We get a system of equations:
a+b=12
a2+b2=90.
Replace 12-a for b, and we get:
a2+(12-a)2=90
2a2-24a+144=90
2a2-24a+54=0
a2-12a+27=0
a1=9
a2=3
Because of symmetry, we get two equivalent solutions:
a=9, b=3
or
a=3,b=9
Let the numbers be l & m .
Hence
Sum l + m = 12
Sum of the squares is l^2 + m^2 = 90
Hemce
l = 12 - m
l^2 = ( 12 - m)^2
l^2 = 144 - 24m + m^2
Substitute
144 - 24m + m^2 + m^2 = 90
2m^2 - 24m = -54
m^2 - 12m = -27
m^2 - 12m + 27 = 0
Factor
(m - 3)(m - 9) =0
m = 3 & m = 9 are the two numbers.
5
They are 8 and 13.
3 and 5
7 and 9 72+92 = 130
x2 + y2 = (x + y)2 => x2 + y2 = x2 + 2xy + y2 => 2xy = 0 => xy = 0 So, one of x and y must be 0.
split 10 in two parts such that sum of their squares is 52. answer in full formula
The two numbers are 6 and 9
The two numbers are 9 and 13.
The sum of their squares is 10.
Sum of squares? Product?
There is no single number here. The two seed numbers are 5 and 6; their squares sum to 61.
Not unless at least one of the numbers is zero.
85
5
5
5
They are 8 and 13.