ကိုအံ့မင္းညိဳ၏ ေမးခြန္းကို ေျဖၾကားျခင္း

Inequation နဲ႔ ပါတ္သက္ၿပီး ကိုအံ့မင္းညိဳက အခုလိုေမးပါတယ္။
x² + 2hx + c² ကုိရွင္းရင္ ရမယ့္ညီမွ်ျခင္းကုိ သိလုိပါတယ္...။ quadratic ကုိသံုးမယ္ဆုိရင္ တူညီပါ သလား..။ ဒါမွမဟုတ္ x = - h ± √(h²-c²) ကုိရပါသလား..။ ရွင္းျပေစခ်င္ပါတယ္ ခင္ဗ်ာ...။

ကိုအံ့မင္းညိဳ ေမးတာ x² + 2hx + c² = 0 ရဲ့ roots ကိုေမးတာလို႔ ထင္ပါတယ္။ ဒါကို ကၽြန္ေတာ္ completing square method ကိုသံုးၿပီး ရွင္းပါတယ္။

x² + 2hx + c² = 0 h,c Є R=the set of real numbers
x² + 2hx = - c²
x² + 2hx + h² = h²- c²
(x + h)² = h²- c²x + h = ± √(h²-c²)
x = - h± √(h²-c²)
If h² - c² 0, the roots are real.
If h²-c² the roots are complex.

Wolfram Alpha ၏ Result

Input:



Geometric figure:

Circular Cone

Integer Solution:









Solutions for the variable x:






Implicit derivatives:

Arithmetic Progression

Arithmetic Progression (A . P)

In a sequence the difference of any two consecutive terms is constant, then the sequence is called an arithmetic progression. That difference is called the common difference and is denoted by d.



Sequence တစ္ခုတြင္ ကပ္လွ်က္ရွိေသာ မည့္သည့္ကိန္းႏွစ္ခုမဆို ျခားနားျခင္း(ႏႈတ္ျခင္း) သည္ ကိန္းေသျဖစ္ လွ်င္ ၎ sequence ကို arithmetic progression ဟုေခၚသည္။



If u₁ , u₂ , u₃ , u₄ , - - -, u_n-1 , u_n is an A . P,then u₂ -u₁ = u₃ -u₂= u₄ -u₃ = - - - = u_n -u_n-1 = constant.

u_n -u_n-1 = d.

u_n = u_n-1 + d

Generally the first term (u₁₎ of a sequence is denoted by a.

Therefore ....

u₁ = a

u_n = u_n-1 + d

u₂ = u₁ + d = a + d

u₃ = u₂ + d = 2a + d

u₄ = u₃ + d = 3a + d

From above expression the n^th term of an arithmetic progression can be expressed as

un = a + (n-1)d

where ...

u_n = the n^th term

a = the first term

d = the common difference

n = number of term

Sequences and Series

Sequence



ေအာက္မွာေပးထားတဲ့ ကိန္းတန္းေလးကို ေလ့လာၾကည့္ရေအာင္။



1,4,9,16,25,...



အခုကိန္းတန္းမွာပါ၀င္တဲ့ ကိန္းလံုးေတြဟာ ေရးခ်င္သလို ေရးထားျခင္း (random) မဟုတ္ပါဘူး။ ကိန္းလံုးေတြ ေျပာင္းလဲမႈမွာ စနစ္တစ္ခု (တနည္းေျပာရရင္ function တစ္ခု) အရ ေျပာင္းလဲသြားတာပါ။ ဒါကို functional notation နဲ႔ေျပာရမယ္ဆိုရင္...



f(1) = 1 = 1²

f(2) = 4 = 2²

f(3) = 9 = 3²

f(4) = 16 = 4²

f(5) = 25 = 5² လို႔ဆိုႏိုင္တာေပါ။့



ဒါကိုၾကည့္ျခင္းအားျဖင့္ function ရဲ့ Domain ဟာ {1,2,3,4,5,...n}=the set of natural numbers ဆိုရပါမယ္။ ဒါဆိုရင္ အထက္ပါေျပာင္းလဲမႈကို ၾကည့္ရံုနဲ႔ function ရဲ့ general formula ကို အလြယ္တကူ ေျပာႏိုင္ပါၿပီ။

