HOW TO KNOW THE DIVISIBILITY TEST OF A NUMBER || DIVISIBILITY RULES FOR NUMBERS

How to know whether a given number, however large is divisible by 2, divisible by 3,4,5,6 and 10 .There are fixed divisibility test and divisibility rules for checking the divisibility of any given numbers. For different numbers there are  different rules to divide with . So in this post  we are going to discuss these divisibility rules with examples  one by one. Although there are divisibility test  for fraction also ,but this post will be restricted to natural numbers.

Divisibility rules for 2

To check whether the given number, however large  is divisible by 2 ,we have to check its right most digit/Unit place digit, if it is even number or zero.Then the given number is definitely divisible by 2

For Example

12 is divisible by 2 as its right most digit is 2 which is Even .

3548 is divisible by 2 as its right most digit is 8 which is Even 

999998 is divisible by 2 as its right most digit is 8 which is Even .

65989564 is divisible by 2 as its right most digit is 4 which is Even. 

22222229 is not divisible by 2 as its right most digit is 9 which is not A Even number.

589423100780      is divisible by 2 as its right most digit is 0 .

357913579571536   is divisible by 2 as its right most digit is 6 which is Even although all the remaining digits are odd.

Divisibility rules for 3

To check whether the given number, however large  is divisible by 3,we have to check the SUM of all its digits, if its sum is divisible by 3 then the  given number is divisible by 3, If sum of  all the digits of given number is again a large number then add the result so obtained and apply the rule again which is said earlier.

For Example

15 is divisible by 3 as sum of all its digits 1 + 5 = 6 is divisible by 3 .

833 is not divisible by 3 as sum of all its digits 8 + 3 + 3 = 14 = 1 + 4 = 5 is not divisible by 3 .

2678 is not divisible by 3 as sum of all its digits 2 + 6 + 7 + 8 = 23 = 3 + 3 = 5 is not divisible by3 .

98784552 is not divisible by 3 as sum of all its digits 9 + 8+ 7+ 8+ 4 + 5 + 5 + 2 = 48 = 4 + 8 = 12 = 1 + 2 = 3 is not divisible by 3.

359875269 is divisible by 3 as sum of all its digits 3 + 5 + 9 +8 +7 + 5 + 2 + 6 + 9 = 54 = 5 + 4 = 9 is divisible by 3.

3597841137 is divisible by 3 as sum of all its digits 3 + 5 + 9 + 7 + 8 + 4 + 1 + 1 + 3 + 7 = 48 = 12   is divisible by 3 .

Divisibility rules for 4

If the last two digits of a number is divisible by 4 ,Then the number is divisible by 4 . The number having two or more zeros at the end is also divisible by 4.

For Example

568928 : Here in this number last two digits are 28 ,which are divisible by 4,Hence the given number is divisible by 4.

134826900 : As ther are two zeros at the end,so the given number is divisible by 4.

13444255452 : As the last two digits number (52) is divisible by 4 ,the given number is divisible by 4.

35888875698549 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4

97971349999567776 : As the last two digits number (76) is divisible by 4 ,the given number is divisible by 4.

44444444444444444449 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4.

Divisibility rules for 5

To check whether the given number, however large is divisible by 5 ,we have to check its right most digit/Unit place digit,if it is 5 or zero. Then the given number is definitely divisible by 5.I.e if the given number ends with 0 or 5 then it is divisible by by 5.

For Example

35 is divisible by 5 as its right most digit is 5.

97835 is divisible by 5 as its right most digit is 5.

6854940 is divisible by 5 as its right most digit is 0.

35000000355 is divisible by 5 as its right most digit is 5.

3579515465855 is divisible by 5 as its right most digit is 5.

12345678888880 is divisible by 5 as its right most digit is 0.

568954975311525 is divisible by 5 as its right most digit is 5.


55555555555556 is not divisible by 5 as its right most digit is 6.

66666666666666665 is divisible by 5 as its right most digit is 5.

Divisibility rule for 6

For any number to be divisible by 6 ,it must be divisible by both 2 and 3 ,then the given number is divisible by 6, Therefore
1) The number should ends up with an even digits or zero and
2) The sum of its digit should be divisible by 3.


