Question

if the zeroes of the polynomial x3-3x+x+1 are a-b,a,a+b,find a and b

Answer 1

here is your answer buddy.

Answer 2

Answer:

Let P(x) = x3 - 6x2 + 3x + 10

And (a), (a + b) and (a + 2b) are the zeroes of P(x).

We know,  Sum of the zeroes = - (coefficient of x2) ÷ coefficient of x3

α + β + γ = - b/a

a + (a + b) + (a + 2b) = - (- 6)

3a + 3b = 6

a + b = 2

⇒ a = 2 - b      (1)

Product of all the zeroes = - (constant term) ÷ coefficient of x3

αβγ = - 10

a(a + b)(a + 2b) = - 10

(2 - b) (2) (2 + b) = - 10

(2 - b) (2 + b) = - 5

4 - b2 = - 5

⇒ b2 = 9

⇒ b = +-3

When b = 3, a = 2 - 3 = - 1 (from (1)

⇒ a = - 1

When b = - 3,a = 2 - (- 3) = 5 (from(1)

⇒ a = 5

Case 1: when a = - 1 and b = 3

The zeroes of the polynomial are:

a = - 1 a + b = - 1 + 3 = 2 a + 2b = - 1 + 2(3) = 5

⇒ - 1, 2 and 5 are the zeroes

Case 2: when a = 5,b = - 3

The zeroes of the polynomial are:

a = 5 a + b = 5 - 3 = 2 a + 2b = 5 - 2(3) = - 1

⇒ - 1, 2 and 5 are the zeroes

By both the cases the zeroes of the polynomial are -1,2,5

Sorry but I don't read the full question but it is some what like that , am i right?> <

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