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Sum

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

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#### Solution

Given polynomial is x^{2} - (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(2"k" - 1))/(1)`

⇒ `"k" + 6 = 2"k" - 1`

⇒ `6 + 1 = 2"k" - "k"`

⇒ `"k" = 7`

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