Mathematics Dictionary

Factor theory method. Example : Solve x^2 - 4*x - 5 = 0
- Let F(x) = x^2 - 4*x - 5.
- Since F(-1) = (-1)^2 - 4*(-1) - 5 = 0.
- Hence (x + 1) is a factor of F(x). and x + 1 = 0 and x = -1.
- And other root is 5.
Factorize method.
- Find factors : (x - 5)*(x + 1) = 0.
- Hence x = 5 or x = -1.
Quadratic formula. Example Solve x^2 - 6*x + 8 = 0
- Formula :
  - x = (-b + Sqr(b^2 - 4*a*c))/(2*a).
  - x = (-b - Sqr(b^2 - 4*a*c))/(2*a).
- Submite a = 1, b = -6 and c = 8 into formula
  - x = (-(-6) + Sqr((-6)^2 - 4*1*8))/(2*1) =(6 + Sqr(4))/2 = 4.
  - x = (-(-6) - Sqr((-6)^2 - 4*1*8))/(2*1) =(6 - Sqr(4))/2 = 2.
Synthetic division. Example : Solve x^2 + x - 6 = 0.
- +1 +1 -6 | +2 ...... (Trail by 1, 2, 3, 6, -1, -2,-3, -6)
- +0 +2 +6 ........... (Add this line to the line above)
- ---------
- +1 +3 +0 ........... (Remainder is 0)
- Hence (x - 2) is factor. Othe factor is (x + 3).
- Hence (x - 2)*(x + 3) = 0. Hence x = 2 or x = -3.

Equation theory

Let r and s be te roots of a*x^2 + b*x + c = 0
Since a*x^2 + b*x + c = (x - r)*(x - s)
Hence sum of roots is r + s = -b/a.
Hence product of roots is r*s = c/a.

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AL 11 02. Solve cubic equations

1. Method to solve

Factor theory. If F(a) = 0 then (x - a) is a factor of F(x).
Synthetic division. See example 1 below.
Cubic formula method : It is not simple as quadratic formula.

2. Roots of cubic equations

1. It must have one real root with 2 complex roots.
2. It can have 3 different real roots.
3. It can have 3 same real roots.
4. It can have 1 real with 2 same real roots.

3. Cubic equation roots theory

Let r, s, t be roots of cubic equation a*x^3 + b*x^2 + c*x + d = 0.
Sum of roots : r + s + t = -b/a.
Combination of 2 roots : r*s + r*t + s*t = c/a.
Product of roots : r*s*t = -d/a.

Example 1 : Solve 11*x^3 + 32*x - 58*x + 5 = 0
Hint to use synthetic division

Start trail with 1/11, 5/11, 1 and 5

Solution

11 32 -58 +5 | 1/11
00 01 +03 -5
-------------
11 33 -55 0
Hence x1 = 1/11 is one root.
Hence (x - 1/11)*(11*x^2 + 33*x - 55) = 0
Hence 11*x^2 + 33*x - 55 = 0 or x^2 + 3*x - 5 = 0
Using quadratic formula we have
x2 = (-3 + Sqr(29))/2 and x3 = (-3 - Sqr(29))/2

Verify using equation theory

Sum of roots : x1 + x2 + x3 = -32/11
Product of roots : x1*x2*x3 = -5/11

Example 2 : x^3 - 18*x^2 + k*x + 91 = 0 has one rational roots and k is integer. Find k.

Since k = -x^2 + 18*x - 91/x and k is integer.
Hence x should be a factor of 91. That is x = -1, 1, -7, 7, or -13, 13 are roots.
Hence k has six answers. If x = 13 then k = -169 + 234 - 7 = 58.

Example 3 : Solve x^3 - 1 = 0.

Since x^3 - 1 = (x - 1)*(x^2 + x + 1) = 0.
Hence x - 1 = 0 or x = 1.
Or x^2 + x + 1 = 0. Using quadratic formula, we have
x = (-1 + Sqr(-3))/2 = (-1 + i*Sqr(3))/2
x = (-1 - Sqr(-3))/2 = (-1 - i*Sqr(3))/2

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AL 11 03. Solve quartic equation

Text and examples

Study subject | Quartic Equations
Objectives
- Equation theory of quartic equation
- Solve quartic equation in polynomial form
- Solve quartic equation in factor form
- Solve special equation : x^4 + x^3 + x^2 + x + 1

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AL 11 04. Solve quaint equation

Method to solve

Factor theory.

Synthetic division.

