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Educational Level Secondary school, High school and College
Program Purpose Provide step by step solutions of your problems using online calculators (online solvers)
Problem Source Your textbook, etc

1 . Addition, subtraction, multiplication, division of two polynomials eg.`(X^3-4X^2+4X-8)` `xx` `(X-2)`
2 . Factor of an equation eg. (1) `25x^2-36`, (2) `4x^2+12xy+9y^2`, (3) `x^3-3x^2-6x+8`, (4) `a^2(b-c)+b^2(c-a)+c^2(a-b)`
3 . Expand of an algebraic expression eg. (1) `(X+2)(X+3)`, (2) `(2X+3Y)^2`, (3) `(2X+3Y+4Z)^2`, (4) `102^2`, (5) `102 xx 106` (Expand using the identity)

4 . Complete square, Is perfact square, Find missing term
1. Convert the given equation into perfact square form eg. `9x^2+6x+1= 9( x+1/3 )^2`
2. Check a given equation is perfact square or not eg. (1) `x^2-4xy+4y^2`, (2) `3x^2+5x+2`
3. Find the missing term so the given equation is perfect square eg. (1) `9x^2` - __ + 16, (2) __ + `12x^2` + 9, (3) `49x^2` + 56 xy + __

5 . GCD-LCM of Polynomials eg. Find GCD, LCM of `(2X^2-4X), (3X^4-12X^2), (2X^5-2X^4-4X^3)`

6 . Rational Expressions
1. Reduced terms eg. `(4X^2-25)/(8X^3-125)`
2. Addition, Subtraction, Multiplication and Division of Rational Polynomials eg. `(X-3)/(X+1)-(X-6)/(X)`

7 . Solve or simplify linear equation
eg. (1) `(4x+1)/(3x-2)=3/2`, (2) `(2x-1)/(3x+1)+(4x-1)/(6x)=0`

8 . Quadratic Equation
1.1 Roots by factorization, eg. (1) `25x^2-30x+9=0`, (2) `x^2+10x-56=0`
1.2 Roots by Perfact Square, eg. (1) `25x^2-30x+9=0`, (2) `x^2+10x-56=0`
1.3 Discriminant & Nature of Roots eg. (1) `25x^2-30x+9=0`, (2) `x^2+10x-56=0`
2. Form quadratic equation using given two roots eg. (1) `alpha=3, beta=-4`, (2) `alpha=1+3sqrt(2), beta=1-3sqrt(2)`
3. Roots for non-zero denominator eg. (1) `(5x-18)/(x+2)=(2x-6)/(x-1)`, (2) `(x)/(x+1)+(x+1)/(x)=5/2`, (3) `4((4x+1)/(4x-1))^(2)+(4x+1)/(4x-1)=3`, (4) `(4x+1)/(4x-1)+(4x-1)/(4x+1)=3`
4. Roots of non-quadratic equation eg. (1) `6(x^2+1/x^2)-25(x-1/x)+12=0`, (2) `(x^2+1/x^2)-8(x+1/x)+14=0`

9 . Solve linear equation in two variables using (eg. Solve `7y+2x-11=0` and `3x-y-5=0` using Substitution method)
1. Substitution method
2. Elimination method
3. Cross multiplication method
4. Addition-Subtraction method
5. Inverse matrix method
6. Cramer's Rule method
7. Graphical method

10 . Variation
1. Find value of variation using given value
(1) x `prop` y and x=6 when y=3. Find y=? when x=18, (2) x `prop` `y/z`. x=8 when y=4 and z=3. Find x=? when y=6 and z=4.
2. Prove results for given variation
(1) If x `prop` y then prove that `x^3+y^3 prop x^2y-xy^2`, (2) If 3x-5y `prop` 5x+6y then prove that x `prop` y.

11 . Expansion: `X+1/X` to `X-1/X`
1. If `x-1/x=6` then find (1) `x^2+1/x^2` (2) `x+1/x` (3) `x^2-1/x^2`
2. If `x+y=5` and `xy=6` then find `x^2+y^2`
3. If `x^2+y^2+z^2=29` and `xy+yz+zx=-14` then find `x+y+z`

12 . Interval notation and Builder set notation eg. (1) `3 <= x <=7`, x is odd. (2) `|x^3-2| <= 25`, x in Z. (3) `x in[2,8)`

13 . Set Theory eg. `A={x<=5; x in N}, B={2<=x<=8; x in N}, C={x^3-3x^2-4x=0},` Find
1. Union eg. `A uu (B uu C)=(A uu B) uu C`
2. Intersection eg. `A nn (B uu C)=(A nn B) uu (A nn C)`
3. Complement eg. `(A uu B)'=A' nn B'`
4. Power set(Proper Subset) eg. `P(A)`
5. Minus eg. (1)`A-B`, (2) `A-(B uu C)=(A-B) nn (A-C)`
6. Cross Product eg. `A xx B`
7. Prove that any two expression is equal or not eg. `A-(B uu C)=(A-B) nn (A-C)`
8. Cardinality of a set eg. `n(A)`
9. is Belongs To a set eg. `2inB` ?
10. is Subset Of a set eg. `AsubB` ?
11. is two set Equal or not eg. `A=B` ?

14 . Functions
1. Find Range of `f:A->B` eg. 1. `f(x)=5x+2` where `A={1<=x<5}`, 2. `f(x)=sqrt(x)` where `A={1,4,16,36}`
2. Composite functions eg. `f(x)=2x+1`, `g(x)=x+5`. Find `fog(x)`, `gof(x)`, `fog(2)`, `f(-3)`, `g(4)`
3. Find value eg. 1. `f(x)=x(x+1)(2x+1)`. Find `f(x)-f(x-1)`, 2. `f(x)=x^2-2^x`. Find `f(2)-f(0)`

15 . Descartes' rule of signs eg. `x^5-x^4+3x^3+9x^2-x+5`

16 . Solve linear equation of any number of variables Inverse matrix method, Gauss elimination method eg. Solve `2x+y+z=5, 3x+5y+2z=15, 2x+y+4z=8` using Inverse matrix method

17 . Mathematical Logic eg. Prepare the truth table `p^^(qvvr)=(p^^q)vv(p^^r)`
18 . Boolean Algebra eg. `D_(6)`, `D_(9)` is a boolean algebra ?
1. Factor of an equation
Type-1 (Taking common) Examples...
(1) ax + a + 2x + 2
    =(ax + a) + (2x + 2)
    =a(x + 1) + 2(x + 1)
    =(x + 1)(a + 2)

