Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)
And given, the zeros are in A.P.
So, let’s consider the roots as
α = a – d, β = a and γ = a + d
Where, a is the first term and d is the common difference.
From given f(x), a = 2, b = -15, c = 37 and d = 30
⇒ Sum of roots = α + β + γ
= (a – d) + a + (a + d)
= 3a
= (\(\frac{-b}{a}\))
= -(\(\frac{-15}{2}\))
= \(\frac{15}{2}\)
So, calculating for a, we get 3a = \(\frac{15}{2}\)
⇒ a = \(\frac{5}{2}\)
⇒ Product of roots = (a – d) x (a) x (a + d)
= a(a2 –d2)
= \(\frac{-d}{a}\)
= \(\frac{-(30)}{2}\)
= 15
⇒ a(a2 –d2) = 15
Substituting the value of a, we get
⇒ (\(\frac{5}{2}\))[(\(\frac{5}{2}\))2 –d2] = 15
⇒ 5[(\(\frac{25}{4}\)) –d2] = 30
⇒ (\(\frac{25}{4}\)) – d2 = 6
⇒ 25 – 4d2 = 24
⇒ 1 = 4d2
∴ d = \(\frac{1}{2}\) or -\(\frac{1}{2}\)
Taking d = \(\frac{1}{2}\) and a = \(\frac{5}{2}\)
We get,
the zeros as 2, \(\frac{5}{2}\) and 3
Taking d = -\(\frac{1}{2}\) and a = \(\frac{5}{2}\)
We get, the zeros as 3, \(\frac{5}{2}\) and 2.