If the equations x^2 -ax+b=0 and x^2-ex+f =0 have a root in common and the second equation has equal roots show that ae =2(b+f)

question on quadratic equations

If the equations (x^2) -ax+b=0 and (x^2)-ex+f =0 have a root in common and the second equation has equal roots show that ae =2(b+f)

Let alpha be the common root

Therefore  (alpha) ,(alpha) are the equal roots of the second equation

[(x^2)-ex+f =0]

using sum of roots formula

(alpha) + (alpha) ={ -(-e)} / {1}

2(alpha) = e

(alpha)  ={e}/{2} -----------------(1)

using product of roots formula

 (alpha) * (alpha) = {f}/{1}

{alpha^2} = f -----------------------(2)

Therefore let 

(alpha) ,(beta) be the roots of the first equation [(x^2) -ax+b=0]

again using product of roots formula

 (alpha) * (beta) ={b}/{1}

 (alpha) * (beta) = b

beta = b/{alpha} -------------------(3)

using sum of roots formula

 (alpha) + (beta) ={-(-a)}/{1}

 (alpha) + (beta) = a

using equation(3)

(alpha) +{ {b}/{alpha}} = a

{{alpha}^2 + b } / {alpha} = a

{{alpha}^2 + b } = a *{alpha}

use eqn(2) in LHS and  eqn(1) in RHS

{f +b}=a*{{e}/{2}}

therfore

2*(f+b) = ae

hence

ae =2(b+f)