Q.1Q.1 If alpha,betaα,β are the roots of the equation x^2-2x+3=0x2−2x+3=0 obtain the equation whose roots are alpha^3-3 alpha^2+5 alpha -2α3−3α2+5α−2 and beta^3-beta^2+beta+5β3−β2+β+5?
Answer
given equation x^2-2x+3=0x2−2x+3=0
=>x=(2pmsqrt(2^2-4*1*3))/2=1pmsqrt2i⇒x=2±√22−4⋅1⋅32=1±√2i
Let alpha=1+sqrt2i and beta=1-sqrt2iα=1+√2iandβ=1−√2i
Now let
gamma=alpha^3-3 alpha^2+5 alpha -2γ=α3−3α2+5α−2
=>gamma=alpha^3-3 alpha^2+3 alpha -1+2alpha-1⇒γ=α3−3α2+3α−1+2α−1
=>gamma=(alpha-1)^3+alpha-1+alpha⇒γ=(α−1)3+α−1+α
=>gamma=(sqrt2i)^3+sqrt2i+1+sqrt2i⇒γ=(√2i)3+√2i+1+√2i
=>gamma=-2sqrt2i+sqrt2i+1+sqrt2i=1⇒γ=−2√2i+√2i+1+√2i=1
And let
delta=beta^3-beta^2+beta+5δ=β3−β2+β+5
=>delta=beta^2(beta-1)+beta+5⇒δ=β2(β−1)+β+5
=>delta=(1-sqrt2i)^2(-sqrt2i)+1-sqrt2i+5⇒δ=(1−√2i)2(−√2i)+1−√2i+5
=>delta=(-1-2sqrt2i)(-sqrt2i)+1-sqrt2i+5⇒δ=(−1−2√2i)(−√2i)+1−√2i+5
=>delta=sqrt2i-4+1-sqrt2i+5=2⇒δ=√2i−4+1−√2i+5=2
So the quadratic equation having roots gamma and deltaγandδ is
x^2-(gamma+delta)x+gammadelta=0x2−(γ+δ)x+γδ=0
=>x^2-(1+2)x+1*2=0⇒x2−(1+2)x+1⋅2=0
=>x^2-3x+2=0⇒x2−3x+2=0
Q.2Q.2 If one root of the equation ax^2+bx+c=0ax2+bx+c=0 be the square of the other,
Prove that b^3+a^2c+ac^2=3abcb3+a2c+ac2=3abc
Let one root be alphaα then other root will be alpha^2α2
So alpha^2+alpha=-b/aα2+α=−ba
and
alpha^3=c/aα3=ca
=>alpha^3-1=c/a-1⇒α3−1=ca−1
=>(alpha-1)(alpha^2+alpha+1)=c/a-1=(c-a)/a⇒(α−1)(α2+α+1)=ca−1=c−aa
=>(alpha-1)(-b/a+1)=(c-a)/a⇒(α−1)(−ba+1)=c−aa
=>(alpha-1)((a-b)/a)=(c-a)/a⇒(α−1)(a−ba)=c−aa
=>(alpha-1)=(c-a)/(a-b)⇒(α−1)=c−aa−b
=>alpha=(c-a)/(a-b)+1=(c-b)/(a-b)⇒α=c−aa−b+1=c−ba−b
Now alpha α being one of the roots of the quadratic equation ax^2+bx+c=0ax2+bx+c=0 we can write
aalpha^2+balpha+c=0aα2+bα+c=0
=>a((c-b)/(a-b))^2+b((c-b)/(a-b))+c=0⇒a(c−ba−b)2+b(c−ba−b)+c=0
=>a(c-b)^2+b(c-b)(a-b)+c(a-b)^2=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⇒ac2−2abc+ab2+abc−ab2−b2c+b3+ca2−2abc+b2c=0
=>b^3+a^2c+ac^2=3abc⇒b3+a2c+ac2=3abc
proved
Alternative
aalpha^2+balpha+c=0aα2+bα+c=0
=>aalpha+b+c/alpha=0⇒aα+b+cα=0
=>a(c/a)^(1/3)+b+c/((c/a)^(1/3))=0⇒a(ca)13+b+c(ca)13=0
=>c^(1/3)a^(2/3)+c^(2/3)a^(1/3)=-b⇒c13a23+c23a13=−b
=>(c^(1/3)a^(2/3)+c^(2/3)a^(1/3))^3=(-b)^3⇒(c13a23+c23a13)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⇒(c13a23)3+(c23a13)3+3c13a23×c23a13(c13a23+c23a13)=(−b)3
=>ca^2+c^2a+3ca(-b)=(-b)^3⇒ca2+c2a+3ca(−b)=(−b)3
=>b^3+ca^2+c^2a=3abc⇒b3+ca2+c2a=3abc
