If the ratio of the roots of #lx^2+nx+n = 0# is #p/q# then how do you prove that #(p/q)^(1/2)+(q/p)^(1/2)-(n/l)^(1/2) = 0# ?

1 Answer
George C.
Sep 10, 2017

See explanation...

Explanation:

Suppose the roots are #alpha# and #beta#.

Then:

#lx^2+nx+n = l(x-alpha)(x-beta)#

#color(white)(lx^2+nx+n) = lx^2-l(alpha+beta)x+lalphabeta#

So equating coefficients, we find:

#{(alpha+beta = -n/l), (alphabeta = n/l) :}#

Then:

#((alpha/beta)^(1/2)+(beta/alpha)^(1/2))^2#

#=alpha/beta+2+beta/alpha#

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

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

#=(-n/l)^2 / (n/l)#

#=n/l#

So taking the principal square root of both sides, we have:

#(alpha/beta)^(1/2)+(beta/alpha)^(1/2) = (n/l)^(1/2)#

So if #alpha/beta = p/q# then:

#(p/q)^(1/2)+(q/p)^(1/2)-(n/l)^(1/2) = 0#

#color(white)()#
Notes

I rushed through "taking the principal square root of both sides" without explaining it properly.

There are a couple of cases to consider:

If #n/l > 0# then either #alpha, beta# are both negative or they are a complex conjugate pair. If they are both negative then #alpha/beta > 0#, #beta/alpha > 0#. Hence #(alpha/beta)^(1/2)+(beta/alpha)^(1/2) > 0#. So we need the positive, principal square root of #n/l#. If they are both complex, then both #(alpha/beta)^(1/2)# and #(beta/alpha)^(1/2)# have positive real part and are complex conjugates of one another. So their sum is a positive real numbers. So again we need the positive, principal square root of #n/l#.

If #n/l < 0# then #alpha, beta# are both real, with opposite signs. As a result #alpha/beta < 0# and #beta/alpha < 0#. So their principal square roots are pure imaginary with a positive imaginary coefficient. So #(alpha/beta)^(1/2)+(beta/alpha)^(1/2)# is pure imaginary with positive imaginary coefficient and we need the principal square root of #n/l# in order to match it.

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