SOLUTION: The sum of two numbers is 25 and the sum of their squares is 313. Find the numbers. (Express one number as x, the other in terms of x).
Thanks.
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Question 117635This question is from textbook
: The sum of two numbers is 25 and the sum of their squares is 313. Find the numbers. (Express one number as x, the other in terms of x).
Thanks.
This question is from textbook
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
x+y=25 or x=25-y
x^2+y^2=313 substitute (25-y) for x in this equation & solve for y
(25-y)^2+y^2=313
625-50y+y^2+y^2=313
2y^2-50y+625-313=0
2y^2-50y+312=0
2(y^2-25y+156)=0
2(y-13)(y-12)=0
y-13=0
y=13 answer.
y-12=0
y=12 answer.
proof
12^2=13^2=33
144+169=313
313=313
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of two numbers is 25 and the sum of their squares is 313. Find the numbers. (Express one number as x, the other in terms of x).
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Let one of the numbers be "x" ; the other number is "25-x".
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EQUATION:
x^2 + (25-x)^2 = 313
x^2 + 625 - 50x + x^2 = 313
2x^2 -50x + 212 = 0
x^2-25x+106 = 0
x = [25 +- sqrt(625-4*106)]/2
x = [25 +- sqrt(201)]/2
x = (25+sqrt(201))/2 = 19.5887
OR
x = (25-sqrt(201)/2 = 5.4113
======================
Cheers,
Stan H.
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