The roots of equation 2x^2-5x+3=0 are alpha and beta. Find an equation with integral coefficients where roots are alpha/beta
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2x^2 -5x+3 =0
Let the roots be x1 and x2
We know that:
x1+x2= -b/a= 5/2
x1x2= c/a= 3/2
For the equation whose roots are x1/x2 and x2/x1 :
x1/x2 + x2/x1= (x1^2 + x2^2)/x1x2 = [(x1+x2)^2 -2x1x2]/ x1x2
= (5/2)^2 -2(3/2) / (3/2)
= 25/4 -3 / (3/2)= 13/4 / (3/2)=
= 13/6 = -b/a
==> b= -13 c= 6
x1/x2 * x2/x1= 1 = c/a
==> c=a =6
Then the equation is:
6x^2 -13x + 6= 0
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2x^2-5x+3 = 0. Alpha =a and beta=b are the roots of the equation.
to find the equation whose roots are a/b .
Solution:
By the relation of roots and coefficients,
Sum of the roots = a+b = -(-5)/2 =5/2 and Product of the roots = ab = 3/2.
For the new equation, a/b and b/a are roots.
So sum of roots = a/b+b/a = (a^2+b^2)/(ab) = {(a+b)^2 -2ab}/(ab) = ((5/2)^2-2(3/2))/(3/2) = (25/2-6)/3 = (25-12)/6 = 13/6
Product of the roots = (a/b) and (b/a) = ab/ab =1.
Therefore the equation whose roots are alpha/beta and beta / alpha is
x^2 -(13/6)x+1 = 0. To make the coefficients integral we multiply the equation by 6.
6x^2 - 13x + 6 = 0.
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