For Example :

56898 is divisible by 6 as sum of its digits is 5 + 6 + 8 + 9 + 8 = 36 = 3 + 6 = 9 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

3578952  As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 +7 + 8 + 9 + 5 + 2 = 39 = 4 + 2 = 6  which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

25689879798 is divisible by 6 as sum of its digits is 2 + 5+ 6 + 8 + 9 + 8 + 7 + 9 + 7 + 9 + 8 =78 = 15 = 1 + 5 = 6 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

35658999962 is not divisible by 6 as sum of its digits is 3 + 5 + 6 +  5 + 8 + 9 + 9 + 9 + 9 + 6 + 2 = 71 = 7 + 1 = 8 so it is not divisible by 3 and last digit is not even ,so it is not divisible by 6.

35789248956 As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 + 7+ 8+ 9+ 2+ 4 + 8+ 9 + 5 + 6 = 66 =  6 + 6 = 12 = 1 + 2 = 3 which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

Divisibility rule for 10

It is the most easiest number to identify whether it is divisible by 10 , if the given number however large ends up with 0 then it is divisible by 0.

For Example

4546546546540 ,44545454560, 445474456110 5555598959550 are divisible by 10 as all the given numbers ends with 0.

and 445645489,454545,456445555 ,454545577 and 545454 are not divisible by 10 as all these numbers do not ends up with 0.

Conclusion

Thanks for giving your valuable time to read this post of the divisibility rules and divisibility test rules for  divisibility rule for 2 , divisibility rule for 3,    divisibility rule of 4 etc . If  you liked this post  , Then share it with your friends and family members . You can also read my others articles on Mathematics Learning and understanding Maths in easy ways.

Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

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HOW TO MULTIPLY TWO DIFFERENT NUMBERS IN FASTEST AND QUICKEST WAYS

        

                     Whenever we are to multiply two numbers then most of the time we undergo long calculations , and if we have to do easy calculation then we were lucky.

         But what to do when we have to multiply two numbers in very short time. Suppose we have to multiply 32 with 11 then it is easy, because just put right most digit as it is and then increase every digit by one to left and at last put the left most digit as it is and put 352 as answer.

Example

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HOW TO UNDERSTAND RELATIONS AND FUNCTIONS ,INVERSE OF A FUNCTION

Hello Friends Welcome 
                              Today we are going to  discuss Relations and Functions , "How to understand  Relations and  Functions, Inverse of a Function" under the topic  Relations and Functions.

Ordered-Pair Numbers :-

Ordered-pair number is written within a set of parentheses and separated by a comma.

For example, (5, 6) is an ordered-pair number; the order is designated by the first element 5 and the second element 6. The pair (3, 6) is not the same as (6,3) because they have different order. Sets of ordered-pair numbers can represent relations or functions.

Example of ordered pair :

(3,8),(2,1),(7,6)

Relation

A relation is a  set of ordered-pair numbers.

consider the following table

Number of Students	1	2	3	4	5	6
Marks Obtained	96	98	97	78	77	86

In the above table the numbers of students and marks obtained by them  is a relation and can be written as a set of ordered-pair numbers.

A= {(1, 96), (2, 98), (3, 97), (4, 88),(5,77),(6,86)}

When we collect all the elements written in 1st column of the ordered pairs and placed in a set then the set so formed is called  Domain of the relation.
The domain of A= {1, 2, 3, 4,5,6}

As all the elements written in 2nd column of the ordered pairs and placed in a set then the set so formed is called  Range of the relation.

The range of A = { 96, 98, 97,88 , 77, 86}

Function

A function is a relation in which every first element in ordered pairs have unique second element associated with them. Second  elements may or may not be same.

we can better understand this concept with the help of this video

Example

 {(1, 2), (2, 3), (3, 4), (4, 5),(5,6)}  is an example of function 

 { (1, 2), (2, 3), (3, 4), (4, 5),(5,6) } is a function because all the  first elements are different.

Example

{(1, 3), (3, 3), (2, 1), (4, 2)}  is an example of function 

 {(1, 3), (2, 3), (2, 1), (4, 2)}  is a function because all the first elements are different.

Example

{ (1, 6), (2, 5), (1, 9), (4, 3) }  is not an  example of function 

As in  {(1, 6), (2, 5), (1, 9), (4, 3)}  the element "1 "   appeared twice .

