Question

if alpha and beta are the zeroes of the quadratic polynomial p(x) =x^2-(k-6)x+(2x+1), find the value of K, if alpha +beta =alpha ×beta.

Answer 1

$\\ {x}^{2} - kx + 6x + 2x + 1 \\ {x}^{2} - (k - 8)x + 1 \\ \alpha + \beta = \alpha \times \beta \\ - \frac{b}{a} = \frac{c}{a} \\ k - 8 = 1 \\ k = 1 + 8 \\ k = 9$

Answer 2

Hey

Here is your answer,

p(x) =x^2-(k-6)x+(2x+1)

Sum of zeroes = -b/a
Alpha + beta = -(-(k-6)) / 1
Alpha + beta = k-6

Product of zeroes = c/a
Alpha x beta = 2k+1/1
Alpha x beta = 2k+1

Given that,

Alpha + beta = alpha x beta
k-6 = 2k+1
k-2k=1+6
-k=7
k=-7

Hope it helps you!