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Mathematics Question on Quadratic Equations

If α,β,γ\alpha , \beta , \gamma are the roots of x3−6x2+11x−6=0x^3 - 6x^2 + 11x - 6 = 0 , then the equation having the roots α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 and γ2+α2\gamma^2 + \alpha^2 is

A

x3−28x2+245x−650=0x^3 - 28x^2 + 245x - 650 = 0

B

x3−28x2+245x+650=0x^3 - 28x^2 + 245 x + 650 = 0

C

x3+28x2−245x−650=0x^3 + 28x^2 - 245x - 650 =0

D

x3+28x2+245x−650=0x^3 + 28x^2 + 245x - 650 = 0

Answer

x3−28x2+245x−650=0x^3 - 28x^2 + 245x - 650 = 0

Explanation

Solution

x3−6x2+11x−6=0…x^{3}-6 x^{2}+11 x-6=0 \dots(i)
⇒(x−1)(x−2)(x−3)=0\Rightarrow(x-1)(x-2)(x-3)=0
⇒x=1,2,3\Rightarrow x=1,2,3
∵α,β,γ\because \alpha, \beta, \gamma are the roots of the E(i), so
α=1,β=2,γ=3\alpha=1, \beta=2, \gamma=3
Therefore, α2+β2=(1)2+(2)2=5=α′\alpha^{2}+\beta^{2}=(1)^{2}+(2)^{2}=5=\alpha' (say)
β2+γ2=(2)2+(3)2=13=β′(\beta^{2}+\gamma^{2}=(2)^{2}+(3)^{2}=13=\beta'( say ))
and γ2+α2=(3)2+1=10=γ′(\gamma^{2}+\alpha^{2}=(3)^{2}+1=10=\gamma'( say ))
Equation of the having the roots α′,β′\alpha', \beta' and γ′\gamma' ,
x3−(α′+β′+γ′)x2+(α′β′+β′γ′+γ′α′)xx^{3}-\left(\alpha'+\beta'+\gamma'\right) x^{2}+\left(\alpha' \beta'+\beta'\gamma'+\gamma' \alpha'\right) x
−α′β′γ′=0-\alpha' \beta' \gamma'=0
⇒x3−(5+13+10)x2+(5×13+13×10+10×5)x\Rightarrow x^{3}-(5+13+10) x^{2}+(5 \times 13+13 \times 10+10 \times 5) x
−5×13×10=0-5 \times 13 \times 10=0
⇒x3−28x2+245x−650=0\Rightarrow x^{3}-28 x^{2}+245 x-650=0