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Question
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 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(2"k" - 1))/(1)`
⇒ `"k" + 6 = 2"k" - 1`
⇒ `6 + 1 = 2"k" - "k"`
⇒ `"k" = 7`
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