f(n) = n² ေပါ့...။



ဒါေၾကာင့္ sequence ဆိုတာဟာ special function လို႔ဆိုႏိုင္ၿပီး သူရဲ့ domain ကေတာ့ အၿမဲတမ္း သဘာ၀ကိန္း မ်ား ပါ၀င္ေသာအစု (the set of natural numbers) ျဖစ္ပါတယ္။ image ေတြကိုေတာ့ ဒီေနရာမွာ terms လို႔ ေျပာင္းလဲေခၚပါမယ္။ အေခၚအေ၀ၚ ေျပာင္းလဲမႈနဲ႔အတူ အသံုးျပဳမယ့္ symbols ေတြကိုလည္း ေျပာင္းလည္း သတ္မွတ္ပါတယ္။ အထက္မွာေပးထားတဲ့ ကိန္းလံုးေတြကို အခုလိုေခၚေ၀ၚ သတ္မွတ္ပါမယ္။

first term	=	u1	=	1
second term	=	u2	=	4
third term	=	u3	=	9
fourth term	=	u4	=	16
fifth term	=	u5	=	25
- - - - - - - -	- -	- -	- -	- -
nth term	=	un	=	n2

ဒီေနရာမွာ n^th term ကို general term သို႔မဟုတ္ general formula လို႔ဆိုႏိုင္ပါတယ္။ ဒါေၾကာင့္ sequence ကို အခုလို definition ဖြင့္ဆိုႏိုင္ပါတယ္။



Sequence

A sequence is a function whose domain is either the set of all or part of natural numbers. The values (images) of function are called terms .



Example 1

Find the first four terms of the sequence defined by u_n = 3n - 5.



Solution

u_n = 3n - 5

u₁ = 3(1) - 5 =-2

u₂ = 3(2) - 5 = 1

u₃ = 3(3) - 5 = 4

u₄ = 3(4) - 5 = 7

Therefore the fist four terms are -2, 1, 4, 7.



Exercises

Find the first four terms of the sequence defined by

(a) un = 2n + 3

(b) un = 3n2 - 2

(c) un = 4n2+ 3n - 5

Example2

Which term of the sequence defined by u_n = 4n - 23 is 25?



Solution

u_n = 4n - 23

Let the n^th term be 25.

Therefore u_n = 25

4n - 23 = 25

4n = 48

n = 12

Therefore the 12^th term is 25.

Algebraic Method to Find the Solution Set of a Quadractic Equation

Algebraic Method ကိုသံုးမယ္ဆိုရင္ ေအာက္ပါေပးထားတဲ့ logical rule ကို သိရပါမယ္။

(i) ab>0 then there are two different possibilities that

(ii) Similarly, for ab<0 the possibilities may be

(i)ab>0 ဆိုသည္မွာ a ႏွင့္ b ေျမႇာက္ျခင္း သည္ အေပါင္းတန္ဘိုး ျဖစ္သည္။ ထိုသို႔ျဖစ္ရန္ a ႏွင့္ b သည္ အေပါင္းတန္ဘိုး ျဖစ္ရမည္။ သို႔မဟုတ္ a ႏွင့္b ႏွစ္ခုလံုးသည္ အႏႈတ္တန္ဘိုး ျဖစ္ရမည္။

(ii) ab<0 ဆိုသည္မွာ a ႏွင့္ b ေျမႇာက္ျခင္း သည္ အႏႈတ္တန္ဘိုး ျဖစ္သည္။ ထိုသို႔ျဖစ္ရန္ a သည္ အေပါင္းတန္ဘိုးဟု ယူဆလွ်င္ b သည္ အႏႈတ္တန္ဘိုး ျဖစ္ရမည္။ သို႔မဟုတ္ a သည္ အႏႈတ္တန္ဘိုးဟု ယူဆလွ်င္ b သည္ အေပါင္းတန္ဘိုး ျဖစ္ရမည္။

Example 1
Use algebraic method, to find the solution set of the inequation 12 - 5x - 2x² ≥ 0 and illustrate it on the number line.

Solution
12 - 5x - 2x² ≥ 0
(4 + x)(3 - 2x) ≥ 0
အထက္တြင္ ေဖၚျပခဲ့ေသာ rule(i)ကို သံုး၍ ေျဖရွင္းပါမည္။

There are two possibilities for (4 + x)(3 - 2x) ≥ 0
(i)(4 + x) ≥ 0 and (3 - 2x) ≥ 0
x ≥ -4 and x ≤ 3/2
ႏွစ္မ်ိဳးလံုးကိုယူရန္မလို၊ အေျခအေန ႏွစ္ရပ္လံုးကို ေျပလည္ေစေသာ အေျဖတစ္ခုကိုသာ ယူမည္။ မည္သည္ကို ယူမည္ ဆိုသည္ကို number line ဆြဲ၍ စဥ္းစားႏိုင္သည္။

အေျခအေန ႏွစ္ရပ္လံုးတြင္ ဘံုျဖစ္ေသာ -4 ≤ x ≤ 3/2ကိုသာ ယူပါမည္။




(ii)(4 + x) ≤ 0 and (3 - 2x) ≤ 0
x ≤ -4 and x ≥ 3/2
အထက္ပါနည္းအတိုင္း number line ဆြဲ၍ စဥ္းစားမည္။

အေျခအေန ႏွစ္ရပ္လံုးတြင္ ေျပလည္ေစေသာ ဘံုအေျဖ မရွိပါ။



The solution set of 12 - 5x - 2x² ≥ 0 is {x/-4 ≤ x ≤ 3/2}.