Type-2 (Difference of squares) Examples...
(1) 25x2 - 36
    =(5x)2 - (6)2
    =(5x - 6)(5x + 6)

Type-3 (Sum and Difference of cubes) Examples...
(1) x3 + 27
    =(x)3 + (3)3
    =(x + 3)(x2 - (x)(3) + (3)2)
    =(x + 3)(x2 - 3x + 9)

Type-4 (Whole square of a bionomial) Examples...
(1) 4x2 + 12xy + 9y2
    =(2x)2 + 2(2x)(3y) + (3y)2
    =(2x + 3y)2

Type-5 (Splitting the middle term of a Quadratic Equation) Examples...
(1) x2 + 10x + 24
    =x2 + 4x + 6x + 24
    =x(x + 4) + 6(x + 4)
    =(x + 4)(x + 6)

Type-6 (Whole square of a trinomial) Examples...
(1) 4x2 + y2 + 1 + 4xy + 4x + 2y
    =(2x)2 + (y)2 + (1)2 + 2(2x)(y) + 2(2x)(1) + 2(y)(1)
    =(2x + y + 1)2

Type-7 (Factorization with the help of factor theorem) Examples...
(1) x3 - 3x2 - 6x + 8
    Here p(x)=x3 - 3x2 - 6x + 8
    sum of coefficient of all the terms of p(x) = 1 - 3 - 6 + 8 = 0
    \ (x-1) is a factor of p(x).
    Now, p(x) = x3 - 3x2 - 6x + 8
    =x3 - x2 - 2x2 + 2x - 8x + 8
    =x2(x - 1) - 2x(x - 1) - 8(x - 1)
    =(x - 1)(x2 - 2x - 8)
    =(x - 1)(x - 4)(x + 2)

Type-8 Cyclic Expressions Examples...
(1) a2(b - c) + b2(c - a) + c2(a - b)
     = a2b - a2c + b2c - b2a + c2a - c2
     = a2b - a2c + cb2 - ab2 + ac2 - bc2 
     = a2b - a2c - ab2 + ac2 + b2c - bc2 
     = a2(b - c) - a(b2 + c2) + bc(b - c) 
     = a2(b - c) - a(b - c)(b + c) + bc(b - c) 
     = (b - c)(a2 - a(b + c) + bc) 
     = (b - c)(a2 - ab - ac + bc) 
     = (b - c)(a(a - b) - c(a - b)) 
     = (b - c)(a - b)(a - c) 
     = -(a - b)(b - c)(c - a)
2. Expand of an algebraic expression
Option-1 => Expression Like
1. (X + 2)(X + 3)
2. (X + Y)(X2 - XY + Y2)
3. (2X + 3Y + 4Z)2
4. (3Y - 2X)3
5. (2X + 3Y)2
6. (X + Y)2 - (X - Y)2
7. (X + Y + Z)2 + (X + Y - Z)2
8. (3X - 5Y)3 + (5Y - 9Z)3 + (9Z - 3X)3

1. Expand an Algebraic Expressions (X+2)(X+3)

Using The Identity,
(X + A)(X + B) = X2 + (A + B)X + AB
Here X=X,A=2, B=3
= (X)2 + (2 + 3)X + (2)×(3)
= X2+5X+6

2. Expand an Algebraic Expressions (2X+3Y+4Z)2

Using The Identity,
(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CA
Here A = 2X, B = 3Y, C = 4Z
= (2X)2 + (3Y)2 + (4Z)2 + 2(2X)(3Y) + 2(3Y)(4Z) + 2(4Z)(2X)
= 4X2+9Y2+16Z2+12XY+24YZ+16XZ


Option-2 => Numeric Expression Like
1. 1022, 0.232, 9.82
2. 123, 12.23
3. 93, 8.93
4. 102 × 106
5. 342 - 322, 3.42 - 3.22
6. 103 - 73 - 33
7. 43 + 33, 3.43 - 2.93

1. Find Value of 102 × 106

Using, The Identity (X+A)(X+B) = X2 + (A+B)X + AB
Here X = 100, A = 2 and B = 6
102 × 106 = (100 + 2)(100 + 6)
= 1002 + (2 + 6)100 + 2 × 6
= 10000 + 800 + 12
= 10812

2. Find Value of 342 - 322

Using, The Indentity A2 - B2 = (A-B)(A+B)
Here A = 34 and B = 32
Here 342 - 322 = (34-32)(34+32)
= (2)(66)
= 132

 
3. Expansion: `X + 1/X` to `X - 1/X`
1. If `x + 1/x = 2` then find `x^2 + 1/x^2`

2. If `x - 1/x = 6` then find
(1) `x^2 + 1/x^2` (2) `x^4 + 1/x^4` (3) `x^3 + 1/x^3` (4) `x + 1/x`
(5) `x^2 - 1/x^2` (6) `x^4 - 1/x^4` (7) `x^3 - 1/x^3`

3. If `x^2 + 1/x^2 = 23` then find (1) `x + 1/x` (2) `x^2 + 1/x^2`

4. If `x + y = 5` and `xy = 6` then find
(1) `x^2 + y^2` (2) `x^3 + y^3` (3) `x^4 + y^4` (4) `x - y`
(5) `x^2 - y^2` (6) `x^3 - y^3` (7) `x^4 - y^4`

5. If `x + y = 3` and `x - y = 15` then find `x^2 + y^2` and `xy`

6. If `x + y + z = 9` and `x^2 + y^2 + z^2 = 29` then find `xy + yz + zx`

7. If `x^2 + y^2 + z^2 = 29` and `xy + yz + zx = -14` then find `x + y + z`


1. If `X+1/X = 2`, then find `X-1/X`

Here `X + 1/X = 2`


Now, We know that
`( X - 1/X )^2 = ( X + 1/X )^2 - 4`

`( X - 1/X )^2 = 2^2 - 4`

`( X - 1/X )^2 = 4 - 4`

`( X - 1/X )^2 = 0`

`X - 1/X = 0`



2. If `X+Y = 5` and `X-Y = 1`, then find `X^2+Y^2`

Here `X + Y = 5` and `X - Y = 1`


Now, We know that
`4 XY = ( X + Y )^2 - ( X - Y )^2`

`4 XY = 5^2 - 1^2`

`4 XY = 25 - 1`

`4 XY = 24`

`XY = 24/4`

`XY = 6`


Now, We know that
`X^2 + Y^2 = ( X + Y )^2 - 2 XY`

`X^2 + Y^2 = 5^2 - 2 * 6`

`X^2 + Y^2 = 25 - 12`

`X^2 + Y^2 = 13`



3. If `X+Y+Z = 1` and `X^2+Y^2+Z^2 = 29`, then find `XY+YZ+ZX`

Here `X+Y+Z = 1` and `X^2 + Y^2 + Z^2 = 29`


Now, We know that
`(X+Y+Z)^2 = (X^2 + Y^2 + Z^2) + 2(XY+YZ+ZX)`

`2 (XY+YZ+ZX) = (X+Y+Z)^2 - (X^2 + Y^2 + Z^2)`

`2 (XY+YZ+ZX) = 1^2 - (29)`

`2 (XY+YZ+ZX) = 1 - (29)`

`2 (XY+YZ+ZX) = -28`

`(XY+YZ+ZX) = -28/2`

`(XY+YZ+ZX) = -14`



4. Addition, subtraction, multiplication, division of two polynomials
Find addition, Subtraction, multiplication, division 4X3-3X2+2X-4 and X+1

