If the roots of l(m-n)x²+m(n-l)x+n(l-m)=0 are equal then prove that m=2ln/(l+n) ?

1 Answer
Steve M
Sep 10, 2017

If we consider the general case quadratic:

# ax^2 + bnx+ c = 0 #

Then the quadratic has equal roots is the discriminant is zero, ie:

# Delta = b^2-4ac = 0#

We have:

# l(m-n)x²+m(n-l)x+n(l-m)=0 #

So the discriminant is given by:

# Delta = {m(n-l)}^2-4{l(m-n)}{n(l-m)} #

We require #Delta=0#, and so we have:

# {m(n-l)}^2-4{l(m-n)}{n(l-m)} = 0#
# :. m^2(n-l)^2-4l n(m-n)(l-m) = 0#

# :. m^2(n^2-2nl+l^2) - 4l n(ml-m^2-nl+mn) = 0#

# :. m^2n^2 - 2lm^2n + l^2m^2 - 4l^2mn+4lm^2n+4l^2n^2-4lmn^2 = 0#

# :. m^2n^2 + l^2m^2 +2lm^2n- 4l^2mn-4lmn^2+4l^2n^2 = 0#

# :. (n^2 + l^2 +2l n)m^2 + (-4l^2n-4l n^2)m +4l^2n^2 = 0#

# :. (l+n)^2m^2 - 4l n(l+ n)m +4l^2n^2 = 0#

This is a quadratic in #m#, so we can solve for #m# using the quadratic formula:
# m = (4l n(l+ n) +- sqrt{ (4l n(l+ n))^2 - 4((l+n)^2) (4l^2n^2) } )/ (2(l+n)^2) #

# \ \ = (4l n(l+ n) +- sqrt{ (4l n(l+ n))^2 - (4l n(l+n))^2 } )/ (2(l+n)^2) #

# \ \ = (4l n(l+ n) +- 0 )/ (2(l+n)^2) #

# \ \ = (4l n(l+ n) )/ (2(l+n)^2) #

# \ \ = (2l n )/(l+n) # QED

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