AP EAPCET 21 May 2025 Shift 1 Paper

We have, f(x)=4x4+4x3−23x2−12x+36=0
Using concept a polynomial f(x) has a multiple root r, then f(r)=0 and
‌f′(r)=0
‌f′(x)=16x3+12x2−46x−12
Now, using hit and trial
By taking x=−2
f(−2)=4(−2)4+4(−2)3−23(−2)2
−12(−2)+36
=64−32−92+24+36=0
⇒f′(−2)=16(−2)3+12(−2)2
−46(−2)−12
=−128+48+92−12=0
∵x=−2 is a multiple root.
By dividing the equation by (x+2)2, we get another multiple root x=‌

3

2

Given, α>β
‌∴‌‌α=‌

3

2

‌ and ‌β=−2
‌∴2α−β=3+2=5