College Algebra

Question

Three consecutive odd integers are such that the square of the third integers is 105 less than the sum of the squares of the first two. One solution is -9,-7,and -5. Find three other consecutive odd integers that also satisfy the given conditions.

Answer 1

The other solution is 13, 15, and 17.  THe square of 13 is 169 while 15 squared is 225. These two squares sum to 394 which is also the sum of 105 and 289. 289 is the square of 17.

Answer 2

Three consecutive odd integers are such that the square of the third integers is 105 less than the sum of the squares of the first two. One solution is -9,-7,and -5. Find three other consecutive odd integers that also satisfy the given conditions.

 

Let the three consecutive odd integers be:

     x

    x + 2

    x + 4

 

Translate:

  "square of the third integers is 105 less than the sum of the squares of the first two"     means

              (x + 4)²                  =  -105                   +    (    x²  + (x + 2)²   )

 

Now, the algebra:

     x² + 8x + 16     =   -105   +  x²   + x² + 4x + 4             [use FOIL to expand]

     0 = x² -4x - 115

    0 = (x + 9)(x - 13)                     [either factor or use quadratic formula]

  Either x=-9  or x=13                    [-9 was given, so the other answer is x=13]

The three consecutive odd integers are 13, 15, 17

 

Checking (very important):

  Is 17² = 13² + 15² - 105   ?

      289 = 169 + 225 - 105   ?

      289  =  289   ?  yes