1. Addition

(X4 - 4X3 - 4X + 4) + (X2 + 2X - 2)
= X4 - 4X3 - 4X + 4 + X2 + 2X - 2
= X4 - 4X3 + X2 - 4X + 2X + 4 - 2
= X4 - 4X3 + X2 - 2X + 2

OR
X4-4X3-4X+4
+
X2+2X-2

X4-4X3+X2-2X+2


2. Subtraction

(4X3 - 3X2 + 2X - 4) - (X + 1)
= 4X3 - 3X2 + 2X - 4 - X - 1
= 4X3 - 3X2 + 2X - X - 4 - 1
= 4X3 - 3X2 + X - 5

OR
4X3-3X2+2X-4
-
-X+-1

4X3-3X2+X-5


3. Multiplication

(4X3 - 3X2 + 2X - 4) (X + 1)
= + 4X3(X + 1) - 3X2(X + 1) + 2X(X + 1) - 4(X + 1)
= 4X4 + 4X3 - 3X3 - 3X2 + 2X2 + 2X - 4X - 4
= 4X4 + 4X3 - 3X3 - 3X2 + 2X2 + 2X - 4X - 4
= 4X4 + 4X3 - 3X3 - 3X2 + 2X2 + 2X - 4X - 4
= 4X4 + X3 - X2 - 2X - 4

4. Division

  
4X2-7X+9
  
  
  
X + 1|
 4X3- 3X2+ 2X- 4
  
 |
-4X3+-4X2
 (X + 1) (4X2)
 |
  
 |
- 7X2+ 2X- 4
  
 |
-+7X2-+7X
 (X + 1) (- 7X)
 |
  
 |
 9X- 4
  
 |
-9X+-9
 (X + 1) (9)
 |
  
 |
- 13
  

 
Here, Divisor = X + 1
Dividend = 4X3 - 3X2 + 2X - 4
Quotient = 4X2 - 7X + 9
Remainder = - 13
 
5. GCD-LCM of Polynomials
Gives you Greatest Common Divisor (Or Highest Common Factor) and Least Common Multiple.

1. Find GCD, LCM of 30(X^2-3X+2) and 50(X^2-2X+1)

Factor of 30(X^2-3X+2)
= 30(X2 - 3X + 2)
= 30(X2 - X - 2X + 2)
= 30(X(X - 1)+(- 2)(X - 1))
= 30(X - 2)(X - 1)
= 2 × 3 × 5 (X - 2)(X - 1)
= 2 × 3 × 5 (X - 2)(X - 1)

Factor of 50(X^2-2X+1)
= 50(X2 - 2X + 1)
= 50(X - 1)(X - 1)
= 50(X - 1)2
= 2 × 5 × 5 (X - 1)2
= 2 × 5 × 5 (X - 1)2


GCD = 2 × 5 (X - 1)
GCD = 10(X - 1)

LCM = 2 × 3 × 5 × 5 (X - 2)(X - 1)2
LCM = 150(X - 2)(X - 1)2

2. Find GCD, LCM of (2X^2-4X), (3X^4-12X^2), (2X^5-2X^4-4X^3)

Factor of (2X^2-4X)
= (2X2 - 4X)
= 2X(X - 2)
= 2 X(X - 2)

Factor of (3X^4-12X^2)
= (3X4 - 12X2)
= 3X2(X2 - 4)
= 3X2(X - 2)(X + 2)
= 3 X2(X - 2)(X + 2)

Factor of (2X^5-2X^4-4X^3)
= (2X5 - 2X4 - 4X3)
= 2X3(X2 - X - 2)
= 2X3(X2 + X - 2X - 2)
= 2X3(X(X + 1)+(- 2)(X + 1))
= 2X3(X - 2)(X + 1)
= 2 X3(X - 2)(X + 1)


GCD = X(X - 2)

LCM = 2 × 3 X3(X - 2)(X + 2)(X + 1)
LCM = 6X3(X - 2)(X + 2)(X + 1)

6. Rational Expressions
1. Find Reduced Terms of Rational Expressions
(1) `(X^2+2X+1)/(2(X+1))` (2) `(2(X^2-Y^2))/(3(X^3-Y^3))`

(3) `(X^4-1)/(X^2+1)` (4) `(X^2-9)/(X^3-27)`


2. Find addition, Subtraction, multiplication, division of two and three terms of rational polynomials.

(1) `(X-3)/(X+1) - (X-6)/(X)`

(2) `(4x+1)/(4x-1) - (4x-1)/(4x+1) - 3`

(3) `(1)/(6X^2-13X+6) + (1)/(8X^2-14X+3) - (1)/(12X^2-11X+2)`

(4) `(X+2)/(X^2-4X+3) xx (X+3)/(X^2-3X+2) - (X+1)/(X^2-5X+6)`

(5) `(X^3-1)/(X-1) + (X^3+1)/(X+1) - (2(X^4-1))/(X^2-1)`



1. Find Reduced Term of `(4X^2-25)/(8X^3-125)`

` = ((4X^2 - 25))/((8X^3 - 125))`

` = ((2X-5)(2X+5))/((2X-5)(4X^2+10X+25))`

`"Now cancel the common factor " (2X - 5)`

` = ((2X+5))/((4X^2+10X+25))`

2. Find `(X+2)/(X^2-4X+3) - (X+3)/(X^2-3X+2)`

` = ((X + 2))/((X^2 - 4X + 3)) - ((X + 3))/((X^2 - 3X + 2))`

` = ((X+2))/((X-3)(X-1)) - ((X+3))/((X-2)(X-1))`

` = ((X + 2) × (X - 2) - (X + 3) × (X - 3))/((X-1)(X-3)(X-2))`

` = ((X^2 - 4) - (X^2 - 9))/((X-1)(X-3)(X-2))`

` = (5)/((X-1)(X-3)(X-2))`

 
7. Complete Square,
Is Perfact Square,
Find Missing Term
1. Convert the given equation into perfact square form
(1) 3x2+5x+2
(2) 9x2+6x+1
(3) x2+6x+9

