SOLUTION: Find 3 consecutive odd numbers where the product of the smaller two numbers is 46 less than the square of the largest number.

Question 884520: Find 3 consecutive odd numbers where the product of the smaller two numbers is 46 less than the square of the largest number.
Answer by algebrapro18(249) (Show Source): You can put this solution on YOUR website!
Well consecutive numbers come one after the next. So 3 consecutive odd numbers would be like 1,3,5 or 3,5,7. Notice the difference between the numbers is two. So to go from 1 to the next we just take the first number and add 2. Writing this algebraically we get:

Number 1 = x
Number 2 = x+2
Number 3 = x+2+2 = x+4

So now we need to find 3 consecutive odd numbers such that the product of the smaller two numbers is 46 less than the square of the largest number. The smallest two numbers above are x and x+2 and the largest number would be x+4. Writing this algebraically we get:

x(x+2)=(x+4)^2-46

Now we just plug and chug to solve for x.

x(x+2)=(x+4)^2-46
x^2+2x = (x+4)(x+4)-46
x^2+2x = x^2+4x+4x+16-46
x^2+2x = x^2+8x+16-46
x^2+2x = x^2+8x-30
0 = x^2-x^2+8x-2x-30
0 = 6x-30
30 = 6x
5 = x

So we know that the first number is 5. Plugging back into our equations gives us the final answer of:

Number 1 = x = 5
Number 2 = x+2 = 5+2 = 7
Number 3 = x+2+2 = x+4 = 5+4 = 9

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