential Equation is y = A + B |
Therefore y(x)= A + B |
Example 2 : Find the solution to the equation: . |
Solution : |
The auxiliary equation is:βm 2 + 20 m += 0 |
Current Method, |
If = - |
= ( - |
= x + y - 2 |
Comparing the both sides, |
Weβveββx + y = 100 |
βββββy = 100 - x β¦.. i) |
alsoββx y = 475ββ¦.β¦.ii) |
From i) and ii),βx (100 - x) = 475 |
ββββ - 100x +475 = 0 |
βββββββ(x - 95) (x - 5) = 0 |
x = 95, 5ββββββ |
From i), Ifβx = 95, y = 100 - 95 = 5 |
ββββIf x = 5, y = 100 - 5 = 95 |
ββββ΄ = - |
iii)βPeter Chew Theorem, |
βCauseββ- 100x + 475 = 0, then x = 95, 5 |
βββ΄ = - |
Fromβ |
βββ |
- - 10 , + -10 |
For , Solution To The Differential Equation is y = A + B |
Therefore y(x) = A + B |
Example 3 : Find the solution to the equation: . |
Solution : |
The auxiliary equation is: m 2 + 20 m += 0 |
ββββββββββββ |
ββββββββββββ = |
ββββββββββββ = |
Current Method, |
Ifβ= - |
= ( - |
ββββββ = x + y - 2 |
Comparing the both sides, |
Weβveβx + y = 3 844 |
ββββy = 3 844 - x β¦.. i) |
alsoββx y = 118 203ββ¦.β¦.ii) |
from i) and ii),βx (3 844 - x) = 118 203 |
ββ - 3 844 x + 118 203 = 0 |
(x β 3 813) (x - 31) = 0 |
x = 3 813, 31βββββ |
From i), Ifβx = 3 813, y = 3 844 - 3813 = 31 |
βββIfβx = 31 , y = 3 844 β 31 = 3 813 |
ββ β΄ = - |
iii)βPeter Chew Theorem, |
Causeβ- 100x + 475 = 0, then x = 95, 5 |
βββ΄ - 3 844 x + 118 203 = 0 = = - |
Fromβ |
- |
- - 62 , - + - 62 |
Forββ , Solution To The Differential Equation is y = A + B |
Thereforeβy(x) = A + B |
Example 4 : Find the solution to the equation: . |
Solution : |
The auxiliary equation is: 2m 2 + m += 0 |
ββββ |
= ββββ |
Current Method, |
Letββbe - |
β= ( - |
βββββ= x + y - 2 |
Comparing the both sides, |
βWeβveβx + y = 2 |
βββββy = 2 - x β¦.. i) |
βalsoβββx y = 5ββ¦.β¦.ii) |
Fromβi) and ii),ββx (2 - x) = 5 |
ββββββ - 2 x + 5 = 0 |
ββββββ |
ββββββ |
From i), Ifβx = 1+2i, y = 2 - (1+2i) = 1 - 2i |
βββIfβx = 1-2i , y = 2 - (1-2i) = 1 + 2i |
βββ΄ = - |
iii)βPeter Chew Theorem, |
Causeβ - 2 x + 5 = 0, then x = |
βββ΄ = - |
Fromβ= |
βββββ= |
βββββ= |
Forββm1 = Ξ± + Ξ² i, m2 = Ξ± - Ξ² i . let Ξ±= , Ξ² = |
SolutionβToβThe Differential Equation is y = ( A cos Ξ² x + B sin Ξ² x ) |
Thereforeβy(x) = y = [ A cos( x) + B sin) x ) ] |
Example 5 : Find the solution to the equation: . |
Solution : |
The auxiliary equation is: 2m 2 + m += 0 |
βββββ |
= βββββ |
Current Method, |
Letβbe - |
β= ( - |
βββββ= x + y - 2 |
Comparing the both sides, |
Weβveββx + y = 6 |
βββββy = 6 - x β¦.. i) |
alsoβββx y = 13ββ¦.β¦.ii) |
From i) and ii), x (6 - x) = 13 |
βββββββ6 x - = 13 |
βββββ - 6 x + 13 = 0 |
βββββ |
βββββ |
From i), Ifβx = 3+2i, y = 6 - (3+2i) = 3 - 2i |
βββIfβx = 3-2i , y = 6 - (3-2i) = 3 + 2i |
ββββ΄ = - |
iii)βPeter Chew Theorem, |
Causeβ - 6 x + 13 = 0, then x = |
βββ΄ = - |
Fromβ= |
βββββ= |
βββββ= |
Forββm1 = Ξ± + Ξ² i, m2 = Ξ± - Ξ² i . let Ξ±= , Ξ² = |
SolutionβToβThe Differential Equation is y = ( A cos Ξ² x + B sin Ξ² x ) |
Therefore ,βy(x) = y = [ A cos( x) + B sin) x ) ] |
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