Find two consecutive of the odd natural numbers in the sum of the whose squares is 202

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Let the two consecutive odd numbers be x and x+2 .

It is given that the sum of their squares is equal to 202. Therefore we have x^2 + (x+2)^2 = 202

=> x^2 + x^2 + 4x + 4 = 202

=> 2x^2 + 4x + 4 = 202

subtract 202 from both the sides

=> 2x^2 + 4x - 198 =0

divide all terms by 2

=> x^2 + 2x - 99 = 0

=> x^2 + 11x - 9x - 99 =0

=> x(x+11) -9(x+11) =0

=> (x -9) (x+11) =0

From x -9 =0 we get x = 9

and from x +11 = 0 we get x = -11 which is rejected.

Therefore the numbers are ( 9, 11).

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let the first odd number be x

Then the next odd number is (x+2)

Given that the sum of the squares of the numbers is 202.

Then we will write the expression as follows.

x^2 + (x+2)^2 = 202

We will open brackets.

==> x^2 + x^2  + 4x + 4 = 202

Combine like terms.

==> 2x^2 + 4x + 4 = 202

Divide by 2:

==> x^2 + 2x + 2 = 101

==> x^2 + 2x - 99 = 0

==> ( x +11)( x- 9) = 0

==> Then, the numbers are : 11 and 9

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We know the odd number are of the form 2m+1  Or 2m-1. And the consecutive odd numbers  could be at a difference of 2.

So let the two odd natural number be 2m-1 and 2m+1, for m in natural number set.

Then their sum is equal to 2m-1+2m+1 = 4m.

But given that the sum of the 2 odd numbers is 202.

So 4m = 202.

m = 202/4 = 50.5 which is not a natural number.

Therefore  2m+1 = 2*50.5+1 = 101+1 = 102.

2m-1 = 2*50.5-1 =101 -1 = 100.

100+102 = 202.

But 100 and 102 are even consecutive numberand not natural consecutive odd numbernumber .

There is solution set in odd cosecutive natural number.

Therefore

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The odd number is 2k + 1.

The other consecutive odd number is 2k + 3.

The sum of their squares is 202:

(2k+1)^2 + (2k+3)^2 = 202

We'll raise to square:

4k^2 + 4k + 1 + 4k^2 + 12k + 9 = 202

We'll combine like terms:

8k^2 + 16k - 192 = 0

We'll divide by 8:

k^2 + 2k - 24 = 0

We'll apply the quadratic formula:

k1 = [-1 + sqrt(4 + 96)]/2

k1 = (-1 + sqrt 100)/2

k1 = (-1+10)/2

k1 = 9/2

k2 = -11/2

The first odd number is:

2k + 1 = 2*9/2  = 9

The second odd number is 11.

Since the numbers are natural, the second solution of k is rejected.

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