α, β, γ are zeroes of cubic polynomial x³– 12x² + 44x + c. If α, β, γ are in AP, find the value of c.
Answers
Step-by-step explanation:
α,β and γ are zeros of cubic polynomial and are in AP.
So, Let β=a ; α=a−d & γ=a+d
Polynomial=x
3
−12x
2
+44x+c
Sum of roots=
1
−(−12)
=12
So,a−d+a+a+d=12
3a=12
a=4
Sum of products of two consecutive roots=44.
a(a−d)+a(a+d)+(a−d)(a+d)=44
a
2
−ad+a
2
+ad+a
2
−d
2
=44
3a
2
−d
2
=44
3(4)
2
−d
2
=44
d
2
=48−44=4
d=±2
So, α=a−d=
4+2
4−2
=
6
2
β=4
β=a+d=
4−2
4+2
=
2
6
So,Product (−c)=2×4×6
=−48.
Answer:
The given polynomial is
p (x) = x³ + 12x² + 44x + c
Since, α, β and γ are the roots of p (x),
α + β + γ = - 12 ...(i)
αβ + βγ + γα = 44 ...(ii)
αβγ = - c ...(iii)
Given that, α, β and γ are in AP,
β - α = γ - α
or, 2α = β + γ ...(iv)
Now, (i) × 2 ⇒
2α + 2 (β + γ) = - 24
⇒ (β + γ) + 2 (β + γ) = - 24
⇒ 3 (β + γ) = - 24
⇒ β + γ = - 8 ...(v)
Now, putting β + γ = - 8 in (i), we get
α - 8 = - 12
⇒ α = - 12 + 8
⇒ α = - 4
Since α (= - 4) is a zero of p (x),
p (α) = 0
⇒ p (- 4) = 0
⇒ (- 4)³ + 12 (- 4)² + 44 (- 4) + c = 0
⇒ - 64 + 192 - 176 + c = 0
⇒ c = 48
Therefore, the value of c is 48.