Question 658079: Suppose that the sum of two numbers is 19 and the sum of their squares is 221. Find the numbers. Found 2 solutions by ewatrrr, aaronwiz:Answer by ewatrrr(24785) (Show Source):
Hi,
the sum of two numbers& is 19 and the sum of their squares is 221
Question states***
x + y = 19 0r
x^2 + y^2 = 221
x^2 + (19-x)^2 = 221
x^2 + 361-38x + x^2 = 221
2x^2 - 38x + 140 = 0
x^2 - 19x + 70 = 0
factoring
(x - 14)(x-5) = 0
(x - 14)= 0, x = 14 and y = 5
0r(x-5) = 0, x = 5 and y = 14
numbers are 5 and 14
You can put this solution on YOUR website! Hi, my name is Aaron I am in 10th grade and im in honors trig/pre-calc.
let
x=first number
y=second number
x+y=19
x^2+Y^2=221
x=19-y
x^2+Y^2=221
Now simply substitute
(19-y)^2+y^2=221
foil
(361-38y+y^2)+y^2=221
put into standard form
2y^2-38y+140=0
simplify
y^2-19y+70=0
factor
(y-14)(y-5)=0
y=14 or 5
plug y=5 back in to x^2+Y^2=221
25+x^2=221
x^2=196
thus x= plus or minus 14 when y=5
plug y=14 back in to x^2+Y^2=221
x^2+196=221
x^2=25
thus x= plus or minus 5 when y=14
However when negative numbers are plugged into x+y=19 it is never true. That means only positive solutions work. So one of the numbers is 5 and the other one is 14.
Hope I helped. I would really appreciate your recommendation, their are plenty of other things I could be doing rather then spending time on this site helping people. Good luck! :)
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