Question

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.

Answer 1

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.