find the value of K such that 3x²+2kx+x-k -5has the sum of their zeroes as half of their product
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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
.........hope that this may help you......
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
.........hope that this may help you......
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