Question 871953
If the two roots of the equation: X^2+X-3=0 are M and N:

sum of roots = -b/a  in equation ax^2+bx +c =0

product of roots = c/a

Therefore in x^2+x-3 =0
M+N = =-1 
MN = -3

The roots of the required equation are 1/M & 1/N

First we find the value of 1/M + 1/N and 1/MN and substitute (M+N) and MN

1/M + 1/N = (M+N)/MN = -1/-3 = 1/3

1/M * 1/N = 1/MN = 1/-3= -1/3

The required equation is 
x^2-(sum of roots)x + ( product of roots =0

{{{x^2-(1/3)x + (-1/3)=0}}}

mmultiply equation by 3

{{{3x^2-x-1=0}}}


M^ N^2 are the roots

{{{M^2+N^2= (M+N)^2- 2MN =  (1/3)^2 - (-3)= 1/9 +3 = 13/9}}}

{{{M^2N^2 = (MN)^2 = (-3)^2 = 9}}}

frame the equation as done above

{{{x^2- (13/9)x+9 =0}}}

multiply by 9

{{{9x^2-13x+81 =0}}}