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If both zeroes of quadratic polynomial (k+2)x^2-(k-2)x-5 are equal in magnitude but opposite in sign find k

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Answer:

k=2

Explanation:


(k + 2) {}^(2) - (k - 2)x - 5 = 0

If we have two zeroes that are equal in size, but different signs then we have a different of squares,


(a + b)(a - b) = {a}^(2) - b {}^(2)

So in order to do this we must make the middle term, x 0 so


k - 2 = 0


k = 2

User Gordon K
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