2. Check the given equation is perfact square or not
(1) 4x2 + 4x + 4
(2) x2 - 4xy + 4y2
(3) 30xy - 25x2y2 + 9

3. Find the missing term so the given equation is perfect square
(1) x2 + ____ + 4
(2) 9x2 - ____ + 16
(3) 81x2 + ____ + 49
(4) x2 + ____ + 9
(4) ____ + 12x2 + 9
(5) 49x2 + 56 xy + ____
(6) 25x2 - ____ + 121y2
(7) 9x2 + ____ + y2
(8) x4 + 6x2 + ____
(9) x4y2 - 10x2yz + ____


1.1 Convert the given equation `3X^2+5X+2` into perfact square form

`= (3X^2+5X+2)`

`= 3 (X^2+5/3X+2/3)`

`= 3 (X^2+5/3X + 25/36 - 25/36 + 2/3)`

`= 3 [(X^2+5/3X+25/36) -1/36]`

`= 3 [( X + 5/6 )^2 -1/36]`


2. Check the given equation `3X^2+5X+2` is perfact square or not

` 3X^2+5X+2`

` "Here "F.T. = 3X^2, M.T. = 5X and L.T. = 2 `

` (M.T.)^2 = (5X)^2 = 25X^2`

` 4(F.T.)(L.T.) = 4 * 3X^2 * 2 = 24X^2`

` :. (M.T.)^2 != 4(F.T.)(L.T.)`

` :. "given polynomial is not a perfect square."`


3. Find Missing First Term of a given equation `X^2+4`

` X^2+4`

` "Here "L.T. = 4, M.T. = X^2 and F.T. = ? `

` (M.T.)^2 = 4(F.T.)(L.T.)`

` F.T. = (M.T.)^2 / (4(L.T.))`

` F.T. = (X^2 * X^2) / (4 * 4)`

` F.T. = (X^4) / (16)`


8. Solve or Simplify Linear Equation
Solve linear equation

1. `5 - 3/5 + 7/(3*3)`

2. `4 + a^2 * 3 * (x+y)` where `a=4,x=4`

3. `x/5 + 3/5 - 8/3`

4. `(2x-1)/(3x+1) + (4x-1)/(6x)`

5. `(2x+8)/(3x+1) + 3 - (x+4)/(7x+1) - 17/4`

6. `(3x+1)/2 - (3x-1)/3 - (5(x+2))/6`
Simplify linear equation

1. `4 + 3 * (x+y) = 0` where `x=4`

2. `(4x+1)/(3x-2) = 3/2`

3. `(2x-1)/(3x+1) = (4x-1)/(6x)`

4. `(3x+1)/2 - (3x-1)/3 = (5(x+2))/6`

1. Solve linear equation `(3x+1)/5-(3x-1)/3`

`(3x+1)/5-(3x-1)/3`

` = ((3x + 1))/(5) - ((3x - 1))/(3)`

` = ((3x+1))/(5) - ((3x-1))/(3)`

` = ((3x + 1) × 3 - (3x - 1) × 5)/(15)`

` = ((9x + 3) - (15x - 5))/(15)`

` = ((-6x+8))/(15)`

2. Solve linear equation `4+3*(x+y)=0` and given values are `x=4`

`4+3*(x+y)=0` where `x=4`

`=> (4+3(4+y)) = 0`

`=> (3y+16) = 0`

`=> 3y = -16`

`=> y = -16/3`

 
1. Substitution Method
Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Substitution Method

`2X+7Y-11 = 0 ->(1)`

`3X-Y-5 = 0 ->(2)`

Taking Eqn. `(2)`, we have

`3X-Y-5 = 0`

`=> Y = 3X-5 ->(3)`

Putting `Y = 3X-5` in Eqn. `(1)`, we get

`2X+7(3X-5)-11 = 0`

`=> 2X+21X-35-11 = 0`

`=> 23X-46 = 0`

`=> 23 (X-2) = 0`

`=> X-2 = 0`

`=> X = 2 ->(4)`

Now, Putting `X = 2` in Eqn. `(3)`, we get

`Y = 3(2)-5`

`=> Y = 6-5`

`=> Y = 1`

`:. Y = 1" and "X = 2`
2. Elimination Method
Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Elimination Method

`2X+7Y = 11 ->(1)`

`3X-Y = 5 ->(2)`

Eqn.`(1) xx 1 => 2X+7Y = 11`

Eqn.`(2) xx 7 => 21X-7Y = 35`

Adding ` => 23X = 46`

`=> X = 46 / 23`

`=> X = 2`

Putting `X = 2` in Eqn. `(2)`, we have

`3(2)-Y = 5`

`=> -Y = 5-6`

`=> -Y = -1`

`=> Y = 1`

`:. X = 2" and "Y = 1`
 
3. Cross Multiplication Method
Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Cross Multiplication Method

`2X+7Y-11 = 0 ->(1)`

`3X-Y-5 = 0 ->(2)`

Here,
`a_1=2, b_1=7, c_1=-11`

`a_2=3, b_2=-1, c_2=-5`

`X = (b_1*c_2-b_2*c_1)/(a_1*b_2-a_2*b_1)`

`= ((7)(-5)-(-1)(-11))/((2)(-1)-(3)(7))`

`= ((-35)-(11))/((-2)-(21))`

`= (-46)/(-23)`

`= 2`

`Y = (c_1*a_2-c_2*a_1)/(a_1*b_2-a_2*b_1)`

`= ((-11)(3)-(-5)(2))/((2)(-1)-(3)(7))`

`= ((-33)-(-10))/((-2)-(21))`

`= (-23)/(-23)`

`= 1`

`:. X = 2" and "Y = 1`
4. Addition-Subtraction Method
Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Addition-Substriaction Method

`2X+7Y-11 = 0 ->(i)`

`3X-Y-5 = 0 ->(ii)`

Adding Equation `(i)` and `(ii)`, we get

`5X+6Y-16 = 0`

`5X+6Y-16 = 0 ->(iii)` (On simplifying)

