Math, asked by monikagupta1086, 4 months ago

if a,b are the zeros of x^-6x+k=0 what is the value of k if 3a + 2b =20​

Answers

Answered by snehitha2
14

Appropriate Question :

If a, b are the zeroes of x²-6x+k=0, then what is the value of k if 3a + 2b =20​

Answer :

The required value of k is -16

Step-by-step explanation :

Given :

  • a, b are the zeroes of x² - 6k + k = 0
  • 3a + 2b = 20

To find :

the value of k

Solution :

To solve this question, we must know the relation between sum of zeroes and coefficients of quadratic equation.

Sum of zeroes = -(x coefficient)/x² coefficient

For the given quadratic equation, x² - 6x + k = 0

x² coefficient = 1

x coefficient = -6

constant term = k

Sum of zeroes = -(-6)/1

a + b = 6 ➙ [1]

3a + 2b = 20 ➙ [2]

Multiplying the equation [1] by 3, we get

3(a + b) = 3(6)

3a + 3b = 18 ➙ [3]

Now, subtract equation [2] from equation [3],

3a + 3b - (3a + 2b) = 18 - 20

 3a + 3b - 3a - 2b = -2

   b = -2

So, one of the zeroes = -2

Since it's a zero, when we substitute x = -2, the result is zero.

x² - 6x + k = 0

(-2)² - 6(-2) + k = 0

4 + 12 + k = 0

 16 + k = 0

  k = -16

∴ The value of k is -16

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