Question

Find the value of k such that the quadratic polynomial 3xsqure + 2kx +x-k-5 has the sum of zeroes as half of their products

Answer 1

Answer:

Step-by-step explanation:

Polynomial given

3x² + 2kx +x-k-5

= 3x² + (2k + 1)x + (-k-5)

a = 3, b = 2k+1 c = -(k+5)

Let α and β be the zeroes of the equation.

Sum of zeroes = α + β = -b/a = - (2k+1)/3.

Product of zeroes = αβ = c/a = - (k + 5)/3.

Given sum of zeroes = 1/2(product of zeroes)

=> α + β = 1/2 (αβ)

=> - (2k+1)/3 = 1/2(- (k + 5)/3)

= - (2k+1)/3 =  -1/3((k+5)/2)

// -1/3 can be striked out on both sides.

=> (2k + 1) = (k + 5)/2

=> 2(2k+1) = k + 5

=> 4k + 2 = k + 5

=> 4k - k = 5 - 2

=> 3k = 3

=> k = 1