Question

If the sum of the zeroes of the polynomials f(x)=2x3-3kx2+4x-5=6 then the value of x=

Answer 1

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This question is from Physics bolo of class 11, CBSE board.

(It is a numerical) ...

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This question is from Physics bolo of class 11, CBSE board.

(It is a numerical) ...

https://brainly.in/question/19308007?utm_source=android&utm_medium=share&utm_campaign=question$\huge\red{\tt{QUESTION :-}}$

This question is from Physics bolo of class 11, CBSE board.

(It is a numerical) ...

https://brainly.in/question/19308007?utm_source=android&utm_medium=share&utm_campaign=question

Answer 2

Given :

Sum of the zeroes of the polynomial f(x) = 2x³– 3kx² + 4x – 5 is 6.

To find :

Value of k .

Concept :

For a cubic polynomial is in the form of ax³ + bx² + cx + d :

∵ Sum of zeros = -b/a  where, b is coefficient of x² and a is the coefficient of x³

Solution :

Now , f(x) = 2x³ - 3kx² + 4x - 5

Or, f(x) = 2x³ + (-3kx²) + 4x + (-5)

Now according to concept :

⇒ -coefficient of x²/coefficient of x³ = 6

⇒ -(-3k)/2 = 6

⇒ 3k/2 = 6

⇒ 3k = 6 × 2

⇒ 3k = 12

⇒ k = 12/3

⇒ k = 4

∴ Required value of k = 4

Answer 3

Given :

Sum of the zeroes of the polynomial f(x) = 2x³– 3kx² + 4x – 5 is 6.

To find :

Value of k .

Concept :

For a cubic polynomial is in the form of ax³ + bx² + cx + d :

∵ Sum of zeros = -b/a  where, b is coefficient of x² and a is the coefficient of x³

Solution :

Now , f(x) = 2x³ - 3kx² + 4x - 5

Or, f(x) = 2x³ + (-3kx²) + 4x + (-5)

Now according to concept :

⇒ -coefficient of x²/coefficient of x³ = 6

⇒ -(-3k)/2 = 6

⇒ 3k/2 = 6

⇒ 3k = 6 × 2

⇒ 3k = 12

⇒ k = 12/3

⇒ k = 4

∴ Required value of k = 4

Answer 4

Given :

Sum of the zeroes of the polynomial f(x) = 2x³– 3kx² + 4x – 5 is 6.

To find :

Value of k .

Concept :

For a cubic polynomial is in the form of ax³ + bx² + cx + d :

∵ Sum of zeros = -b/a  where, b is coefficient of x² and a is the coefficient of x³

Solution :

Now , f(x) = 2x³ - 3kx² + 4x - 5

Or, f(x) = 2x³ + (-3kx²) + 4x + (-5)

Now according to concept :

⇒ -coefficient of x²/coefficient of x³ = 6

⇒ -(-3k)/2 = 6

⇒ 3k/2 = 6

⇒ 3k = 6 × 2

⇒ 3k = 12

⇒ k = 12/3

⇒ k = 4

∴ Required value of k = 4

Answer 5

Given :

Sum of the zeroes of the polynomial f(x) = 2x³– 3kx² + 4x – 5 is 6.

To find :

Value of k .

Concept :

For a cubic polynomial is in the form of ax³ + bx² + cx + d :

∵ Sum of zeros = -b/a  where, b is coefficient of x² and a is the coefficient of x³

Solution :

Now , f(x) = 2x³ - 3kx² + 4x - 5

Or, f(x) = 2x³ + (-3kx²) + 4x + (-5)

Now according to concept :

⇒ -coefficient of x²/coefficient of x³ = 6

⇒ -(-3k)/2 = 6

⇒ 3k/2 = 6

⇒ 3k = 6 × 2

⇒ 3k = 12

⇒ k = 12/3

⇒ k = 4

∴ Required value of k = 4