Study subject | Solve x^5 +3*x^4 - 5*x^3 - 15*x^2 + 4*x + 12 = 0

Equation theory

Sum of roots = p + q + r + s + t = -b/a.
Combination of 2 roots = p*q + p*r + p*s + p*t + q*r + q*s + q*t
+ r*s + r*t + s*t = +c/a.
Combination of 3 roots = p*q*r + p*q*s + p*q*t + p*r*s + p*r*t + p*s*t
+ q*r*s + q*r*t + q*s*t + r*s*t = -d/a.
Combination of 4 roots = p*q*r*s + p*q*r*t + p*q*s*t + p*r*s*t + q*r*s*t = e/a
Product of roots = p*q*r*s*t = -f/a.

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AL 11 05. Equations with absolute operation

Samples on internet

Study subject Linear function with absolute oberation.
Study subject Quadratic function with absolute oberation.

Application program on internet

Program ABH
Example : Solve Abs(x^2 - 6*Abs(x) + 8) = 1.
- Cilck start.
- Click subject 7 in the upper box.
- Click program number 5 in lower box.
- Type in 1, -6, 8, 1.

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AL 11 06. Special polynomial equations : Solve x^n + 1 = 0

DeMoivre's theory

Study Subject | DeMoivre's Theory

Example : Solve x^5 + 1 = 0 using DeMovire's theory .

Since x^5 = -1 = cos(pi) + i*sin(pi)
Hence x1 = cis((2*0*pi+pi)/5) = cis(036)
Hence x2 = cis((2*1*pi+pi)/5) = cis(108)
Hence x3 = cis((2*2*pi+pi)/5) = cis(180)
Hence x4 = cis((2*1*pi+pi)/5) = cis(252)
Hence x5 = cis((2*1*pi+pi)/5) = cis(324)

Example : Solve x^5 + 1 = 0 by construction.

Draw a large unit circle with axese 0x and 0y and O is the center.
For example let 10 cm = one unit.
Draw an angle 036 degrees with x-axis and meet circle at A(a1,b1).
Measure a1 and b1. Hence x1 = a1 + i*b1.
Draw an angle 108 degrees with x-axis and meet circle at B(a2,b2).
Measure a2 and b2. Hence x2 = a2 + i*b2.
Draw an angle 180 degrees with x-axis and meet circle at B(a3,b4).
Measure a3 and b3. Hence x3 = -1.
Draw an angle 252 degrees with x-axis and meet circle at B(a4,42).
Measure a4 and b4. Hence x4 = a4 + i*b4.
Draw an angle 324 degrees with x-axis and meet circle at B(a5,b5).
Measure a5 and b5. Hence x5 = a5 + i*b5.

Using equation theory prove that

cos(36) + cos(108) + cos(180) + cos(252) + cos(324) = 0
sin(36) + sin(108) + sin(180) + sin(252) + sin(324) = 0

Solve x^4 - x^3 + x^2 + x - 1 = 0

Since (x^5 + 1) = (x + 1)*(x^4 - x^3 + x^2 + x - 1)
Hence the solutions are given in x^5 + 1 = 0.
The roots are cis(36), cis (108), cis(252) and cis(324).

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AL 11 07. Special polynomial equations : Solve x^n - 1 = 0

DeMoivre's theory

Study Subject | DeMoivre's Theory

Example : Solve x^5 - 1 = 0 using DeMovire's theory .

Since x^5 = 1 = cos(0) + i*sin(0)
Hence x1 = cis((2*0*pi+0)/5) = cis(000)
Hence x2 = cis((2*1*pi+0)/5) = cis(072)
Hence x3 = cis((2*2*pi+0)/5) = cis(144)
Hence x4 = cis((2*1*pi+0)/5) = cis(216)
Hence x5 = cis((2*1*pi+0)/5) = cis(288)

Example : Solve x^5 - 1 = 0 by construction.

Draw a large unit circle with axese 0x and 0y and O is the center.
For example let 10 cm = one unit.
Draw an angle 000 degrees with x-axis and meet circle at A(a1,b1).
Measure a1 and b1. Hence x1 = 1.
Draw an angle 072 degrees with x-axis and meet circle at B(a2,b2).
Measure a2 and b2. Hence x2 = a2 + i*b2.
Draw an angle 144 degrees with x-axis and meet circle at B(a3,b4).
Measure a3 and b3. Hence x3 = a3 + i*b3.
Draw an angle 216 degrees with x-axis and meet circle at B(a4,42).
Measure a4 and b4. Hence x4 = a4 + i*b4.
Draw an angle 288 degrees with x-axis and meet circle at B(a5,b5).
Measure a5 and b5. Hence x5 = a5 + i*b5.

