Question

Find two consecutive odd positive integers, sum of whose squares is 290
​

Answer 1

Answer:

11 and 13

Given:

• The two numbers are consecutive odd positive integers.
• Sum of their squares is 290

To Find:

• The two integers.

Solution:

The integers are odd consecutive, so, there difference should be 2.

let the first integer = x

The second integer = x + 2

According to the question

⟹ ( x )² + ( x + 2)² = 290

By opening the whole squares,

⟹ x² + x² + 4x + 4 = 290

⟹ x² + x² + 4x + 4 - 290 = 0

⟹ 2x² + 4x - 286 = 0

⟹ 2( x² + 2x - 286) = 0

⟹ x² + 2x - 286 = 0

By splitting the middle term,

⟹ x² + 13x - 11x - 143 = 0

⟹ x ( x + 13 ) - 11 ( x + 13 ) = 0

⟹ ( x + 13 ) ( x - 11 ) = 0

Now,

⟹ (x + 13) = 0

⟹ x = -13

- 13 is not possible as the integers should be positive

OR

⟹ (x - 11 ) = 0

⟹ x = 11

First integer = x = 11

Second integer = (x + 2) = 13

11 and 13 are the integers.

Verification:

Sum of their squares should be 290

⟹ (11)² + (13)² ≟ 290

⟹ 121 + 169 ≟ 290

⟹ 290 = 290

LHS = RHS

verified !