Example 2

Find the solution set of the inequation 12x² ≥ 10 - 7x by graphical method and illustrate it on the number line.

Solution
12x² ≥ 10 - 7x
12x² + 7x - 10 ≥ 0
(4x + 5)(3x - 2)≥ 0

There are two possibilities for (4 + x)(3 - 2x) ≥ 0
(i)(4x + 5) ≥ 0 and (3x - 2) ≥ 0
Therefore x ≥ - 5/4 and x ≥ 2/3


The solution which satisfies both conditions is
x ≥ 2/3.





(ii)(4x + 5) ≤ 0 and (3x - 2) ≤ 0
Therefore x ≤ - 5/4 and x ≤ 2/3

The solution which satisfies both conditions is
x ≤ - 5/4.

The solution set of 3x² < x² - x + 3 is {x/ x ≤- 5/4 or x ≥ 2/3}.

Graphical Method to Find the Solution Set of a Quadractic Equation

1. ေပးထားေသာ Function ကို y ဟုထားပါ။
2. y=0 ထားၿပီး X-axis ျဖတ္မွတ္မ်ားကို ရွာပါ။
3. x=0 ဟုထားၿပီး Y-axis ျဖတ္မွတ္ကို ရွာပါ။
4. ျဖတ္မွတ္မ်ားကို အသံုးျပဳၿပီး smooth parabola ဆြဲပါ။
5. ေပးေထားေသာ inequation sign ကို ၾကည့္ၿပီး solution set ကို ဆံုးျဖတ္ပါ။

Example 1

Use a graphical method, to find the solution set of the inequation 12 - 5x - 2x² ≥ 0 and illustrate it on the number line.





Solution

Let y=12 - 5x - 2x²

When y = 0,

12 - 5x - 2x² = 0

(4 + x)(3 - 2x) = 0

x = -4 or x = 3/2

Therefore, the graph cuts the X-axis at (-4,0) and (3/2,0).

When x = 0, y = 12

Therefore, the graph cuts the Y-axis at (0,12).



The solution set of 12 - 5x - 2x² ≥ 0 is {x/-4 ≤ x ≤ 3/2}.









Example 2

Find the solution set of the inequation 3x² < x² - x + 3 by graphical method and illustrate it on the number line.



Solution

3x² < x² - x + 3

2x² + x - 3 <0

Let y=2x² + x - 3

When y = 0,

2x² + x - 3 = 0

(2x + 3)(x - 1) = 0

x = - 3/2 or x = 1

Therefore, the graph cuts the X-axis at (-3/2,0) and (1,0).

When x = 0, y = -3

Therefore, the graph cuts the Y-axis at (0,-3).



The solution set of 3x² < x² - x + 3 is {x/-3/2 < x <1}.>







Example 3

Find the solution set of the inequation 12x² ≥ 10 - 7x by graphical method and illustrate it on the number line.



Solution

12x² ≥ 10 - 7x

12x² + 7x - 10 ≥ 0

Let y=12x² + 7x - 10

When y = 0,

12x² + 7x - 10= 0

(4x + 5)(3x - 2) = 0

x = - 5/4 or x = 2/3

Therefore, the graph cuts the X-axis at

(- 5/4,0) and (2/3,0).

When x = 0, y = -10

Therefore, the graph cuts the Y-axis at (0,-10).



The solution set of 3x² < x² - x + 3 is {x/ x ≤- 5/4 or x ≥ 2/3}.

Inequations

ဒီအခန္းမွာေတာ့ Quadratic Inequations ေတြရဲ့ ေပးထားေသာ အေျခအေနကို ေျပလည္ေစမယ့္ solution set ကိုရွာမွာ ျဖစ္ပါတယ္။ Quadratic Inequations ေတြကို မေျဖရွင္းမီ Quadratic Function ဆိုတာကို သိရပါမယ္။

Quadratic Function

An expression f(x) = ax² + bx + c, where a, b, and c are numbers with a≠0 is

called a quadratic function.