Subtracting Equation `(ii)` from `(i)`, we get

`-X+8Y-6 = 0`

`X-8Y+6 = 0 ->(iv)` (On simplifying)

`5X+6Y = 16 ->(1)`

`X-8Y = -6 ->(2)`

Eqn.`(1) xx 1 => 5X+6Y = 16`

Eqn.`(2) xx 5 => 5X-40Y = -30`

Substracting `=> 46Y = 46`

`=> Y = 46 / 46`

`=> Y = 1`

Putting `Y = 1` in Eqn. `(2)`, we have

`X-8(1) = -6`

`=> X = -6+8`

`=> X = 2`

`=> X = 2`

`:. X = 2" and "Y = 1`
 
5. Inverse Matrix Method
Solve linear equations 12x+5y=7 and x+y=7 using Using Matrix Method

`=> (12x+5y-7) = 0`

`=> (x+y-7) = 0`

Here `12x+5y=7`, `x+y=7`

Now converting given equations into matrix form
`[[12,5],[1,1]] [[ x ],[ y ]] = [[7],[7]]`

Now, A = `[[12,5],[1,1]]`, X = `[[ x ],[ y ]]` and B = `[[7],[7]]`

`:. AX = B`

`:. X = A^-1 B`

`| A |=|[12,5],[1,1]|`

= 12 × 1 - 5 × 1
= 12 - 5
= 7
`"Here, " | A | = 7 != 0 `

`:. A^(-1)" is possible."`

`Adj(A)=Adj[[12,5],[1,1]]`

`=[[+(1),-(1)],[-(5),+(12)]]^T`

`=[[1,-1],[-5,12]]^T`

`=[[1,-5],[-1,12]]`

`"Now, "A^(-1)=1/| A | × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X =1/| A | × Adj(A) × B`

`=1/7 × [[1,-5],[-1,12]] × [[7],[7]]`

` =1/7 ×[[1*7 + -5*7],[-1*7 + 12*7]]`

` =1/7 ×[[-28],[77]]`

` =[[-4],[11]]`



`:.[[ x ],[ y ]] = [[-4],[11]]`

`:. x = -4, y = 11`
6. Cramer's Rule Method
Solve linear equations 12x+5y=7 and x+y=7 using Using Cramer's Rule Method

The equations can be expressed as
`12x+5y-7=0`

`x+y-7=0`

Use Cramer’s Rule to find the values of x, y, z.
`(x)/D_x=(-y)/D_y=(1)/D`

` D_x =|[5,-7],[1,-7]|`

` = 5 × -7 - -7 × 1`

` = -35 + 7`

` = -28`


` D_y =|[12,-7],[1,-7]|`

` = 12 × -7 - -7 × 1`

` = -84 + 7`

` = -77`


` D =|[12,5],[1,1]|`

` = 12 × 1 - 5 × 1`

` = 12-5`

` = 7`


`(x)/D_x=(-y)/D_y=(1)/D`

`:. (x)/-28=(-y)/-77=(1)/7`

`:. (x)/-28=(1)/7,(-y)/-77=(1)/7`

`:. x=(-28)/(7),y=(77)/(7)`

`:. x=-4,y=11`
 
7. Graphical Method
Solve linear equations 12x+5y=7 and x+y=7 using Graphical Method

`12x+5y=7`

`x+y=7`

The point of intersection of the linear equations
Consider `12x+5y=7`

`x`-intercept: put `y = 0,` we get `x = 0.5833` `:. A(0.5833,0)`

`y`-intercept: put `x = 0,` we get `y = 1.4` `:. B(0,1.4)`


Consider `x+y=7`

`x`-intercept: put `y = 0,` we get `x = 7` `:. C(7,0)`

`y`-intercept: put `x = 0,` we get `y = 7` `:. D(0,7)`

Intersection Point = `P(-4,11)`


 
Solve linear equation of any number of variables Inverse matrix method, Gauss elimination method
Solve equations like
1. 2x + y + z = 10, 3x + 2y + 3z = 18, x + 4y + 9z = 16
2. 2x + 5y = 16, 3x + y = 11
3. 2x + 5y = 21, x + 2y = 8
4. 2x + y = 8, x + 2y = 1
5. 2x + 3y - z = 5, 3x + 2y + z = 10, x - 5y + 3z = 0
6. x + y + z = 3, 2x - y - z = 3, x - y + z = 9
7. x + y + z = 7, x + 2y + 2z = 13, x + 3y + z = 13
8. 2x - y + 3z = 1, -3x + 4y - 5z = 0, x + 3y - 6z = 0
1. Inverse matrix method
1. Solve Equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Inverse Matrix method

Here `2x+y+z=5`, `3x+5y+2z=15`, `2x+y+4z=8`

Now converting given equations into matrix form
`[[2,1,1],[3,5,2],[2,1,4]] [[ x ],[ y ],[ z ]]=[[5],[15],[8]]`

Now, A = `[[2,1,1],[3,5,2],[2,1,4]]`, X = `[[ x ],[ y ],[ z ]]` and B = `[[5],[15],[8]]`

`:. AX = B`

`:. X = A^-1 B`

`| A |=|[2,1,1],[3,5,2],[2,1,4]|`

` = 2 (5 × 4 - 2 × 1) - 1 (3 × 4 - 2 × 2) + 1 (3 × 1 - 5 × 2)`

` = 2 (20 - 2) - 1 (12 - 4) + 1 (3 - 10)`

` = 2 (18) - 1 (8) + 1 (-7)`

` = 36 - 8 - 7`

` = 21`


`"Here, " | A | = 21 != 0 `

`:. A^(-1)" is possible."`

`Adj(A)=Adj[[2,1,1],[3,5,2],[2,1,4]]`

`=[[+(5 × 4 - 2 × 1),-(3 × 4 - 2 × 2),+(3 × 1 - 5 × 2)],[-(1 × 4 - 1 × 1),+(2 × 4 - 1 × 2),-(2 × 1 - 1 × 2)],[+(1 × 2 - 1 × 5),-(2 × 2 - 1 × 3),+(2 × 5 - 1 × 3)]]^T`

`=[[18,-8,-7],[-3,6,0],[-3,-1,7]]^T`

`=[[18,-3,-3],[-8,6,-1],[-7,0,7]]`

`"Now, "A^(-1)=1/| A | × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X =1/| A | × Adj(A) × B`

