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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

Sagot :

Answer:

k=2

Step-by-step explanation:

[tex](k + 2) {}^{2} - (k - 2)x - 5 = 0[/tex]

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

[tex](a + b)(a - b) = {a}^{2} - b {}^{2} [/tex]

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

[tex]k - 2 = 0[/tex]

[tex]k = 2[/tex]

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