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?
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"How to prove this ?
#2|k^2# then #2|k# for some #k\inZZ#"
# acx^2-(b^2-2ac)x+ac=0#.
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#.
The equation is #acx^2-(b^2-2ac)x+ac=0#
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#