`=1/21 × [[18,-3,-3],[-8,6,-1],[-7,0,7]] × [[5],[15],[8]]`

` =1/21 ×[[18*5 + -3*15 + -3*8],[-8*5 + 6*15 + -1*8],[-7*5 + 0*15 + 7*8]]`

` =1/21 ×[[21],[42],[21]]`

` =[[1],[2],[1]]`



`:.[[ x ],[ y ],[ z ]]=[[1],[2],[1]]`

`:. x=1, y=2, z=1`
2. Gauss elimination method
1. Solve Equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Gauss Elimination method

Total Equations are `3`

`2 x + y + z = 5`

`3 x + 5 y + 2 z = 15`

`2 x + y + 4 z = 8`


Converting given equations into matrix form
`[[2,1,1,|,5],[3,5,2,|,15],[2,1,4,|,8]]`

Dividing `R_1` by `2`

`[[1,1/2,1/2,|,5/2],[3,5,2,|,15],[2,1,4,|,8]]`

`R_2 larr R_2 - 3 * R_1`

`[[1,1/2,1/2,|,5/2],[0,7/2,1/2,|,15/2],[2,1,4,|,8]]`

`R_3 larr R_3 - 2 * R_1`

`[[1,1/2,1/2,|,5/2],[0,7/2,1/2,|,15/2],[0,0,3,|,3]]`

Dividing `R_2` by `7/2`

`[[1,1/2,1/2,|,5/2],[0,1,1/7,|,15/7],[0,0,3,|,3]]`

`R_1 larr R_1 - 1/2 * R_2`

`[[1,0,3/7,|,10/7],[0,1,1/7,|,15/7],[0,0,3,|,3]]`

Dividing `R_3` by `3`

`[[1,0,3/7,|,10/7],[0,1,1/7,|,15/7],[0,0,1,|,1]]`

`R_1 larr R_1 - 3/7 * R_3`

`[[1,0,0,|,1],[0,1,1/7,|,15/7],[0,0,1,|,1]]`

`R_2 larr R_2 - 1/7 * R_3`

`[[1,0,0,|,1],[0,1,0,|,2],[0,0,1,|,1]]`

Solution By Gauss Elimination Method.
`x = 1`

`y = 2`

`z = 1`
 
1. Roots By Factorization
1. Find the roots of Quadratic Equation x^2+8x+12=0 by the method of factorization

` x^2+8x+12=0`

` => x^2+8x+12 = 0`

` => (x^2+8x+12) = 0`

` => (x^2+2x+6x+12) = 0`

` => x(x+2)+6(x+2) = 0`

` => (x+6)(x+2) = 0`

` => (x+6) = 0" or "(x+2) = 0`

` => x = -6" or "x = -2`

2. Find the roots of Quadratic Equation X^2-25=0 by the method of factorization

` X^2-25=0`

` => X^2-25 = 0`

` => (X^2-25) = 0`

` => (X-5)(X+5) = 0`

` => (X-5) = 0" or "(X+5) = 0`

` => X = 5" or "X = -5`

2. Roots For Non-Zero Denominator
Method Example
1. `(5X-18)/(X+2) = (2X-6)/(X-1)` `(X-3)/(X+1) = (X-6)/(X)`
2. `(x)/(x+1) + (x+1)/(x) = 5/2`
3. `4((4x+1)/(4x-1))^(2) + (4x+1)/(4x-1) = 3`
4. `(4x+1)/(4x-1) + (4x-1)/(4x+1) = 3`

