Let the first odd integer be x. So, second one will be x+2.

ATQ

x^2+(x+2)^2=290
x^2+x^2+4x+4=290
2x^2+4x=286
x^2+2x-143=0
x^2+13x-11x-143=0
x(x+13)-11(x+13)=0
(x-11)(x+13)=0
x-11=0, x+13=0
x=11 , x=-13

we have given both are positive integers so that numbers must be 11 and 13.

Thank you.. :-)

To find two consecutive odd positive integers whose sum of squares is 290, we will use algebraic equations that represent the given conditions.


1. Let x be the first odd positive integer.
2. Since we are looking for two consecutive odd positive integers, the second odd positive integer will be x + 2.
3. According to the problem, the sum of the squares of these two integers is 290. Therefore, we can set up the following equation:

x^2 + (x + 2)^2 = 290

4. Expanding the equation gives us:

x^2 + x^2 + 4x + 4 = 290

5. Simplifying the equation gives us:

2x^2 + 4x - 286 = 0

6. Dividing both sides by 2 gives us:

x^2 + 2x - 143 = 0

7. We can now solve for x using the quadratic formula:

x = (-2 ± √(2^2 - 4(1)(-143)))/2(1)

x = (-2 ± √(576))/2

x = (-2 ± 24)/2

x = -13 or x = 11

8. Since we are looking for positive integers, we can ignore the negative value of x and conclude that the first odd positive integer is 11.
9. The second odd positive integer is x + 2 = 11 + 2 = 13.
10. To check if these numbers are correct, we can plug them back into the original equation:

11^2 + 13^2 = 121 + 169 = 290

11 and 13 are indeed two consecutive odd positive integers whose sum of squares is 290.


We have successfully found two consecutive odd positive integers whose sum of squares is 290 using algebraic equations. The first odd positive integer is 11 and the second odd positive integer is 13.