Let's solve the problem step-by-step.
Problem:
Find two consecutive odd positive integers such that the sum of their squares is 290.
Step 1: Define variables
Let the first odd positive integer be xxx.
Since the integers are consecutive odd numbers, the next consecutive odd
integer will be x+2x+2x+2.
Step 2: Write the equation
The sum of their squares is given as 290:
x2+(x+2)2=290x^2+(x+2)^2=290x2+(x+2)2=290
Step 3: Expand and simplify
x2+(x2+4x+4)=290x^2+(x^2+4x+4)=290x2+(x2+4x+4)=290
2x2+4x+4=2902x^2+4x+4=2902x2+4x+4=290
2x2+4x+4−290=02x^2+4x+4-290=02x2+4x+4−290=0
2x2+4x−286=02x^2+4x-286=02x2+4x−286=0
Divide the entire equation by 2 to simplify:
x2+2x−143=0x^2+2x-143=0x2+2x−143=0
Step 4: Solve the quadratic equation
Use the quadratic formula x=−b±b2−4ac2ax=\frac{-b\pm \sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac, where a=1a=1a=1, b=2b=2b=2, and c=−143c=-143c=−143:
x=−2±22−4×1×(−143)2×1x=\frac{-2\pm \sqrt{2^2-4\times 1\times (-143)}}{2\times 1}x=2×1−2±22−4×1×(−143)
x=−2±4+5722x=\frac{-2\pm \sqrt{4+572}}{2}x=2−2±4+572
x=−2±5762x=\frac{-2\pm \sqrt{576}}{2}x=2−2±576
x=−2±242x=\frac{-2\pm 24}{2}x=2−2±24
Two possible solutions:
- x=−2+242=222=11x=\frac{-2+24}{2}=\frac{22}{2}=11x=2−2+24=222=11
- x=−2−242=−262=−13x=\frac{-2-24}{2}=\frac{-26}{2}=-13x=2−2−24=2−26=−13
Since the problem states positive integers, we take x=11x=11x=11.
Step 5: Find the consecutive odd integer
The next consecutive odd integer is:
x+2=11+2=13x+2=11+2=13x+2=11+2=13
Step 6: Verify the solution
112+132=121+169=29011^2+13^2=121+169=290112+132=121+169=290
This matches the given condition.
Final Answer:
The two consecutive odd positive integers are 11 and 13.
