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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.
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Solution
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.
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