AP EAPCET 23 May 2025 Shift 2 Paper

We are given the equation x3+6x2+5x−42=0.
Let's check if x=2 is a root by plugging it in:
(2)3+6×(2)2+5×2−42=8+24+10−42=42−42=0
Since the equation equals zero, x=2 is a real root. So, α=2.
Now, we substitute α=2 into the matrix:
[

α−1	α+1	α+2
α−2	α+3	α−3
α+4	α−4	α+5

]=[

1	3	4
0	5	−1
6	−2	7

]
Next, let's find the determinant of this matrix:
‌=|

1	3	4
0	5	−1
6	−2	7

|
‌=1×(5×7−(−1)×(−2))−3×(0×7−(−1)×6)+4×(0×(−2)−5×6)
‌=1×(35−2)−3×(0+6)+4×(0−30)
‌=33−18−120
‌=−105