Example

{(2, 15), (3, 15), (4, 15), (5, 13),(6,18)}  is  an  example of function 

As in  {(2, 15), (3, 15), (4, 15), (5, 15)}   all the first elements are different.

Example

{(1, 1), (-1, 1),(2,4),(-2,4), (3, 9), (-3, 9),(4,16),(-4,16)}  is an  example of function although   the element "1" and "-1" ,"2" and "-2" , "3" and "-3"  ,"4", "-4" have same images. This is an example of many one function.

Question:-   Find x and y if: 

(i) (5x + 3, y) = (4x + 5,  2)

(ii) (x – y, x + y) = (8, 12)
(iii) ( 2x-y , y+5 ) = ( -2,3 )

Solution

(1)  Given  (5x + 3 , y) = (4x + 5, 2)
So By the equality of ordered pair elements,
1st element of the ordered number written on the left hand side will be equal to the 1st element of the ordered pair number written on the  right hand side . Therefore 

5x + 3 = 4x + 5   and y =  2 
5x-4x = 5 - 3   and y = 2 
x = 2 and y = 2

(ii) So By the equality of ordered pair elements
x – y = 8 and  x + y = 12

Solving these two equations for x and y 

 2x =20  and    10+ y =12 

x=10   y = 2

(iii) So By the equality of ordered pair elements
2x-y  =-2  , y+5 = 3 
2x = -2+y  , y = 3-5
2x = -2+y  , y = -2
Putting the value of y in 1st Equation ,we get
2x = -2 - 2
2x = -4
x = -2
so x= -2 and y =-2

Types of Relations

A relation R in a set A is called

(i) reflexive, if (a, a) ∈ R, for every a ∈ A,

(ii) symmetric, if (a, b) ∈ R implies that (a, b) ∈ R, for all a,b ∈ A.

(iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b,c ∈ A.

If you want to Read  HOW  TO  UNDERSTAND BINARY  OPERATIONS , RELATIONS  AND  FUNCTIONS || COMMUTATIVE || ASSOCIATIVE  click here 

Equivalence Relation

A relation R in a set A is said to be an equivalence  relation if R is reflexive, symmetric and transitive.

1 ) Let B be the set of all triangles in a plane with R a relation in B given by

R = {(T1, T2) : T1 is congruent to T2}. Then R is an equivalence relation.

2 ) Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7}  by

R = {(a, b) : both a and b are either odd or even}. Then R is an equivalence

one-one Function

A function f : X → Y is defined to be one-one (or injection ), if the images of distinct elements of X under f are distinct, i.e., for every x, y ∈ X, f (x) = f (y) implies x = y. Otherwise, f is called many-one.

Onto Function

A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an

element x in X such that f (x) = y.

Example

1   Function f : R → R, given by f (x) = 2x, is one-one and Onto As all the elements  have only one and uniqe image under f.

2  Function f : N → N, given by f (x) = 2x, is one-one but not onto.Because  the elements  have only one and unique image under f Therefore it is one one function .But not all elements of N have image under f 

e. g .  1,3,5,7... are not the image of any elements of N under f so it is not onto function

Example

The function f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,

for every x > 2, is onto but not one-one.

Solution

Since f is Not one-one, as f (1) = f (2) = 1. 

But f is Onto, as given any y ∈ N, y ≠ 1,

Choose x = y + 1 s.t.

 f (y + 1) = y + 1 – 1

f (y + 1)  = y. 

Also for 1 ∈ N, 

we are given  f (1) = 1

Inverse of a Function

A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f –1

Example

Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f , if it exists.

(a) f = {(1, 1), (2, 2), (3, 3)}

(b) f = {(2, 2), (3, 1), (4, 1)}

(c) f = {(1, 5), (3, 4), (2, 1)}

Solution

(a) It is to  proved that  f is one-one and onto Hence f is invertible with the inverse f –1 of  f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f.

(b) Since f (3) = f (4) = 1, f is not one-one, so that f is not invertible.

(c) Here  f   is one-one and onto, so that f is invertible with

 f –1 = {(5, 1), (4, 3), (1, 2)}.

