If one zero of the polynomial 3x^2-8x (2k+1) is seven time the other find both zeroes and K ?
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Hello
Please recheck the question, i thought that the polynomial should be 3x² - 8x + (2k+1) or 3x² - 8x - (2k-1) instead of 3x² - 8x(2k+1)
So i am solving for first one, if i am correct then the solution is given below.
3x² - 8x + (2k + 1)
Let the roots be a and b
then a = 7b
a + b = -b/a
8b = 8/3
b = 1/3
now
ab = 2k + 1/3
7b² = 2k + 1/3
7/3 = 2k + 1
7/3 - 1 = 2k
k = 2/3
Hope this helps
Please recheck the question, i thought that the polynomial should be 3x² - 8x + (2k+1) or 3x² - 8x - (2k-1) instead of 3x² - 8x(2k+1)
So i am solving for first one, if i am correct then the solution is given below.
3x² - 8x + (2k + 1)
Let the roots be a and b
then a = 7b
a + b = -b/a
8b = 8/3
b = 1/3
now
ab = 2k + 1/3
7b² = 2k + 1/3
7/3 = 2k + 1
7/3 - 1 = 2k
k = 2/3
Hope this helps
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