Question 871953

If the two roots of the equation: x^2+x-3=0 are m and n, then


(x-m)(x-n) = x^2+x-3
x^2-xm-xn+mn = x^2+x-3
x^2-(m+n)x+mn = x^2+x-3


Equate coefficients to get

-(m+n) = 1 which leads to m+n = -1
mn = -3


Now square both sides and isolate m^2 + n^2


m+n = -1
(m+n)^2 = (-1)^2
(m+n)^2 = 1
m^2 + 2mn + n^2 = 1
m^2 + 2(-3) + n^2 = 1 ... plug in mn = -3
m^2 - 6 + n^2 = 1
m^2 + n^2 = 1 + 6
m^2 + n^2 = 7


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a) If the roots are 1/m and 1/n, then...


(x - 1/m)(x - 1/n)
x^2 - x/m - x/n + 1/mn
x^2 - (1/m + 1/n)x + 1/mn
x^2 - (n/mn + m/mn)x + 1/mn
x^2 - ((n + m)/mn)x + 1/mn
x^2 - ((m + n)/mn)x + 1/mn
x^2 - ((-1)/mn)x + 1/mn
x^2 - ((-1)/(-3))x + 1/(-3)
x^2 - (1/3)x - 1/3



If the roots are 1/m and 1/n, then the polynomial is x^2 - (1/3)x - 1/3


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b) If the roots are m^2 and n^2, then 


(x - m^2)(x - n^2)
x^2 - xn^2 - xm^2 + m^2n^2
x^2 - (n^2 + m^2)x + (mn)^2
x^2 - (m^2 + n^2)x + (mn)^2
x^2 - (7)x + (-3)^2
x^2 - 7x + 9


If the roots are m^2 and n^2, then the polynomial is x^2 - 7x + 9