Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two.

Question

This question has been stumping me for the past 24 hours i need help.Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. one solution is negative −5, negative −3, and negative −1. Find three other consecutive odd integers that also satisfy the given conditions.

Answer 1

1. Let A, A+2 and A+4 be the three consecutive odd integers
2. (A+4)² = (A)² + (A+2)² - 33
3. A² + 8A + 16 = A² + A² +4A + 4 -33
4. A² + 8A + 16 = 2A² + 4A -29
5. A² -4A - 45 = 0
6. Use quad formula where a=1, b=-4 and c=-45
7. A = 4 +/-  sqrt[16- (4)(1)(-45)] / 2(1) =  (4 +/- 14) / 2 =  -10/2 and 18/2 = -5 and 9
8. You already provided the case where A = -5, so the other answer is A =9.  When A=9, the three numbers are 9, 11, 13.
9. Check the answer :  Does 13² = 11² + 9² - 33?  yes, both sides of equation equal 169

Answer 2

 

1st integer = X

2nd integer = X+2

3rd integer = x + 4

 

(X+4)^2 = -33 +  x^2 + (x+2)^2  

 

x^2 + 8x + 16 = -33 + x^2 + x^2 + 4x + 4 <--- FOIL 

 

x^2 + 8x + 16 = 2x^2 + 4x - 29  <--- combines like terms on right side

 

0 = 2x^2 + 4x - 29 - x^2 - 8x - 16 <---- moves everything to right side, signs change

 

0 = x^2 -4x - 45  <---- combines like terms again

 

0 = (x   -  9)(x +   5 ) <--- factors

 

x-9 = 0   OR x +5 = 0 <--- zero product property; one of them must be zero

 

x = 9  or x = -5

 

SO there indeed are 2 sets of integers that meet this condition:

 

9,11,and 13  --->   13^2 = 169 = 202 -33 = 81 + 121 - 33 = 9^2 + 11^2 - 33

 

-5,-3,-1  -->   (-1)^2 = 1 = -33 + 34 = 25 + 9 - 33 = (-5)^2 + (-3)^2 - 33