Using equation theory prove that

cos(72) + cos(144) + cos(216) + cos(288) = -1
sin(72) + sin(144) + sin(216) + sin(288) = 0

Solve x^4 + x^3 + x^2 + x + 1 = 0

Since (x^5 - 1) = (x - 1)*(x^4 + x^3 + x^2 + x + 1)
Hence the solutions are given in x^5 - 1 = 0.
The roots are cis(72), cis (144), cis(216) and cis(288).

Note

cis(A) = cos(A) + i*sin(A)

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AL 11 08. Solve two linear equations

Method

Ellimination
- Equations
  - a1*x + b1*y = c1 ........ (1)
  - a2*x + b2*y = c2 ........ (2)
- Solution
  - Elliminate y : (1)*b2 - (2)*b1
  - b2*a1*x - b1*a2*x = b2*c1 - b1*c2
  - Hence x = (b2*c1 - b1*c2)/(b2*a1 - b1*a2)
  - Elliminate x : (1)*a2 - (2)*a1
  - a2*b1*y - a1*b2*y = a2*b1 - a1*b2
  - Hence y = (a2*c1 - a1*c2)/(a2*b1 - a1*b2)
Substitution
- Equations
  - a1*x + b1*y = c1 ........ (1)
  - a2*x + b2*y = c2 ........ (2)
- Solution
  - From (1) y = (c1 - a1*x)/b1 ....... (3)
  - Sub (3) into (2) : a2*x + b2*(c1 - a1*x)/b1 = c2
  - Hence x*(a2 - a1*b2/b1) = c2 - b2*c1
  - Hence x*(a2*b1 - a1*b2) = b1*c2 - b2*c1
  - Hence x = (b1*c2 - b2*c1)/(a2*b1 - a1*b2)
  - Similary, we can find y.

Example 1.

Question
- F(x) = x^3 + 2*x^2 + a*x + b.
- F(x) divide by (x - 1) and remainder is 3.
- F(x) divide by (x + 2) and remainder is 1.
- Find a and b.
Solution
- F(+1) = 3. Hence 1^3 + 2*(1)^2 + a*1 + b = 3. Hence a + b = 0.
- F(-2) = 1. Hence (-2)^3 + 2*(-2)^2 + a*(-2) + b = 1. Hence -2*a + b = 1.
- Eliminate b we have -3*a = 1 and a = -1/3.
- Hence b = 1/3.
Verify : F(x) x^3 + 2*x^2 - x/3 + 1/3
- F(+1) = 1^3 + 2*(1)^2 - 1/3 + 1/3 = 3 which is satisfied.
- F(-2) = (-2)^3 + 2*(-2)^2 - (-2)/3 + 1/3 = 1 which is satisfied.

Example : Solve

Equations
- x + y = 3 ......... (1)
- x - y = 1 ......... (2)
Solution
- (1) + (2) we have 2*x = 4 and x = 2.
- Hence y = 1.
Graphic solution
- The answer is the intersection of y = 3 - x and y = x - 1.
- The intersection is (2,1).

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AL 11 09. Solve three linear equations

Ellimination

Equations
- a1*x + b1*y + c1*z = d1 ........ (1)
- a2*x + b2*y + c2*z = d2 ........ (2)
- a3*x + b3*y + c3*z = d3 ........ (3)
Solution
- Elliminate z from (1) and (2) : c2*(1) - c1*(2)
- Hence (c2*a1 - c1*a2)*x + (c2*b1 - c1*b2)*y = c2*d1 - c1*d2
- Hence A1*x + B1*y = C1 ......... (4)
- Elliminate z from (2) and (3) : c3*(2) - c2*(3)
- Hence (c3*a2 - c2*a3)*x + (c3*b2 - c2*b3)*y = c3*d2 - c2*d3
- Hence A2*x + B2*y = C2 ......... (5)
- From (4) and (5), we elliminate y
  - Hence x = (B2*C1 - B1*C2)/(B2*A1 - B1*A2)
  - Hence y = (A2*C1 - A1*C2)/(A2*B1 - A1*B2)
  - Sub x and y into (1), we can find z
  - Where A1 = c2*a1 - c1*a2, B1 = c2*b1 - c1*b2, C1 = c2*d1 - c1*d2
  - Where A2 = c3*a2 - c2*a3, B2 = c3*b2 - c2*b3, C2 = c3*d2 - c2*d3
Exercises
- Exercise : Find x in terms a1,a2,a3,b1,b2,b3,c1,c2,c3,d1,d2,d3.
- Exercise : Find y in terms a1,a2,a3,b1,b2,b3,c1,c2,c3,d1,d2,d3.
- Exercise : Find z in terms a1,a2,a3,b1,b2,b3,c1,c2,c3,d1,d2,d3.

