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Compute the zeros of the polynomial 4x2 - 4x - 8

User Cyberax
by
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1 Answer

1 vote

Answer:

(2, 0) and (-1, 0)

Explanation:


4x^2 - 4x - 8 = 0 \text{ // Divide by 4} \\x^2 - x - 2 = 0\\\\x_(1, 2) = (-(-1) \pm √((-1)^2 - 4 * 1 * (-2)))/(2*1)\\\\x_(1, 2) = \frac{1 \pm √(1 + 8)}2\\\\x_(1, 2) = \frac{1 \pm \sqrt9}2\\\\x_(1, 2) = \frac{1 \pm 3}2\\\\x_1 = \frac{1 + 3}2 = \frac42 = 2\\\\x_2 = \frac{1 - 3}2 = \frac{-2}2 = -1

Therefore, the zeroes are (2, 0) and (-1, 0).

User Miroslav Dzhokanov
by
8.6k points

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