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Question 766155: 1...The sum of the squares of two numbers is 4 and the product of them is 1. Find those two numbers.
Answer by suruman(21)   (Show Source):
You can put this solution on YOUR website! Let the numbers be x and y.
Given :
x^2 + y^2 = 4 ....(a)
x*y=1 ....(b)
Multiplying equation 'b' by 2 on L.H.S and R.H.S,
2*x*y = 2 ....(c)
adding (a) and (c),
x^2 + 2*x*y + y^2 = 6
(x + y)^2 = 6
x + y = sqrt(6) (Taking only the positive square root. if x+y is taken as -sqrt(6), values of x and y would interchange.)
y = sqrt(6) - x
Thus, x*(sqrt(6) -x ) = 1 (since product of the numbers equals 1).
x*sqrt(6) - x^2 = 1
x^2 - sqrt(6)*x + 1 = 0
Roots of the quadratic equation of the type
a*x^2 + b*x + c = 0 is given by :
x = ((-b +- sqrt(b^2 - 4*a*c))/(2*a))
Thus, x = (sqrt(6) + sqrt(2))/2 or (sqrt(6) - sqrt(2))/2
y = (sqrt(6) - sqrt(2))/2 or (sqrt(6) + sqrt(2))/2
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