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# 4. find two consecutive positive integers, sum of whose squares is 365.

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### Mohammed

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## Find two consecutive positive integers, sum of whose squares is 365

Find two consecutive positive integers, sum of whose squares is 365. Two consecutive positive integers the sum of whose square is 365 are x = 13 and x + 1 = 14

## Find two consecutive positive integers, sum of whose squares is 365

Solution:

Let the first integer be x.

The next consecutive positive integer will be x + 1.

According to the given question,

x² + ( x + 1)² = 365

x² +( x + 1)² = 365

x² + (x² + 2x + 1) = 365 [ ∵ (a + b)² = a² + 2ab + b²]

2x² + 2x + 1 = 365

2x² + 2x + 1- 365 = 0

2x² + 2x - 364 = 0 2(x² + x - 182) = 0 x² + x - 182 = 0

x² + 14x - 13x - 182 = 0

x (x + 14) - 13 (x + 14) = 0

(x - 13) (x + 14) = 0

x - 13 = 0 and x + 14 = 0

x = 13 and x = - 14

The value of x cannot be negative (because it is given that the integers are positive).

∴ x = 13 and x + 1 = 14

☛ Check: Class 10 Maths NCERT Solutions Chapter 4Video Solution:

## Find two consecutive positive integers, sum of whose squares is 365

Class 10 Maths NCERT Solutions Chapter 4 Exercise 4.2 Question 4

Summary:

Two consecutive positive integers, the sum of whose squares is 365 are 13 and 14.

☛ Related Questions:

Find the roots of the following quadratic equations by factorization: (i) x^2 - 3x -10 = 0 (ii) 2x^2 + x - 6 = 0 (iii) √2x^2 + 7x + 5√2 = 0 (iv) 2x^2 - x + 1/ 8 = 0 (v) 100x^2 - 20x + 1= 0

Solve the problems given in example 1.(i) John and Jivanti had 45 Both of them lost 5 marbles each and the product of the no. of marbles they now have is 124. We would like to find out how many marbles they had to start with?(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.

Find two numbers whose sum is 27 and product is 182.

स्रोत : www.cuemath.com

## Ex 4.2, 4

Ex 4.2 ,4 Find two consecutive positive integers, sum of whose squares is 365. There is difference of 1 in consecutive positive integers Let First integer = x So, Second integer = x + 1 Also given that Sum of squares = 365 (First number)2 + (Second number)2 = 365 x2 + (x + 1)

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## Ex 4.2, 4 - Chapter 4 Class 10 Quadratic Equations (Term 2)

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### Transcript

Ex 4.2 ,4 Find two consecutive positive integers, sum of whose squares is 365. There is difference of 1 in consecutive positive integers Let First integer = x So, Second integer = x + 1 Also given that Sum of squares = 365 (First number)2 + (Second number)2 = 365 x2 + (x + 1)2 = 365 x2 + x2 + 12 + 2 × x ×1=365 2x2 + 1 + 2x = 365 2x2 + 2x +1 – 365 = 0 2x2 + 2x – 364 = 0 2 (x2 + x – 182) = 0 x2 + x – 182 = 0/2 x2 + x – 182 = 0 We factorize by splitting the middle term method x2 + 14x – 13 x – 182 = 0 x (x + 14) – 13 (x + 14) = 0 (x – 13) (x + 14) = 0 So the root of the equation are x = 13 & x = – 14 Since we have to find consecutive positive numbers We take x = 13 First number = x = 13 Second number = x + 1 = 13 + 1 = 14

Next: Ex 4.2, 5 →

### Davneet Singh

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## Find two consecutive positive integers, sum of whose square is 365.

Find two consecutive positive integers, sum of whose square is 365.. Ans: Hint: Here we assume two consecutive positive integers as two variables and use the concept that consecutive integers have a difference of 1 between them to form the first equa...

## Find two consecutive positive integers, sum of whose square is 365.

Last updated date: 17th Mar 2023

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Hint: Here we assume two consecutive positive integers as two variables and use the concept that consecutive integers have a difference of 1 between them to form the first equation. Then form up an equation of sum of squares of these two numbers and equate it to 365. Use factorization method to find the numbers.

* Factorization method: If ‘p’ and ‘q’ are the roots of a quadratic equation, then we can say the quadratic equation is

(x−p)(x−q)=0 (x−p)(x−q)=0

Complete step-by-step solution:

We know that positive integers are the integers that lie on the right hand side of 0 on the number line. Also, we know consecutive numbers mean numbers that are exactly together i.e. right next to each other.

So, consecutive positive integers can be 1,2 ; 2,3 ; 3,4 ; …..

So if we assume the first positive integer as

x x

Then consecutive positive integer will be

x+1 x+1

Now we know the sum of squares of two consecutive positive integers is 365

⇒ x 2 + (x+1) 2 =365 ⇒x2+(x+1)2=365 Use the identity (a+b ) 2 = a 2 + b 2 +2ab (a+b)2=a2+b2+2ab

to open the value in left hand side of the equation

⇒ x 2 + x 2 +1+2x=365 ⇒x2+x2+1+2x=365

Shift all the constant values to one side

⇒2 x 2 +2x=365−1 ⇒2x2+2x=365−1 ⇒2 x 2 +2x=364 ⇒2x2+2x=364

Cancel 2 from both sides of the equation

⇒ x 2 +x=182 ⇒x2+x=182 ⇒ x 2 +x−182=0 ⇒x2+x−182=0

Now we use factorization method to solve for value of x

Break the coefficient of ‘x’ in such a way that their sum is equal to coefficient of ‘x’ and their product is equal to the product of coefficient of

x 2 x2

and the constant term. We can write

182=14×13 182=14×13 We can write 1=14−13 1=14−13

i.e. the coefficient of x

Substitute the value of

1=14−13 1=14−13 in equation (1) ⇒ x 2 +x−182= x 2 +(14−13)x−182

⇒x2+x−182=x2+(14−13)x−182

Open the bracket in RHS of the equation

⇒ x 2 +x−182= x 2 +14x−13x−182

⇒x2+x−182=x2+14x−13x−182

Take common terms ⇒ x 2

+x−182=x(x+14)−13(x+14)

⇒x2+x−182=x(x+14)−13(x+14)

Combine the factors ⇒ x 2 +x−182=(x+14)(x−13)

⇒x2+x−182=(x+14)(x−13)

Equate factors to 0 x+14=0 x+14=0 and x−13=0 x−13=0

Shift constant values to right hand side

x=−14 x=−14 and x=13 x=13

Since we have both integers positive, we ignore the negative value

⇒x=13 ⇒x=13

Now we calculate the consecutive positive integer i.e.

13+1=14 13+1=14 ∴ ∴

The two positive integers having sum of their squares 365 are 13 and 14.

Note: When shifting the values from one side of the equation to another side, always keep in mind the sign changes from positive to negative and vice versa when we shift a number from one side to another side of the equation.

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Mohammed 8 day ago

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