#Q.1# If #alpha,beta# are the roots of the equation #x^2-2x+3=0# obtain the equation whose roots are #alpha^3-3 alpha^2+5 alpha -2# and #beta^3-beta^2+beta+5#?
Answer
given equation #x^2-2x+3=0#
#=>x=(2pmsqrt(2^2-4*1*3))/2=1pmsqrt2i#
Let #alpha=1+sqrt2i and beta=1-sqrt2i#
Now let
#gamma=alpha^3-3 alpha^2+5 alpha -2#
#=>gamma=alpha^3-3 alpha^2+3 alpha -1+2alpha-1#
#=>gamma=(alpha-1)^3+alpha-1+alpha#
#=>gamma=(sqrt2i)^3+sqrt2i+1+sqrt2i#
#=>gamma=-2sqrt2i+sqrt2i+1+sqrt2i=1#
And let
#delta=beta^3-beta^2+beta+5#
#=>delta=beta^2(beta-1)+beta+5#
#=>delta=(1-sqrt2i)^2(-sqrt2i)+1-sqrt2i+5#
#=>delta=(-1-2sqrt2i)(-sqrt2i)+1-sqrt2i+5#
#=>delta=sqrt2i-4+1-sqrt2i+5=2#
So the quadratic equation having roots #gamma and delta# is
#x^2-(gamma+delta)x+gammadelta=0#
#=>x^2-(1+2)x+1*2=0#
#=>x^2-3x+2=0#
#Q.2# If one root of the equation #ax^2+bx+c=0# be the square of the other,
Prove that #b^3+a^2c+ac^2=3abc#
Let one root be #alpha# then other root will be #alpha^2#
So #alpha^2+alpha=-b/a#
and
#alpha^3=c/a#
#=>alpha^3-1=c/a-1#
#=>(alpha-1)(alpha^2+alpha+1)=c/a-1=(c-a)/a#
#=>(alpha-1)(-b/a+1)=(c-a)/a#
#=>(alpha-1)((a-b)/a)=(c-a)/a#
#=>(alpha-1)=(c-a)/(a-b)#
#=>alpha=(c-a)/(a-b)+1=(c-b)/(a-b)#
Now #alpha # being one of the roots of the quadratic equation #ax^2+bx+c=0# we can write
#aalpha^2+balpha+c=0#
#=>a((c-b)/(a-b))^2+b((c-b)/(a-b))+c=0#
#=>a(c-b)^2+b(c-b)(a-b)+c(a-b)^2=0#
#=>ac^2-2abc+ab^2+abc-ab^2-b^2c+b^3+ca^2-2abc+b^2c=0#
#=>b^3+a^2c+ac^2=3abc#
proved
Alternative
#aalpha^2+balpha+c=0#
#=>aalpha+b+c/alpha=0#
#=>a(c/a)^(1/3)+b+c/((c/a)^(1/3))=0#
#=>c^(1/3)a^(2/3)+c^(2/3)a^(1/3)=-b#
#=>(c^(1/3)a^(2/3)+c^(2/3)a^(1/3))^3=(-b)^3#
#=>(c^(1/3)a^(2/3))^3+(c^(2/3)a^(1/3))^3+3c^(1/3)a^(2/3)xxc^(2/3)a^(1/3)(c^(1/3)a^(2/3)+c^(2/3)a^(1/3))=(-b)^3#
#=>ca^2+c^2a+3ca(-b)=(-b)^3#
#=>b^3+ca^2+c^2a=3abc#
