Find the value of k such that the quadratic polynomial 3x²+2kx+x-k-5 has the sum of zeroes half of their product
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Answered by
3
Answer:
hii dear
Explanation:
l think ans is 2.
Answered by
8
Answer:
3x^2 + 2kx + x - k - 5 = 0
3x^2 + (2k + 1)x + (- k - 5) = 0
a = 3 ... b = (2k + 1) ... c = (- k - 5)
product of roots = c/a ... sum of roots = -b/a
Sum = -b/a = -(2k + 1) / 3
Product = c/a = (- k - 5)/3
sum is 1/2 of product
-(2k + 1) / 3 = (1/2)(- k - 5)/3 <<< 3 cancels
-4k - 2 = -k - 5
- 3k = -3
k = 1
Explanation:
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