Question

Find the zeroes of the following polynomials and verify the relationship between zeroes and coefficients.
a)2x2- 6x
b) x2 - 2x + 7/16

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Answer 1

Step-by-step explanation:

Answer:

First term = 1

Common difference = 6

Given statements about the terms of an AP:

9th term = 7 × 2nd Term

9th term = 7 × 2nd Term12th term = 5 × 3rd term + 2

We have to find the following:

First term, a

Common difference, d

The standard form of an AP is:

a , a + d, a + 2d , a + 3d, ... , a + (n - 1)d

Where,

a = first term of AP

d = common difference of AP

So, According to the formula,

aₙ = a + (n - 1)d

We have 9th term and 2nd term as a + 8d and a + d respectively. So According to the statement given,

⇒ 9th term = 7 × 2nd term

⇒ a + 8d = 7 (a + d)

⇒ a + 8d = 7a + 7d

⇒ 7a - a + 7d - 8d = 0

⇒ 6a - d = 0 ...(i)

Similarly, According to the second statement, we have

⇒ 12th term = ( 5 × 3rd term ) + 2

⇒ a + 11d = { 5(a + 2d) } + 2

⇒ a + 11d = 5a + 10d + 2

⇒ 5a - a + 10d - 11d = -2

⇒ 4a - d = -2 ...(ii)

Subtract eq.(ii) from eq.(i), we get

⇒ 6a - d - (4a - d) = 0 - (-2)

⇒ 6a - d - 4a + d = 2

⇒ 6a - 4a = 2

⇒ 2a = 2

⇒ a = 1

We found the first term to be 1, Hence substitute the value of a in eq.(i), we get

⇒ 6a - d = 0

⇒ 6(1) - d = 0

⇒ 6 - d = 0

⇒ d = 6