AP EAPCET 18 May 2024 Shift 1 Solved Paper

We are given the polynomial equation x4−x3−8x2+2x+12=0 with roots α,β,γ, and δ. It is known that the sum of two of the roots, α and β, is zero, i.e., α+β=0.
From Vieta's formulas, the sum of all roots is:
S1=α+β+γ+δ=1
Given α+β=0, we find:
γ+δ=1
Similarly, from Vieta's formulas, the sum of the products of the roots taken two at a time is:
S2=αβ+αγ+αδ+βγ+βδ+γδ=−8
Using α+β=0, simplify to:
αβ+γδ=−8
From the sum of the products of the roots taken three at a time, we have:
S3=αβγ+αβδ+αγδ+βγδ=−2
Substitute (α+β)γδ+αβ(γ+δ)=−2 and rearrange:
αβ=−2
From Vieta's formulas, the product of the roots is given by:
S4=αβγδ=12
Use αβ=−2 to find:
γδ=−6‌‌...(v)
Next, solve the system of equations obtained from (ii) and (v):
‌γ+δ=1
‌γδ=−6
These are the equations for a quadratic:
t2−(γ+δ)t+γδ=0⟹t2−t−6=0
Solving, we find the roots to be:
t=‌

1±√1+24

2

=‌

1±5

2

Thus, t=3 or t=−2. Since γ>δ, we assign γ=3 and δ=−2.
Finally, compute:
3γ+2δ=3×3+2×(−2)=9−4=5