What is the value of k when the equation x2 − K 6 x 2 2k − 1 0 has sum of the roots equal to half of their product?

Find the value of k such that the polynomial  x2-(k +6)x+ 2(2k - 1) has some of its zeros equal to half of their product.

For given polynomial x2 - (k + 6)x +2( 2k - 1)Here

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

Given that;∴ Sum of zeroes = `1/2` (product of zeroes)⇒ `(-[-(k+6)])/1 = 1/2 xx (2(2k-1))/1`⇒ k + 6=2k-1⇒ 6+1= 2k - k⇒ k = 7

Therefore, the value of k = 7.

Concept: Geometrical Meaning of the Zeroes of a Polynomial

  Is there an error in this question or solution?