Q14 of 39 Page 1

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

Consider the polynomial x2 – (k + 6)x + 2(2k – 1),

Compare it with general equation ax2 + bx + c,

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

We know,



⇒ Sum of zeroes



⇒ Product of zeroes



According to the question,

Sum of zeroes = 1/2 Product of zeroes



⇒ k + 6 = 2k - 1

⇒ 6 + 1 = 2k - k

⇒7= k


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