Composition of Functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by

gof (x) = g(f (x)), ∀ x ∈ A

Example

fof(x) = (16x + 12 + 18x -12 ) / ( 24x + 18 - 24x +16)

fof(x) = (34 x ) / ( 34)

fof(x) =  x  =  I(x)

Example

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g(3) = g(4) = 7 and g(5) = g(9) = 11. Find gof = ?

Solution

We are given

 gof (2) = g (f (2)) 

               = g(3) 

               = 7

 gof (3) = g(f (3)

             = g(4)

              = 7,

gof (4) = g(f (4)) 

           = g(5) 

             = 11 

and  gof (5) = g(f (5))

                   = g (5)                     

                    = 11

So gof ={(2,7),(3,7),(4,11),(5,11)} 

Example

conclusion

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HOW TO FIND THE SOLUTIONS OF QUADRATIC EQUATIONS

Hello Friends welcome to My Blog, 

Today we are going to discuss Quadratic Equations , different methods of solution of Quadratic equation , its roots , Discriminant  and its formation with the help of some examples.

Quadratic Equation 

Any expression of the form   ax² + bx + c = 0,

where x represents a variable , and a, b, and c are constants but a ≠ 0 , is known as Quadratic Equation.The numbers a, b, and c are called  coefficients   of the equation.

ax² + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

Examples

1   x² + 4x − 4 = 0

2   3x² + 4x +4 = 0

3   2x² - 4x − 5 = 0

4   5x² -7x + 3 = 0

5   10 x² + 9x − 8= 0

6   1.5x² + 4.8x − 4.3 = 0

7   3x² + 2x + 1 = 0

8  x² +5x −20 = 0

9  2x²  – 3x + 1 = 0

10  4x – 3x²  + 2 = 0 and 

11  1 – x² + 300x = 0

Discriminant

The number D = b² – 4ac determined from the coefficients of the equation ax² + bx + c = 0 is called the Discriminant of the Quadratic Equation.

Let us consider  

x²  +5x −20 = 0, then a = 1, b = 5 and c = - 20,  D = b² – 4ac = 5² – 4×1×(-20) = 25 + 80 = 105

 2x² + 4x − 4 = 0,  a = 2,b = 4 and c = -4
D = b² – 4ac = 4² – 4×2×(-4) = 16+32 = 48

 x² + 4x + 4 = 0 , a = 1,b = 4 and c = 4
D = b² – 4ac = 4² – 4×1×(4) = 16-16 = 0

3x² + 6x + 4 = 0 , a=3,b=6 and c= 4
D = b² – 4ac = 6² – 4×3×(4) = 36 - 48 = -12

What are  Roots of Quadratic Equation

By roots of the quadratic Equation we means that values which we put in given equation satisfies the given quadratic Equation.

For example in this equation 5x² +20 x - 25 = 0 if we put values "1" and "5" in place of x then these values shall be called the roots of this equation ,Because these values satisfy the given equation.

Similarly the equation 4x2 - 16 = 0 shall have roots 4 and -4 ,because they shall satisfy the given equation

Nature of Roots

The nature of roots of Quadratic Equation can be discussed with the help of Discriminant .

1 When D = 0 Then roots are real and Equal, It means the roots of quadratic equation are without √  and both the roots are equal to each other . Some examples are  (5,5) , (6,6) , (8,8) etc

2 When D <0 Then roots are not real But they are complex conjugate of each other. It means the roots of given equation have roots like √ (-) term in it. Some examples are 2+ √ (-5)  and 2 - √ (-5)

3 When D >0 Then roots are real and Unequal

               a> When D is Positive but not perfect square then roots exists in Quadratic Surd . Some of the examples are 3+√ 5 and 3-√ 5

               b> When D is Positive and perfect square then roots are rational provided a,b and c are rational. Some example are 3 and 2

Note:

If l,m are two roots of any quadratic Equation ax² + bx + c = 0  ,then

Sum of roots     l + m  = - b/a

Product of roots   l × m = c/a

If  sum and the product of roots of any equation is given then, that equation can be written as follows

x² + ( sum of the roots ) x + Product of roots = 0

x² + ( -b/a ) x + (c/a) = 0  

           or

x² +  ( l+m ) x  +  lm = 0

Some Important facts about Quadratic Equation

 1  If sum of all the coefficients of a quadratic equation is equal to Zero ,then one of the root of that  equation is 1 i.e. “ unity”.

