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Question
Given that the zeroes of the cubic polynomial x3 – 6x2 + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.
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Solution
Given that a, a + b, a + 2b are roots of given polynomial x3 – 6x2 + 3x + 10
Sum of the roots
⇒ a + 2b + a + a + b = `(-("coefficient of" x^2))/("coefficient of" x^3)`
⇒ 3a + 3b = `(-(-6))/1` = 6
⇒ 3(a + b) = 6
⇒ a + b = 2 .........(1)
b = 2 – a
Product of roots
⇒ (a + 2b)(a + b)a = `(-("constant"))/("coefficient of" x^3)`
⇒ (a + b + b)(a + b)a = `-10/1`
Substituting the value of a + b = 2 in it
⇒ (2 + b)(2)a = –10
⇒ (2 + b)2a = –10
⇒ (2 + 2 – a)2a = –10
⇒ (4 – a)2a = –10
⇒ 4a – a2 = –5
⇒ a2 – 4a – 5 = 0
⇒ a2 – 5a + a – 5 = 0
⇒ (a – 5)(a + 1) = 0
a – 5 = 0 or a + 1 = 0
a = 5 a = –1
a = 5, –1 in (1) a + b = 2
When a = 5,
5 + b = 2
⇒ b = –3
a = –1, –1 + b = 2
⇒ b = 3
∴ If a = 5 then b = –3
or
If a = –1 then b = 3
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