Question 1107401: The sum of two integers is 8 and the sum of their squares is 19 more than their product. Find the integers.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The sum of two integers is 8
a + b = 8
b = (-a+8)
and the sum of their squares is 19 more than their product.
a^2 + b^2 = ab + 19
replace b with (-a+8)
a^2 + (-a+8)^2 = a(-a+8) + 19
distribute
a^2 + a^2 - 16a + 64 = -a^2 + 8a + 19
combine like terms on the left
a^2 + a^2 + a^2 - 16a - 8a + 64 - 19 = 0
a quadratic equation
3a^2 - 24a + 45 = 0
simplify, divide by 3
a^2 - 8a + 15 = 0
Factors to
(a-5)(a-3) = 0
a = 5, then b = 3, (we know their sum is 8)
and
a = 3, then b = 5
:
Find the integers: 5 & 3
:
:
Check solutions in the statement
" the sum of their squares is 19 more than their product.
25 + 9 = 15 + 19
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