Asked by Joel
                If The Roots Of The Equation X^2-  bx+c are square root of alpha and square root of beta .show that alpa + beta = b^2- 2c and (alpha)^2 + (beta)^2 = (b^2 -2c - sqrt of 2c)(b^2 - 2c + sqrt of 2c)
            
            
        Answers
                    Answered by
            Reiny
            
    equation: x^2 - bx + c
roots: √a and √b
sum of roots = √a + √b = b
product or roots = √a√b = c --> ab = c^2
√a + √b = b
square both sides
(√a + √b)^2 = b^2
a + 2√(ab) + b = b^2
then a+b = b^2 - 2√(ab) = b^2 - 2c
DONE
a^2 + b^2
= (a+b)^2 - 2ab
= (b^2 - 2c)^2 - 2c^2
treat this as a difference of squares
= (b^2 - 2c - √2c)(b^2 - 2c + √2c)
as required
    
roots: √a and √b
sum of roots = √a + √b = b
product or roots = √a√b = c --> ab = c^2
√a + √b = b
square both sides
(√a + √b)^2 = b^2
a + 2√(ab) + b = b^2
then a+b = b^2 - 2√(ab) = b^2 - 2c
DONE
a^2 + b^2
= (a+b)^2 - 2ab
= (b^2 - 2c)^2 - 2c^2
treat this as a difference of squares
= (b^2 - 2c - √2c)(b^2 - 2c + √2c)
as required
                    Answered by
            Damon
            
    alpha^.5 = [b +sqrt(b^2-4c)]/2
beta^.5 = [b -sqrt(b^2-4c)]/2
alpha =(1/4)
[ b^2 +2bsqrt(b^2-4c)+b^2-4c]
beta =(1/4)
[ b^2 -2bsqrt(b^2-4c)+b^2-4c]
alpha + beta = (1/4)
[4 b^2 -8 c]
= b^2 - 2 c
etc ....
    
beta^.5 = [b -sqrt(b^2-4c)]/2
alpha =(1/4)
[ b^2 +2bsqrt(b^2-4c)+b^2-4c]
beta =(1/4)
[ b^2 -2bsqrt(b^2-4c)+b^2-4c]
alpha + beta = (1/4)
[4 b^2 -8 c]
= b^2 - 2 c
etc ....
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