SOLUTION: One number is three more than a second number. The sum of their squares is 37 more than the product of the numbers. Find the numbers.
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Question 228601
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One number is three more than a second number. The sum of their squares is 37 more than the product of the numbers. Find the numbers.
Answer by
rapaljer(4671)
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Let x = second number
x+3 = first number
Sum of squares IS 37 more than the product of the numbers.
x^2 + (x+3)^2= x(x+3) + 37
x^2 + x^2 + 6x +9 = x^2 +3x + 37
Set equation equal to zero by subtracting x^2 +3x + 37 from each side:
2x^2 + 6x + 9-x^2 -3x-37=0
x^2 + 3x - 28=0
This just happens to factor!
(x+7)(x-4)=0
There are two solutions:
First solution:
x=-7
x+3= -4
Second solution:
x=4
x+3= 7
Both solutions do check if you take the time to do it.
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus