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? |