5x² +20 x - 25 = 0  ,  2x² -10 x + 8 = 0

2. when a quadratic equation has one root equal to zero, then it constant term must be equal to zero.

4x² -20 x   = 0 ,  2x² -12 x  = 0 

3. Any quadratic Equation will have reciprocal roots ,if its  coefficients of x² term and constant term are equal  i. e.  a = c.

4x² -20 x +4 = 0 , 2x² -20 x +2 = 0 

4 Any quadratic Equation will have negative reciprocal  roots ,if   coefficients of x² and constant term are equal in magnitude but opposite in sign. i.e  a = - c.

4x² -20 x -4 = 0 ,  2x² -20 x -2 = 0 

5 Any quadratic Equation will have equal in magnitude but opposite in  sign roots , if   coefficients of x equal to zero . i.e. b = 0.

4x²  - 16 = 0 , 5x²  -25 = 0 ,  -4x²  -36 = 0 

Solving the Quadratic Equation

1 Factorisation Method

2 Completing the Square

3 Quadratic Formula

Factorisation Method

To  solve  ax² + bx + c = 0 by Factorisation Method

1 Split the middle term into two terms in such a way that their         product must be equal to the product of a and c.

2 Take whichever is common in 1st two and Last two  factors.

3  Take whichever  is common in the two factors

4  Repeat the process of step 3 

5  Put Both Factors equal  to Zero and calculate the values of x

Example

4x² + 4x - 3 = 0 

4x²  -2x +6x - 3 = 0  Split the middle term into two terms

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

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

Either  2x - 1 = 0  or     2x + 3=0

               x = 1/2   or      x = -3/2

Example

2x² -3 x - 35 = 0 

2x² -10x +7x - 35 = 0 

(  Split the middle term into two terms )

2x (x - 5) +7( x - 5) = 0    Follow  Step 2 

(x-5)(2x+7) = 0                 Follow  Step 2 

Either  x - 5 = 0  or     2x + 7=0

               x = 5   or      x= -7/2

Example

4x² -20 x +25 = 0 

4x² -10x -10x + 25 = 0 
( Split the middle term into two terms )

2x (2x - 5) -5( 2x - 5) = 0   Follow  Step 2 

(2x-5)(2x-5) = 0         Follow  Step 2 

Either  2x - 5 = 0  or     2x -5=0

               x = 5/2   or      x= 5/2

Example

2x2 – 5x + 3 = 0,

2x2 – 2x – 3x + 3=0  
( Split the middle term into two terms )

2x (x – 1) –3(x – 1) = 0 
 ( Taking whichever is common )

(2x – 3)(x – 1)=0

Either  2x - 3 =  0  or x – 1 = 0

 Now, 2x = 3 and   x = 1.

x=3/2   and   x = 1.

Example

10x² + 21 x - 10 = 0 

10x² +25x - 4x - 10 = 0  
( Split the middle term into two terms )

5x (2x + 5) -2 ( 2x + 5) = 0  Follow  Step 2 

(2x+5)(5x-2) = 0    Follow  Step 2 

Either  2x + 5 = 0  or     5x - 2=0

               x = -5/2   or      x= 2/5

Completing the Square

1 Shift the constant term to right hand side of equal sign.

2 Complete the square in left side and add the term which is missing and adjust the added term on the right side.

3 Equate the Left hand term to Right hand term.

Let us consider a Quadratic Equation

9x² – 15x + 6 = 0

To make the complete square add the missing term   (b)²  and  subtract the same term

 (3x)²  – 2*3x *(5/2)+ (5/2)²–  (5/2)² +6 = 0

 {3x-(5/2)}² -(25/4)+6 = 0

{3x - (5/2)}² -(1/4) = 0

{3x-(5/2)}²  = (1/4)                      Taking square roots

3x - (5/2) = (1/2)  or  3x-(5/2) = - (1/2)  

3x = (1/2)+(5/2)  or 3x = -(1/2) + (5/2)

