Consider equation ax^2+bx+c=0 whose roots are x and y.Then find quadratic equation whose roots are x/y and y/x?

2 Answers
Jul 16, 2018

acx^2-(b^2-2ac)x+ac=0acx2(b22ac)x+ac=0.

Explanation:

Given that, x and yxandy are the roots of the quadr. eqn. :

ax^2+bx+c=0ax2+bx+c=0.

:. x+y=-b/a, and, xy=c/a....................(ast).

Let, X=x/y, and, Y=y/x.

Then, X+Y=x/y+y/x=(x^2+y^2)/(xy),

=1/(xy){(x+y)^2-2xy},

=(x+y)^2/(xy)-2,

=((-b/a)^2)/(c/a)-2.

:. X+Y=(b^2-2ac)/(ac).................(ast^1).

Also, X*Y=1..................................(ast^2).

Hence, the desired quadr. eqn. is given by,

x^2-(X+Y)x+XY=0,

i.e., x^2-((b^2-2ac)/(ac))x+1=0,

or, acx^2-(b^2-2ac)x+ac=0.

Jul 16, 2018

The equation is acx^2-(b^2-2ac)x+ac=0

Explanation:

Let the roots of the equation

ax^2+bx+c=0

be

alpha and beta

Then,

alpha+beta=-b/a

and

alphabeta=c/a

The roots of the new equation are

alpha/beta and beta/alpha

Then the sum of the roots are

alpha/beta+beta/alpha=(alpha^2+beta^2)/(alphabeta)

=((alpha+beta)^2- 2alphabeta)/(alphabeta)

=((-b/a)^2-2*c/a)/(c/a)

=(b^2/a^2-2c/a)/(c/a)

=(b^2-2ac)/(ac)

The product of the roots is

alpha/beta*beta/alpha=1

The quadratic equation is

x^2-((b^2-2ac)/(ac))x+1=0

acx^2-(b^2-2ac)x+ac=0