If #alpha != beta, alpha^2= 5alpha - 3, beta^2= 5beta-3# then the quation whose roots are #alpha/beta & beta/alpha# is ?

We are effectively told that #alpha# and #beta# are the two roots of the quadratic equation:

#x^2=5x-3#

Rearranged slightly, that is:

#x^2-5x+3 = 0#

So we have:

#0 = x^2-5x+3#

#color(white)(0) = (x-alpha)(x-beta)#

#color(white)(0) = x^2-(alpha+beta)x+alphabeta#

Equating coefficients, that tells us that:

#{ (alpha+beta = 5), (alphabeta = 3) :}#

In addition, note that:

#alpha^2+beta^2 = (5alpha-3)+(5beta-3)#

#color(white)(alpha^2+beta^2) = 5(alpha+beta)-6#

#color(white)(alpha^2+beta^2) = 5(color(blue)(5))-6#

#color(white)(alpha^2+beta^2) = 19#

A quadratic equation with roots #alpha/beta# and #beta/alpha# is:

#(x-alpha/beta)(x-beta/alpha) = 0#

To clear the denominators and give us integer coefficients, we can multiply both sides by #alphabeta# to find:

#0 = alphabeta(x-alpha/beta)(x-beta/alpha)#

#color(white)(0) = (betax-alpha)(alphax-beta)#

#color(white)(0) = alphabetax^2-(alpha^2+beta^2)x+alphabeta#

#color(white)(0) = 3x^2-19x+3#

So we can write a suitable quadratic equation as:

#3x^2-19x+3 = 0#

#alpha^2=5alpha-3.........(1), &, beta^2=5beta-3.................(2)#

#:. (1)-(2) rArr alpha^2-beta^2=5alpha-3-(5beta-3),#

# rArr (alpha-beta)(alpha+beta)=5(alpha-beta),#

Dividing by #(alpha-beta)!=0,...[ because, alpha!=beta],# we get,

#alpha+beta=5...................(star^1).#

Also, #(1)+(2) rArr alpha^2+beta^2=5(alpha+beta)-6,#

#=25-6,............[because, (star^1)]#, i.e.,

#alpha^2+beta^2=(alpha+beta)^2-2alpha*beta=19,#

#:. (star^1) rArr 5^2-2alpha*beta=19,#

#:. alpha*beta=3.........................(star^2).#

Now, #alpha/beta+beta/alpha=(alpha^2+beta^2)/(alpha*beta),#

#=19/3,# and,

#alpha/beta*beta/alpha=1.#

Hence, the Reqd. Quadr. Eqn., having the roots, #alpha/beta, beta/alpha,# is given by,

#x^2-(alpha/beta+beta/alpha)x+(alpha/beta*beta/alpha)=0,# i.e.,

# x^2-(19/3)x+1=0, or, 3x^2-19x+3=0,# as Respected

George C. Sir has already derived.

Enjoy Maths.!