x²-14x +13 
 x²-13x - x  +13 = 0

  x ( x - 13) - 1 ( x - 13) = 0

  ( x - 13) ( x-1) = 0

x - 13 = 0 and x- 1 = 0

x = 13 and x = 1

Co-prime number is a set of numbers or integers which have only 1 as their common factor i.e. their highest common factor (HCF) will be 1. Co-prime numbers are also known as relatively prime or mutually prime numbers. It is important that there should be two numbers in order to form co-primes.
Example 1: 21 and 22

For 21 and 22:

• The factors of 21 are 1, 3, 7 and 21.
• The factors of 22 are 1, 2, 11 and 22.

Here 21 and 22 have only one common factor that is 1. Hence, their HCF is 1 and are co-prime.

Given : 

To find : 

Consider :   

Squaring both sides , we get

   [Since , ]

   (1)

Consider 

From (1),

i.e. 

Taking square root on both sides , we get

. 

Hence, the value of 

A rational number, in Mathematics, can be defined as any number which can be represented in the form of p/q where q is greater than 0. Also, we can say that any fraction fit under the category of rational numbers, where denominator and numerator are integers and the denominator is not equal to zero.

Let the first zero be x then the another one is 1/x
now , alpha × beta = c/a
x × 1/x = 3k/k+4
1 = 3k/k+4
k+4 =3k
4= 2k
therefore, k =2

25 = 5 × 5

15 = 5 × 3
G.C.F (25,15) = 5

G.C.F (25,15)= 4X+1

4x + 1 = 5
4x = 5 - 1 

4x = 4

x = 4/4

x = 1

Given :

• Zeroes of polynomial = 7 and 5

To Find :

• A quadratic polynomial

Solution :

⇒ x = 7 and x = 5

So,

⇒ ( x - 7 ) ( x - 5 )

⇒ x² - 5x - 7x + 35

⇒ x² - 12x + 35

x² - 12x + 35 is the quadratic polynomial whose zeroes are 7 and 5

An Irrational Number is a real number that cannot be written as a simple fraction . Example: π (Pi) is a famous irrational number. We cannot write down a simple fraction that equals Pi. The popular approximation of 22/7 = 3.1428571428571 ... is close but not accurate. Another clue is that the decimal goes on forever without repeating.

Irrational numbers between 2 and 3

Square of 2 is 4

Square of 3 is 9

Square root of 5,6,7,8 will be irrational number between 2 and 3 since these numbers are not perfect square .

Prime Number Which is Divisible By 1 And Number Itself

Example:2,3,5,7,11, Etc…..

In Prime We Should Have Only One Number

That’s Why it is Called Prime

Co Prime Number Which is Having A Common Factor Among Them

Example:21,22

21-Factors: 1,3,7,21

22 -Factors: 1,2,11,22

In Co -Prime We Should Have Two Numbers

In This We Have Only 1 Common Factor

That’s Why it is Called Co -Prime

let,

5 - 2√3 be a rational number.

.°. 5 - 2√3 = p/q [ where p and q are integer , q ≠ 0 and q and p are co- prime number ]

=> - 2√3 = p/q -5

=>- 2√3 = p - 5q / q

=> √3 = p - 5q / - 2q

we know that p/q is a rational number.

.°. √3 is also a rational number.

This contradicts our assumption

5 - 2√3 is an irrational number

For this let reverse the AP..
now first term = a = -62
common difference = d = 3
now 11th term from this side is...
A₁₁ = a+10d = -62+10(3) =-62 + 30 = -32
hence 11th term from the last is -32

Find the value of k for which the roots of the following equation are real and equal

k^2x^2-2(2k-1)x+4=0

Answer:

Let the unit digit and tens digits of the number be x and y                  

Number = 10y + x                  

Number after reversing the digits = 10x + y    

According to the question,                  

⇒  x + y = 9 ... (i)                  

⇒  9(10y + x) = 2(10x + y)                  

⇒  88y - 11x = 0                  

⇒ -x + 8y =0 ... (ii)                  

Adding equation (i) and (ii), we get                  

⇒  9y = 9                  

⇒  y = 1 ... (iii)                  

Putting the value in equation (i), we get                  

⇒  x = 8                  

Hence, the number is 10y + x = 10 × 1 + 8 = 18.

Let the unit digit and tens digits of the number be x and y                  

Number = 10y + x                  

Number after reversing the digits = 10x + y    

According to the question,                  

⇒  x + y = 9 ... (i)                  

⇒  9(10y + x) = 2(10x + y)                  

⇒  88y - 11x = 0                  

⇒ -x + 8y =0 ... (ii)                  

Adding equation (i) and (ii), we get                  

⇒  9y = 9                  

⇒  y = 1 ... (iii)                  

Putting the value in equation (i), we get                  

⇒  x = 8                  

Hence, the number is 10y + x = 10 × 1 + 8 = 18.

Let us assume Jacob’s current age is x and his son’s current age is y.

Five years hence, Jacob’s age = x + 5 and his son’s age = y + 5

As per question;

x + 5 = 3(y + 5)

Or, x + 5 = 3y + 15

Or, x = 3y + 15 – 5 = 3y + 10

Five years ago, Jacob’s age = x – 5 and his son’s age = y – 5

As per question;

x – 5 = 7(y – 5)

Or, x – 5 = 7y – 35

Or, x = 7y – 35 + 5 = 7y – 30

Substituting the value of x from first equation in this equation, we get;

3y + 10 = 7y – 30

Or, 3y + 10 + 30 = 7y

Or, 7y – 3y = 40

Or, 4y = 40

Or, y = 10

Substituting the value of y in first equation, we get;

x = 3y + 10

Or, x = 3 x 10 + 10 = 40

Hence, Jacob’s age = 40 years and son’s age = 10 years

25 = 5 × 5

15 = 5 × 3
H.C.F (25,15) = 5

H.C.F (25,15)= 4X+1

4x + 1 = 5
4x = 5 - 1 

4x = 4

x = 4/4

x = 1

    let cost(in RS) of one pencil = x
             and cost (in RS) of one pen = y
Therefore , according to question
                      5x+7y = 50  ........ (1)
                     7x + 5y = 46  .........(2)
       Multiply equation 1 by 7 and equation by 5 we get
                         7(5x+7y)= 7*50
                         35x +49y = 350 .......(3)
                   and
                         5(7x +5y) = 5*46
                         35x +25y = 230 ....... (4)

Subtract equation 4 from equation 3 , we get
                            35x + 49y - 35x - 25y = 350 -230
                             49y -25y = 120
                             24y = 120
                                 y = 120 /24
                                 y= 5
Substitute y=5 in equation 1 , we get
                 5x +  7*5 =50
                 5x+35 =50
                 5x = 50 - 35
                 5x =15
                 x= 15/5
                 x=3

Hence Cost of One Pencil = x =3
           Cost  of One Pen =y =5

Let us assume, to the contrary, that 3 √ 2​ is 

rational. Then, there exist co-prime positive integers a and b such that

3 √ 2​= a/b

⇒  √ 2 ​= a/3b 

⇒  √ 2​ is rational       ...[∵3,a and b are integers∴ 1/3b ​is a rational number]

This contradicts the fact that  √ 2​ is irrational. 

So, our assumption is not correct.

Hence, 3 √ 2​ is an irrational number.