if the roots of the equation l(m-n)x^2+m(n-l)x+n(l-m)=0 are equal.Show that m=2ln/l+n.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If the roots of the equation are equal, since the equation is a quadratic, its discriminant has to be zero.

Discriminant = delta = b^2 - 4ac

a,b,c, are the coefficients of the equation:

delta = [m(n-l)]^2 -4l(m-n)*n(l-m)=

We'll expand the square:

m^2*(n^2 - 2nl + l^2) - 4ln(ml - m^2 - nl + mn) = 0

m^2*n^2 - 2m^2*nl + m^2*l^2 - 4mnl^2 + 4lnm^2 + 4l^2*n^2 - 4lmn^2 = 0

m^2*n^2 - 2m^2*nl + m^2*l^2 = 4mnl^2 - 4lnm^2 - 4l^2*n^2 + 4lmn^2 (1)

We'll re-write the constraint:

m=2ln/l+n

m(l+n) = 2ln

We'll raise to square both sides:

m^2*n^2 +2m^2*nl + m^2*l^2 = 4l^2*n^2

m^2*n^2 + m^2*l^2 = 4l^2*n^2 - 2m^2*nl (2)

We'll substitute (2) in (1):

4l^2*n^2 - 2m^2*nl  - 2m^2*nl = 4mnl^2 - 4lnm^2 - 4l^2*n^2 + 4lmn^2

4l^2*n^2 - 4m^2*nl = 4mnl^2 - 4lnm^2 - 4l^2*n^2 + 4lmn^2

We'll divide by 4:

l^2*n^2 = mnl^2 - lnm^2 + lmn^2

2l^2*n^2 = mnl^2 + lmn^2

We'll factorize by m*n*l to the right side:

2l^2*n^2 = m*n*l(l + n)

We'll divide by l*n:

2l*n = m(l+n)

m = (2ln)/(l+n) q.e.d

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