JEE Main 18 March 2021 Shift 1 Solved Paper

Given, α,β,γ are the real roots of x3+ax2+bx+c=0, where a,b,c∈R and a,b≠0
∴α+β+γ= Sum of roots taken one at a time =−a
αβ+βγ+γα= Sum of roots taken two at a time =b
αβγ= Product of roots =−c
Also, given system of equations in u,v,w

αu+βv+γw=0

βu+γv+αw=0

γu+αv+βw=0

}
has non-trivial solution.
∴∆=0⇒|

α	β	γ
β	γ	α
γ	α	β

|=0
⇒α(βγ−α2)−β(β2−γα)+γ(αβ−γ2)=0 (expanding along R1 )
⇒‌‌αβγ−α3−β3+αβγ+αβγ−γ3=0
⇒‌‌α3+β3+γ3=3αβγ
Then (using standard result),
Either α+β+γ=0 or α=β=γ
If α+β+γ=0,‌ then ‌−a=0
⇒a=0 which is not possible according to given condition.
∴α+β+γ=0 (not possible)
Now,
α+β+γ=−a
⇒‌‌α+α+α=−a (∵α=β=γ)
⇒ a=−3α...(i)
αβ+βγ+γα=b
⇒‌‌b=3α2...(ii)
Using Eqs. (i) and (ii),
‌

a2

b

=3