Question

The sum of the squares of two consecutive odd positive integers is 290. Find the numbers.

Answer 1

Solution :

Let x , ( x + 2 ) are two consecutive

odd positive Numbers.

According to the problem given ,

x² + ( x + 2 )² = 290

=> x² + x² + 4x + 4 - 290 = 0

=> 2x² + 4x - 286 = 0

Divide each term by 2 , we get

=> x² + 2x - 143 = 0

=> x² + 13x - 11x - 143 = 0

=> x( x + 13 ) - 11( x + 13 ) = 0

=> ( x + 13 )( x - 11 ) = 0

x + 13 = 0 or x - 11 = 0

x = -13 , x = 11

x is positive integer.

Therefore ,

Required consecutive odd integers

are ,

x = 11 ,

x + 2 = 11 + 2 = 13

••••••

Answer 2

Answer:

• 11 and 13 are the numbers.

Step-by-step explanation:

Given

• Sum of the squares of two consecutive odd positive integers is 290.

To find

• The numbers

Solution

Let the first positive integer be x. Then, the second integer will be x + 2.

According to the question:

• x² + (x + 2)² = 290
• x² + x² + 4x + 4 = 290
• 2x² + 4x - 286 = 0
• x² + 2x - 143 = 0

(Divided by 2)

Using standard quadratic formula:

• x = -b±√b² - 4ac/2

On substituting, we get:

• x = -2±√4 + 572/2
• x = -2±√576/2
• x = -2±24/2
• x = 11 or x = -13

(x cannot be negative, so value of x is 11)

Hence, the numbers are:

• 11
• 13

Since,  11² + 13² = 121 + 169 = 290