3x = 6/2 or 3x = 4/2

x = 1 or    3x = 2

x = 1 or    x = 2/3

The roots of the given equation are 1 and   2/3

Example

9x² + 12 x - 1 = 0 

9x² + 12 x = 1

(3x)² + 2× 3x× 2 + 2² = 1+2² 

(3x+2)² =1+4

(3x+2)² =5      Taking Square Roots

3x+2 =√  5                    or   3x+2 = - (√ 5)

3 x = √ 5  -   2                or     3x = - (√ 5) -2

x = ( √ 5  -  2  )/3      or   x = (  -√  5 - 2 ) /3

Roots are real and Unequal 

Example 

      4x² + 16 x - 9 = 0

      4x² + 16 x = 9

     (2x)² + 2× 2x× (4) + (4)² = 9+(4)² 

      (2x+4)² =9+16

    (2x+4)² =25

     (2x+4)² =5²   Taking Square Roots

2x+4 = 5

2x = 5-4

 x = 1/2

Two  real and Equal roots

Example

 x² -6 x + 11 = 0

  x² -6 x = - 11 

  (x)² -2(x)(3)+ 3²  = -11 + 9

    (x-3)² = -11+9  

     (x-3)² = -2  

As the Square of any real number can not be negative , so this
equation have NO Real Roots , The equation have only complex
roots.

Quadratic Formula

Example

4x² -40 x +100 = 0 

Since  D = b² -4ac 

           D= (-40)² - 4×4×100

           D = 1600-1600

           D= 0

x= -b/2a as second part of the formula vanishes i. e . 

x = -(-40)/(2×4)

x  = 40/8

x = 5

 Here both roots are equal and Real

Example

4x² - 20 x +29 = 0

Since  D = b² -4ac

           

           D = (-20)² - 4×4×29 

           D= 400 - 464= -64

The Squar root of -64 is  -8i

Since Discriminant is negative , so this quadratic Equation have no real roots but two complex roots can be found.

x= {-b+8 i} /(2a)    and  {-b-8 i}/{1/2(a)}

x = {-(-20)+8 i}/{2×4} and {-(-20)-8 i} /{2×4}

x={20+8i}/8  and    {20 - 8 i}/{8}

These are two roots of Quadratic Equation which are complex conjugate of each other .

Example

6x2-6x -1=0

Using the Quadratic Equation formula for the variable $\displaystyle{r}$:

=126±
√  (36+24​​

$\displaystyle=\frac{{{6}\pm\sqrt{{60}}}}{{12}}$

$\displaystyle=\frac{{{6}-{2}\sqrt{{15}}}}{{12}}{\quad\text{or}\quad}\frac{{{6}+{2}\sqrt{{15}}}}{{12}}$

$\displaystyle=\frac{{{3}-\sqrt{{15}}}}{{6}}{\quad\text{or}\quad}\frac{{{3}+\sqrt{{15}}}}{{6}}$

Hence here two roots are Quadratic surd ( Irrational ) 

Example


2x2-4x -5=0

Here a = 2 , b = -4 and c = - 5

Using the Quadratic Equation formula for the x

$\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}$

=44±√ 16+24
​​

$\displaystyle=\frac{{{4}\pm\sqrt{{40}}}}{{4}}$

=44±
√ 40​​

$\displaystyle=\frac{{{4}-\sqrt{{40}}}}{{4}}{\quad\text{or}\quad}\frac{{{4}+\sqrt{{40}}}}{{4}}$

=44−
√ 40  or44+√ 40​​

Roots are Irrational NOT Equal

Example

2x2 + 4x + 2 = 0

Here a = 2 , b = 4 and c = 2

Using the formula ,we get

$\displaystyle=\frac{{{4}\pm\sqrt{{{16}+{24}}}}}{{4}$
$\frac{-\; 4\pm \sqrt{}}{4}$

$\displaystyle=\frac{{{4}\pm\sqrt{{40}}}}{{4}}$

$\displaystyle=\frac{{{4}-\sqrt{{40}}}}{{4}}{\quad\text{or}\quad}\frac{{{4}+\sqrt{{40}}}}{{4}}$

x  = -1    and   -1

Two roots are  Equal and Real

Conclusion

In this post I have discussed different method of solutions of  Quadratic equation , its roots , Discriminant . If this post helped you little bit, then please share it with your friends and like this post  to boost me and to do better, and also follow me on my Blog .We shell meet in next post till then Bye .

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