Determining odd integers.

Question

How would I solve this?

 

Three consecutive odd integers are such that the sum of the squares of the first two integers is 54 more than 20 times the third integer. Determine the three integers.

Answer 1

Start by assigning a variable...  Let x be the smallest of the three integers.

 

Since odd integers skip over the even ones, they are 2 apart, so the next odd integer must be x + 2, and then the third must be x + 4.

 

Now let's see what the problem says:

 

"the sum of the squares of the first two integers" -- that's the first one squared plus the second one squared, or x² + (x + 2)² = x² + (x² + 4x + 4) = 2x² + 4x + 4

 

"is" -- that means "equals"

 

"54 more than 20 times the third integer" -- that 20(x + 4) + 54 = 20x + 80 + 54 = 20x + 134.

 

So now we have an equation we can solve:

 

2x² + 4x + 4 = 20x + 134

2x² -16x - 130 = 0

x² - 8x - 65 = 0

(x - 13)(x + 5) = 0

 

So x is 13 or -5.

 

So the three integers are either:  13,15,17 or -5,-3,-1.  They probably mean the first set, but there's nothing in the problem that says that the integers have to be positive.