If One Zero of Polynomial 3x2-8x+2k+1 in Seven Times other Find the Zero and the Value of the k
Let's assume that one zero of the polynomial 3x2 - 8x + 2k + 1 is seven times the other zero.
Let the two zeros be α and β, with α = 7β.
We know that the sum of the zeros of a quadratic polynomial is given by the formula:
α + β = -b/a
In this case, a = 3 and b = -8, so we have:
α + β = -(-8)/3 = 8/3
Since α = 7β, we can substitute this into the equation:
7β + β = 8/3
Combining like terms:
8β = 8/3
Dividing both sides by 8:
β = 1/3
Substituting this value back into α = 7β:
α = 7(1/3) = 7/3
So the zeros of the polynomial are α = 7/3 and β = 1/3.
To find the value of k, we can substitute one of the zeros into the polynomial equation and solve for k.
Let's substitute α = 7/3 into the polynomial:
3(7/3)2 - 8(7/3) + 2k + 1 = 0
49/3 - 56/3 + 2k + 1 = 0 -7/3 + 2k + 1 = 0 2k - 4/3 = 0 2k = 4/3 k = 2/3
Therefore, the zero of the polynomial is α = 7/3, the other zero is β = 1/3, and the value of k is 2/3.
If one zero of polynomial 3x2-8x+2k+1 in seven times other, then the zero is α = 7/3, the other zero is β = 1/3, and the value of k is 2/3
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