If One Zero of Polynomial 3×2-8x+2k+1 in Seven Times other Find the Zero and the Value of the k

If One Zero of Polynomial 3x²-8x+2k+1 in Seven Times other Find the Zero and the Value of the k

Solution

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

Simplifying:

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.

Answer

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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