Question

if alpha and beta are the roots of the quadratic equation X square - 5 x + 6 is equal to zero find find Alpha square + beta square

Answer 1

given equation x2−2x+3=0

⇒x=2±√22−4⋅1⋅32=1±√2i

Let α=1+√2iandβ=1−√2i

Now let

γ=α3−3α2+5α−2

⇒γ=α3−3α2+3α−1+2α−1

⇒γ=(α−1)3+α−1+α

⇒γ=(√2i)3+√2i+1+√2i

⇒γ=−2√2i+√2i+1+√2i=1

And let

δ=β3−β2+β+5

⇒δ=β2(β−1)+β+5

⇒δ=(1−√2i)2(−√2i)+1−√2i+5

⇒δ=(−1−2√2i)(−√2i)+1−√2i+5

⇒δ=√2i−4+1−√2i+5=2

So the quadratic equation having roots γandδ is

x2−(γ+δ)x+γδ=0

⇒x2−(1+2)x+1⋅2=0

⇒x2−3x+2=0

Q.2 If one root of the equation ax2+bx+c=0 be the square of the other,
Prove that b3+a2c+ac2=3abc

Let one root be α then other root will be α2

So α2+α=−ba
and

α3=ca

⇒α3−1=ca−1

⇒(α−1)(α2+α+1)=ca−1=c−aa

⇒(α−1)(−ba+1)=c−aa

⇒(α−1)(a−ba)=c−aa

⇒(α−1)=c−aa−b

⇒α=c−aa−b+1=c−ba−b

Now α being one of the roots of the quadratic equation ax2+bx+c=0 we can write

aα2+bα+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

⇒ac2−2abc+ab2+abc−ab2−b2c+b3+ca2−2abc+b2c=0

⇒b3+a2c+ac2=3abc

prove