The roots of equation 2x^2-5x+3=0 are alpha and beta. Find an equation with integral coefficients where roots are alpha/beta

2 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

 

 

 

 

We’ve answered 318,909 questions. We can answer yours, too.

Ask a question