Home > Algebra calculators

 AtoZmath.com - Homework help (with all solution steps) 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

 Algebra Calculators 1. Factor of an equation 2. Expand of an algebraic expression 3. Expansion: X + 1/X to X - 1/X 4. Addition, subtraction, multiplication, division of two polynomials 5. GCD-LCM of Polynomials 6. Rational Expressions 1. Reduced terms 2. Addition, Subtraction, Multiplication and Division of Rational Polynomials 7. Complete Square, Is Perfact Square, Find Missing Term 1. Convert the given equation into perfact square form 2. Check a given equation is perfact square or not 3. Find the missing term so the given equation is perfect square 8 Solve or simplify linear equation 9. Solve linear equation in two variables using 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. Solve linear equation of any number of variables Inverse matrix method, Gauss elimination method 11. Quadratic Equation 1. Roots by factorization, Roots by Perfact Square, Discriminant & Nature of Roots 2. Form quadratic equation using given two roots 3. Roots for non-zero denominator 4. Roots of non-quadratic equation 12. 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. 13. Set Theory 1. Union 2. Intersection 3. Complement 4. Power set(Proper Subset) 5. Minus 6. Cross Product 7. Prove that any two expression is equal or not 8. Cardinality of a set 9. is Belongs To a set 10. is Subset Of a set 10. is two set Equal or not 14. Interval notation and Builder set notation (1) 3 <= z <=7, x is odd. (2) |x^3-2| <= 25, x in Z. 14. Descartes' rule of signs 15. Mathematical Logic 16. 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 - c2b       = 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)24. (3Y - 2X)35. (2X + 3Y)26. (X + Y)2 - (X - Y)27. (X + Y + Z)2 + (X + Y - Z)28. (3X - 5Y)3 + (5Y - 9Z)3 + (9Z - 3X)3 1. Expand an Algebraic expression (X+2)(X+3)Using The Identity,(X + A)(X + B) = X2 + (A + B)X + ABHere X=X,A=2, B=3 = (X)2 + (2 + 3)X + (2)×(3) = X2+5X+6 2. Expand an Algebraic expression (2X+3Y+4Z)2Using The Identity,(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CAHere 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, 9.822. 123, 12.233. 93, 8.934. 102 × 1065. 342 - 322, 3.42 - 3.226. 103 - 73 - 337. 43 + 33, 3.43 - 2.93 1. Find Value of 102 × 106Using, The Identity (X+A)(X+B) = X2 + (A+B)X + ABHere X = 100, A = 2 and B = 6102 × 106 = (100 + 2)(100 + 6) = 1002 + (2 + 6)100 + 2 × 6 = 10000 + 800 + 12 = 10812 2. Find Value of 342 - 322Using, The Indentity A2 - B2 = (A-B)(A+B)Here A = 34 and B = 32Here 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 = 2Now, 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 = 0X - 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 = 1Now, We know that4 XY = ( X + Y )^2 - ( X - Y )^24 XY = 5^2 - 1^24 XY = 25 - 14 XY = 24XY = 24/4XY = 6Now, We know thatX^2 + Y^2 = ( X + Y )^2 - 2 XYX^2 + Y^2 = 5^2 - 2 * 6X^2 + Y^2 = 25 - 12X^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 = 29Now, 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

(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)2GCD = 2 × 5 (X - 1)GCD = 10(X - 1)LCM = 2 × 3 × 5 × 5 (X - 2)(X - 1)2LCM = 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 have3X-Y-5 = 0=> Y = 3X-5 ->(3)Putting Y = 3X-5 in Eqn. (1), we get2X+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 getY = 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 = 11Eqn.(2) xx 7 => 21X-7Y = 35Adding  => 23X = 46=> X = 46 / 23=> X = 2Putting X = 2 in Eqn. (2), we have3(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=-11a_2=3, b_2=-1, c_2=-5X = (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)= 2Y = (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 get5X+6Y-16 = 05X+6Y-16 = 0 ->(iii) (On simplifying)Subtracting Equation (ii) from (i), we get-X+8Y-6 = 0X-8Y+6 = 0 ->(iv) (On simplifying)5X+6Y = 16 ->(1)X-8Y = -6 ->(2)Eqn.(1) xx 1 => 5X+6Y = 16Eqn.(2) xx 5 => 5X-40Y = -30Substracting => 46Y = 46=> Y = 46 / 46=> Y = 1Putting Y = 1 in Eqn. (2), we haveX-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) = 0Here 12x+5y=7, x+y=7Now 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=0x+y-7=0Use 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=7x+y=7The point of intersection of the linear equationsConsider 12x+5y=7x-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=7x-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 methodHere 2x+y+z=5, 3x+5y+2z=15, 2x+y+4z=8Now 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 methodTotal Equations are 32 x + y + z = 53 x + 5 y + 2 z = 152 x + y + 4 z = 8Converting 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 = 1y = 2z = 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*YNow, X=4,Y=2=> 4 = K * 2=> K = 2Hence, X=2*YX=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/ZNow, X=8,Y=4,Z=3=> 8 = K * 64/3=> K = 3/8Hence, X=3/8*Y^3/ZX=?,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 CA 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.