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
Ratnaker Mehta
Jul 16, 2018

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

Explanation:

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

#ax^2+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#.

Narad T.
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#

Related questions
Impact of this question
2813 views around the world
You can reuse this answer
Creative Commons License