1. Find roots of the equation `(5X-18)/(X+2) = (2X-6)/(X-1)`

` (5X-18)/(X+2) = (2X-6)/(X-1)`

` => (5X-18)*(X-1) = (2X-6)*(X+2)`

` => 5X^2-23X+18 = 2X^2-2X-12`

` => 3X^2-21X+30 = 0`

` => (3X^2-21X+30) = 0`

` => 3X^2-21X+30 = 0`

` => 3(X^2-7X+10) = 0`

` => 3(X^2-2X-5X+10) = 0`

` => 3(X(X-2)+(-5)(X-2)) = 0`

` => 3(X-5)(X-2) = 0`

` => (X-5) = 0" or "(X-2) = 0`

` => X = 5" or "X = 2`

2. Find roots of the equation `(X)/(X+1) + (X+1)/(X) = (5)/(2)`

` (X)/(X+1) + (X+1)/(X) = (5)/(2)`

` => (X)*(X)*(2) + (X+1)*(X+1)*(2) = (5)*(X+1)*(X)`

` => 2X^2 + (2X^2+4X+2) = (5X^2+5X)`

` => 2X^2 + (2X^2+4X+2) + (-5X^2-5X) = 0`

` => -X^2-X+2 = 0`

` => (-X^2-X+2) = 0`

` => -X^2-X+2 = 0`

` => (-1)(X^2+X-2) = 0`

` => (-1)(X^2-X+2X-2) = 0`

` => (-1)(X(X-1)+2(X-1)) = 0`

` => (-1)(X+2)(X-1) = 0`

` => (X+2) = 0" or "(X-1) = 0`

` => X = -2" or "X = 1`

3. Find roots of the equation `12((2X+1)/(X-1))^2 + -5((2X+1)/(X-1)) + -2 = 0`

` 12((2X+1)/(X-1))^2 - 5((2X+1)/(X-1)) - 2 = 0`

` "Let " (2X+1)/(X-1) = m`

` => (12m^2-5m-2) = 0`

` => 12m^2-5m-2 = 0`

` => (12m^2-5m-2) = 0`

` => (12m^2+3m-8m-2) = 0`

` => 3m(4m+1)+(-2)(4m+1) = 0`

` => (3m-2)(4m+1) = 0`

` => (3m-2) = 0" or "(4m+1) = 0`

` => 3m = 2" or "4m = -1`

` => m = 2/3" or "m = -1/4`

` "Now, " (2X+1)/(X-1) = 2/3`

` => 3(2X+1) = 2(X-1)`

` => 3(2X+1) - 2(X-1) = 0`

` => (3(2X+1)-2(X-1)) = 0`

` => (4X+5) = 0`

` => 4X = -5`

` => X = -5/4`

` "Now, " (2X+1)/(X-1) = -1/4`

` => 4(2X+1) = -1(X-1)`

` => 4(2X+1) + 1(X-1) = 0`

` => (4(2X+1)+(X-1)) = 0`

` => (9X+3) = 0`

` => 9X = -3`

` => X = -3/9`

` => X = -1/3`

4. Find roots of the equation `12((X)/(X-1)) + 12((X-1)/(X)) = 25`

` 12((X)/(X-1)) + 12((X-1)/(X)) = 25`

` "Let " (X)/(X-1) = m`

` => 12m + 12/m = 25`

` => 12m^2 - 25m + 12 = 0`

` => (12m^2-25m+12) = 0`

` => 12m^2-25m+12 = 0`

` => (12m^2-25m+12) = 0`

` => (12m^2-9m-16m+12) = 0`

` => 3m(4m-3)+(-4)(4m-3) = 0`

` => (3m-4)(4m-3) = 0`

` => (3m-4) = 0" or "(4m-3) = 0`

` => 3m = 4" or "4m = 3`

` => m = 4/3" or "m = 3/4`

` " Now," (X)/(X-1) = 4/3`

` => 3(X) = 4(X-1)`

` => 3(X) - 4(X-1) = 0`

` => (3X-4(X-1)) = 0`

` => (-X+4) = 0`

` => X = 4`

` " Now," (X)/(X-1) = 3/4`

` => 4(X) = 3(X-1)`

` => 4(X) - 3(X-1) = 0`

` => (4X-3(X-1)) = 0`

` => (X+3) = 0`

` => X = -3`

 
3. Discriminant & Nature of Roots
Rules:
1. If D > 0 then the roots are real and distinct.
    (i) If D is a perfect square then the roots are rational and distinct.
    (ii) If D is not a perfect square then the roots are irrational and distinct.
2. If D = 0 then the roots are real and equal.
3. If D < 0 then the quadratic equation has no real roots.

Example :
1 Find the discriminant of Quadratic Equation `x^2-5x+6=0` and discuss the nature of its roots

`x^2-5x+6=0`

`=> x^2-5x+6 = 0`

Comparing the given equation with the standard quadratic equation `ax^2 + bx + c = 0,`

we get, `a = 1, b = -5, c = 6.`

`:. Delta = b^2 - 4ac`

` = (-5)^2 - 4 (1) (6)`

` = 25 - 24`

` = 1`

` = (1)^2`

Here, `Delta > 0` and is a perfect square. Also a and b are rational.

Hence, the roots of the equation are distinct and rational.

4. Roots By Perfact Square
1. Find the roots of Quadratic Equation `X^2+10X-56=0` by the method of perfect square

`X^2+10X-56=0`

`=> X^2+10X-56 = 0`

Comparing the given equation with the standard quadratic equation `ax^2 + bx + c = 0,`

we get, `a = 1, b = 10, c = -56.`

`:. Delta = b^2 - 4ac`

` = (10)^2 - 4 (1) (-56)`

` = 100 + 224`

` = 324`

`:. sqrt(Delta) = sqrt(324) = 18`

Now, `alpha = (-b + sqrt(Delta)) / (2a)`

` = (-(10) + 18) / (2 * 1)`

` = 8 / 2`

` = 4`

and, `beta = (-b - sqrt(Delta)) / (2a)`

` = (-(10) - 18) / (2 * 1)`

` = -28 / 2`

` = -14`

 
5. Form Quadratic Equation using given two roots
1. Form the Quadratic Equation whose roots are Alpha = 2, Beta = 5

Let `alpha = 2` and `beta = 5`

Then, the sum of the roots `= alpha + beta = (2)+(5) = 7`

and the Product of the roots `= alpha * beta = (2)*(5) = 10`

The Equation with roots `alpha` and `beta` is given by

`X^2 - (alpha+beta)X + alpha*beta = 0`

`:.` The required equation

`X^2 - (7)X + (10) = 0`

2. Form the Quadratic Equation whose roots are Alpha = -1/2, Beta = +2/3

Let `alpha = -1/2` and `beta = +2/3`

Then, the sum of the roots `= alpha + beta = (-1/2)+(+2/3) = 1/6`

and the Product of the roots `= alpha * beta = (-1/2)*(+2/3) = -1/3`

The Equation with roots `alpha` and `beta` is given by

`X^2 - (alpha+beta)X + alpha*beta = 0`

`:.` The required equation

`X^2 - (1/6)X + (-1/3) = 0`

6. Roots of Non-Quadratic Equation
1. Solve the equation `1 (X^2 + 1/X^2) - 8 ( X + 1/X ) + 14 = 0`

`1 * (X^2+1/X^2) + (-8) * (X+1/X) + (14) = 0`

Let `X + 1/X = m`

`=> (X + 1/X)^2 = m^2`

`=> X^2 + 1/X^2 + 2 = m^2`

`=> X^2 + 1/X^2 = m^2 - 2`

Substituting this values in the given equation, we get

`(m^2 - 2) - 8m + 14 = 0`

`m^2 - 8m + 12 = 0`

`=> (m^2-8m+12) = 0`

`=> m^2-8m+12 = 0`

`=> (m^2-8m+12) = 0`

`=> (m^2-2m-6m+12) = 0`

`=> m(m-2)+(-6)(m-2) = 0`

`=> (m-6)(m-2) = 0`

`=> (m-6) = 0" or "(m-2) = 0`

`=> m = 6" or "m = 2`

Now, `X + 1/X = 6`

`=> X^2 + 1 = 6X`

`=> X^2 - 6X + 1 = 0`

`=> (X^2-6X+1) = 0`

`=> X^2-6X+1 = 0`

Comparing the given equation with the standard quadratic equation `ax^2 + bx + c = 0,`

we get, `a = 1, b = -6, c = 1.`

`:. Delta = b^2 - 4ac`

` = (-6)^2 - 4 (1) (1)`

` = 36 - 4`

` = 32`

`:. sqrt(Delta) = sqrt(32) = 4 * sqrt(2)`



Now, `alpha = (-b + sqrt(Delta)) / (2a)`

` = (-(-6) + 4 * sqrt(2)) / (2 * 1)`

` = (6 + 4 * sqrt(2)) / 2`

` = 3 + 2 * sqrt(2)`



and, `beta = (-b - sqrt(Delta)) / (2a)`

` = (-(-6) - 4 * sqrt(2)) / (2 * 1)`

` = (6 - 4 * sqrt(2)) / 2`

` = 3 - 2 * sqrt(2)`

Now, `X + 1/X = 2`

`=> X^2 + 1 = 2X`

`=> X^2 - 2X + 1 = 0`

`=> (X^2-2X+1) = 0`

`=> X^2-2X+1 = 0`

`=> (X^2-2X+1) = 0`

`=> (X-1)(X-1) = 0`

`=> (X-1) = 0" or "(X-1) = 0`

`=> X = 1" or "X = 1`

 
 