Other methods in Chapter 12 of Computer Mathematics by Dr. K. G. Shih

Inverse matrix method.
Cramer's rule.
Gauss Ellimination.
Gauss Jordan Reduction.

Example 1 : Give 3 points (-1,2), (0,1) and (1,2). Find equation of parabola.

Let parabola be y = a*x^2 + b*x + c.
Point 1 : a*(-1)^2 + b*(-1) + c = 2 and a - b + c = 2 ........ (1)
Point 2 : a*(0^2) + b*0 + c = 1 and c = 1 .................... (2)
Point 3 : a*(1)^2 + b*1 + c = 2 and a + b + c = 2 ............ (3)
Sub. c = 1 into (1) we have a - b = 1 ........................ (4)
Sub. c = 1 into (3) we have a + b = 1 ........................ (5)
(4) + (5), we have 2*a = 2 and a = 2.
Hence b = 0.
Equation of parabola is y = x^2 + 1.

Example 2 : Give 3 points (0,1), (1,0) and (-1,0). Find equation of circle.

Let equation of circle be x^2 + y^2 + a*x + b*y + c = 0.
Point 1 : (0)^2 + (1)^2 + a*0 + b*1 + c = 0 and b + c = -1 ........ (1)
Point 2 : (1)^2 + (0)^2 + a*1 + b*0 + c = 0 and a + c = -1 ......... (2)
Point 3 : (-1)^2 + (0)^2 + a*(-1) + b*0 + c = 0 and -a + c = -1 .... (3)
(2) + (3) we have 2*c = -2 or c = -1 ............................... (4)
Hence b = 0 and a = 0.
Equation of circle is x^2 + y^2 = 1^2.
Center at (0,0) and radius = 1.

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AL 11 10. Special equations

Simultaneous equations

1. Equations

x + y = 3 ......... (1)
x*y = 2 ........... (2)

2. Equations

x - y = 1 ......... (1)
x*y = 2 ........... (2)

3. Equations

x^2 + y^2 = 4 ......... (1)
x*y = 1 ............... (2)

Diagram solutions
- Study subject Find answer in interception of curves

Example : Solve x^2 - 5*x + 2*Sqr(x^2 - 5*x + 3) = 12

(x^2 - 5*x + 3) - 3 + 2*Sqr(x^2 - 5*x + 3) = 12
Let Sqr(x^2 - 5*x + 3) = a
Hence a^2 + 2*a - 15 = 0
Hence (a + 5)*(a - 3) = 0.
For a = 3
- Sqr(x^2 - 5*x + 3) = 3
- Square both sides : x^2 - 5*x + 3 = 9
- (x - 1)*(x + 6) = 0
- Hence x = 1 and x = -6
For a = -5
- Sqr(x^2 - 5*x + 3) = -5
- Square both sides : x^2 - 5*x + 3 = 25
- x^2 - 5*x - 22 = 0
- use quadratic formula
- Hence x = (5 + Sqr(25 + 88))/2 and x = (5 - Sqr(25 + 88))/2

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AL 11 11. Equations including e^x and ln(x)

Solve e^(2*x) - e^x - 2 =0

Formula

e^(2*x) = (e^x)^2.
ln(e^x) = x.

Solution

(e^x)^2 - e^x - 2 = 0.
(e^x + 1)*(e^x - 2) = 0.
(e^x + 1) = 0. Since e^x is always positive, hence there is no real solution.
(e^x - 2) = 0.
- Hence e^x = 2.
- Take logarithm on both sides, we have ln(e^x) = ln(2).
- Hence x = ln(2)

Study subject Exponent

Solve 2*ln(e^x) + x^2 - 8 = 0

Formula
- 2*ln(a) = ln(a^2).
- ln(e^a) = a.
Solution
- 2*ln(e^x) = 2*x.
- Hence we have x^2 + 2*x - 8 = 0.
- (x + 4)*(x - 2) = 0.
- Hence x = 2 or x = -4.
Verify
- x = 2, we have 2*ln(e^2) + 2^2 - 8 = 2*2 + 2^2 - 8 = 0
- x = -4, we have 2*ln(e^(-4)) + (-4)^2 - 8 = -8 + 16 - 8 = 0
Reference
- Study subject Logarithm