Math, asked by educationmaster37, 10 months ago

please answer the red mark questions

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Answered by Anonymous
20

Very Short Answer Type Questions :

Question 1: The sum and product of zeroes of p(x) = 63x² - 7x - 9 are the S and P respectively. Find S and P.

Solution:

Given : p(x) = 63x² - 7x - 9

On comparing this with ax² + bx + c, we get

a = 63, b = - 7, c = - 9

Let α and β be the zeroes of the required polynomial.

Now,

Sum of zeroes, S = α + β = - b/a

→ - (- 7)/63

→ 7/63

1/9

Product of zeroes, P = αβ = c/a

→ - 9/63

- 1/7

Hence, the sum(S) is 1/9 and product(P) is - 1/7.

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Question 2: Write a quadratic polynomial having zeroes 1 and - 2.

Solution:

Let α and β be the zeroes of the required polynomial.

♦ α = 1, β = - 2 (given)

• Sum of zeroes = α + β

→ (1) + (- 2)

→ 1 - 2

- 1

• Product of zeroes = αβ

→ (1)(- 2)

- 2

The required polynomial is :

p(x) = k [ x² - (α + β)x + αβ ]

  • Putting known values.

→ p(x) = k [ x² - (- 1)x + (- 2) ]

→ p(x) = k [ x² + x - 2 ]

  • Putting k = 1.

→ p(x) = x² + x - 2

Hence, the required polynomial is x² + x - 2.

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Short Answer Type Questions :

Question 1: Form a quadratic polynomial p(x) with sum and product of zeroes as 2 and - 3/5 respectively.

Solution:

Let α and β be the zeroes of the required polynomial.

It is given that :

• Sum of zeroes = α + β = 2

• Product of zeroes = αβ = - 3/5

The required polynomial is :

p(x) = k [ x² - (α + β)x + αβ ]

  • Putting known value in it.

→ p(x) = k [ x² - (2)x + (- 3/5) ]

→ p(x) = k [ x² - 2x - 3/5 ]

→ p(x) = k [ (5x² - 10x - 3)/5 ]

→ p(x) = k/5 [ 5x² - 10x - 3 ]

  • Putting k = 5.

→ p(x) = 5x² - 10x - 3

Hence, the required polynomial is 5x² - 10x - 3.

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Question 2: α and β are the zeroes of the quadratic polynomial p(x) = x² - (k + 6)x + 2(2k - 1). Find the value of k, if α + β = (1/2)αβ

Solution:

Given : p(x) = x² - (k + 6)x + 2(2k - 1)

On comparing this with ax² + bx + c, we get

a = 1, b = - (k + 6), c = 2(2k - 1)

It is given that α and β be the zeroes of the required polynomial.

Now,

Sum of zeroes = α + β = - b/a

→ - [ - (k + 6) ] / 1

→ k + 6

Product of zeroes = αβ = c/a

→ 2(2k - 1)/1

→ 2(2k - 1)

Now, it is given that,

→ α + β = (1/2)αβ

  • Putting known values.

→ k + 6 = (1/2)[ 2(2k - 1) ]

→ k + 6 = 2k - 1

→ 2k - k = 6 + 1

→ k = 7

Hence, the value of k is 7.

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Question 3: Form a quadratic polynomial whose one zero is 8 and the product of the zeroes is - 56.

Solution:

Let α and β be the zeroes of the required polynomial.

One zero is given. Let it be α.

So, α = 8

Also, Product of zeroes = - 56

→ αβ = - 56

  • Putting the value of α in it.

→ (8)(β) = - 56

→ β = - 56/8

β = - 7

Now,

• Sum of zeroes = α + β

→ (8) + (- 7)

→ 8 - 7

1

The required polynomial is :

p(x) = k [ - (α + β)x + αβ ]

  • Putting known values.

→ p(x) = k [ x² - (1)x + (- 56) ]

→ p(x) = k [ x² - x - 56 ]

  • Putting k = 1.

→ p(x) = x² - x - 56

Hence, the required polynomial is x² - x - 56.

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