ကိန္းရွင္တစ္ခုရဲ့ ႏွစ္ထပ္ကိန္း ပါ၀င္ေသာ function တစ္ခုကို quadratic function လို႔ေခၚပါတယ္။ Quadratic Function ရဲ့ characteristic က x² ပါ။ x² မပါရင္ quadratic function လို႔ မေျပာႏိုင္ပါဘူး။ ဒါေၾကာင့္ coefficient of x² ဟာ 0 မျဖစ္ရပါဘူး။

Quadratic Equation

ax² + bx + c = 0 is called a quadratic equation.

Solutions or Roots of Quadratic Equation





where a≠0









Graph of a Quadratic Function

The graph of a quadratic function is called a parabola. It is basically a curved shape opening up or down.



Quadratic Function တစ္ခုရဲ့ graph ကို parabola လို႔ေခၚပါတယ္။



When you have a quadratic function in the form f(x) = ax² + bx + c

if a > 0, then the parabola opens up ,

coefficient of x² ဟာ positive(a>0) ျဖစ္မယ္ဆိုရင္ graph ဆြဲတဲ့ အခါ open upward parabola ကိုရပါတယ္။

+x² => open upward parabola

if a <0, then the parabola opens down

coefficient of x² ဟာ positive(a<0) ျဖစ္မယ္ဆိုရင္ graph ဆြဲတဲ့ အခါ open downward parabola ကိုရပါတယ္။

- x² =>open downward parabola

Quadratic Inequations

The open sentences ax² + bx + c>0 and ax² + bx + c<0,a ≠ 0 are quadratic inequations in x.

The solution set of the quadratic inequations in x can be found by

(i) Algebraic method

(ii) Graphical method

The Binomial Theorem

Observe the following expansion.



(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵



This expression can also be expressed as follows:













In general,if n is a positive integer,





















This expression is known as the Binomial Theorem.



Where the general coefficient or the coefficient of (r + 1)^th term is denoted as follows:

Note:

For the expansion of (x + y)ⁿ,where n is a positive integer,



(i) the binomial coefficients are all integers.

(ii) ⁿC₀ =ⁿC_n =1, ⁿC_n =ⁿC_n-1 = n, ⁿC_r =ⁿC_n-r .

The general term or the (r + 1)^th term of the expansion of (x + y)ⁿ is determined as follows:

(r + 1)^th term = ⁿC_r x^n-r y^r

Example 1

If the coefficients of x⁴ and x⁵ in the expansion of (3 + kx)¹⁰ are equal, find the value k.

(3 + kx)¹⁰ ကိုျဖန္႔္လွ်င္ x⁴ ၏ေျမွွာက္ေဖၚကိန္းသည္ x⁵ ၏ေျမွွာက္ေဖၚကိန္းႏွင့္ တူညီလွ်င္ k ၏တန္ဖိုးကို ရွာပါ။



Solution

(3 + kx)¹⁰

(r + 1)^th term = ¹⁰C_r 3^10-r (kx)^r = ¹⁰C_r 3^10-r k^rx^r



Here ¹⁰C_r 3^10-r k^r is the coefficient of x^r.

coefficient of x⁴ = ¹⁰C₄ 3⁶ k⁴

coefficient ofx⁵ = ¹⁰C₅ 3⁵ k⁵



By the problem,

¹⁰C₄ 3⁶ k⁴ = ¹⁰C₅ 3⁵ k<sup>4

10</sup>C₄ 3⁶ k⁴ = ¹⁰C₄ (6/5) 3⁵ k<sup>5

3 = (6/5)k

k = 5/2</sup>

Example 2

In the expansion of ,a x⁷ is four times the coefficient of x¹⁰. Find the value a.

ကိုျဖန္႔္လွ်င္ x⁷ ၏ေျမွွာက္ေဖၚကိန္းသည္ x¹⁰ ၏ေျမွွာက္ေဖၚကိန္း၏ 4 ဆျဖစ္လွ်င္ a ၏တန္ဖိုးကို ရွာပါ။



Solution







(r + 1)^th term =⁸C_r a^rx^16-3r

To get the coefficient of x⁷, 16-3r = 7 and r =3

To get the coefficient of x¹⁰, 16-3r = 10 and r =2



Therefore,coefficient of x⁷ = ⁸C₃ a³

coefficient of x¹⁰ = ⁸C₂ a²



By the problem,

⁸C₃ a³ = 4 ⁸C₂ a<sup>2

8</sup>C₂ (6/3) a³ = 4 ⁸C₂ a²

a = 2