1. Find Value Of Variation
1. `X prop Y and X=4,Y=2.` Find `X=18,Y=?`

`X prop Y`

`=> X=K*Y`

Now, `X=4,Y=2`

`=> 4 = K * 2`

`=> K = 2`

Hence, `X=2*Y`

`X=18,Y=?`

`=> 18 = 2 * Y`

`=> 9 = Y`

`=> Y = 9`

2. `X prop Y^3/Z and X=8,Y=4,Z=3.` Find `X=?,Y=6,Z=4`

`X prop Y^3/Z`

`=> X=K*Y^3/Z`

Now, `X=8,Y=4,Z=3`

`=> 8 = K * 64/3`

`=> K = 3/8`

Hence, `X=3/8*Y^3/Z`

`X=?,Y=6,Z=4`

`=> 1 * X = 3/8 * 54`

`=> X = 81/4`

2. Prove Results For Given Variation
1. If `X prop Y,` then prove that `X^3+Y^3 prop X^2Y-XY^2`

`X prop Y`

`=> X=M*Y` (where constant `M != 0`)

Now `(X^3+Y^3) / (X^2Y-XY^2)`

`= (M^3Y^3+Y^3) / (M^2Y^3-MY^3)`

`= (Y^3(M^3+1)) / (MY^3(M-1))`

`= ((M^3+1)) / (M(M-1))`

`=` non-zero constant

`:. X^3+Y^3 prop X^2Y-XY^2`

2. If `5X-7Y prop 6X+3Y,` then prove that `X prop Y`

`5X-7Y prop 6X+3Y`

`=> 5X-7Y = M(6X+3Y)` (where constant `M != 0`)

`=> 5X-7Y = 6MX+3MY`

`=> 5X-6MX = 7Y+3MY`

`=> X(5-6M) = Y(7+3M)`

`=> X/Y = (7+3M) / (5-6M)`

`=> X/Y = `non-zero constant

`=> X prop Y`

 
 
1. Union
(1) A = {1,2,3,4,5}, B = {3,4,5,6}
Find A U B ...


Here `A = {1,2,3,4,5}, B = {3,4,5,6}`


`A uu B = {1,2,3,4,5} uu {3,4,5,6}`

`= {1,2,3,4,5,6}`
2. Intersection
(1) A = {1,2,3,4,5}, B = {3,4,5,6}
Find A n B ...


Here `A = {1,2,3,4,5}, B = {3,4,5,6}`


`A nn B = {1,2,3,4,5} nn {3,4,5,6}`

`= {3,4,5}`
 
3. Complement
(1) U = {1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4,5}
Find A' ...


Here `U = {1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4,5}`


`A' = {1,2,3,4,5}'`

`= {6,7,8,9,10}`
4. Power set(Proper Subset)
(1) C = {7,8,9}
Find Proper subset of C ...


Here `C = {7,8,9}`

`P(C) = {{7},{8},{9},{7,8},{7,9},{8,9},{7,8,9}}`
 
5. Minus
(1) A = {1,2,3,4,5}, B = {3,4,5,6}
Find A - B ...


Here `A = {1,2,3,4,5}, B = {3,4,5,6}`


`A - B = {1,2,3,4,5} - {3,4,5,6}`

`= {1,2}`
6. Cross Product
(1) A = {1,2,3}, B = {4,5}
Find A × B ...


Here `A = {1,2,3}, B = {4,5}`


`A × B = {1,2,3} × {4,5}`

`= { (1,4), (1,5), (2,4), (2,5), (3,4), (3,5) }`
 
7. Prove that any two expression is equal or not
A = {1,2,3,4,5}, B = {3,4,5,6}, C = {7,8,9}, Prove that A U (B U C) = (A U B) U C ...

Here `A={1,2,3,4,5},B={3,4,5,6},C={7,8,9}`

To find LHS = `(A uu B) uu C`

`A uu B = {1,2,3,4,5} uu {3,4,5,6}`

`= {1,2,3,4,5,6}`

`(A uu B) uu C = {1,2,3,4,5,6} uu {7,8,9}`

`= {1,2,3,4,5,6,7,8,9}`

`:. (A uu B) uu C = {1,2,3,4,5,6,7,8,9} ->(1)`

To find RHS = `A uu (B uu C)`

`B uu C = {3,4,5,6} uu {7,8,9}`

`= {3,4,5,6,7,8,9}`

`A uu (B uu C) = {1,2,3,4,5} uu {3,4,5,6,7,8,9}`

`= {1,2,3,4,5,6,7,8,9}`

`:. A uu (B uu C) = {1,2,3,4,5,6,7,8,9} ->(2)`

From (1) and (2)
`:. (A uu B) uu C = A uu (B uu C)` (proved)
 
 
14. Descartes' rule of signs
1. Find Descartes' rule of signs for `x^5-x^4+3x^3+9x^2-x+5`

Here `f(x)=x^5-x^4+3x^3+9x^2-x+5`

`f(x)=x^5color{red}{-}x^4color{blue}{+}3x^3color{blue}{+}9x^2color{red}{-}xcolor{blue}{+}5`

look first at `f(x)`: (positive case)

`f(x)=``color{blue}{+}``x^5``color{red}{-}``x^4``color{blue}{+}``3x^3``color{blue}{+}``9x^2``color{red}{-}``x``color{blue}{+}``5`
Sign change count`color{blue}{+}` to `color{red}{-}`
1
`color{red}{-}` to `color{blue}{+}`
2
`color{blue}{+}` to `color{blue}{+}`
 
`color{blue}{+}` to `color{red}{-}`
3
`color{red}{-}` to `color{blue}{+}`
4


There are 4 sign changes, so there are 4 or counting down in pairs, 2 or 0 positive roots.


Now look at `f(–x)`: (negative case)

`f(-x)=(-x)^5color{red}{-}(-x)^4color{blue}{+}3(-x)^3color{blue}{+}9(-x)^2color{red}{-}(-x)color{blue}{+}5`

`f(-x)=color{red}{-}x^5color{red}{-}x^4color{red}{-}3x^3color{blue}{+}9x^2color{blue}{+}xcolor{blue}{+}5`

`f(-x)=``color{red}{-}``x^5``color{red}{-}``x^4``color{red}{-}``3x^3``color{blue}{+}``9x^2``color{blue}{+}``x``color{blue}{+}``5`
Sign change count`color{red}{-}` to `color{red}{-}`
 
`color{red}{-}` to `color{red}{-}`
 
`color{red}{-}` to `color{blue}{+}`
1
`color{blue}{+}` to `color{blue}{+}`
 
`color{blue}{+}` to `color{blue}{+}`
 


There is 1 sign change, so there is exactly 1 negative roots.
 

 
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