Question 196350
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In order for


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x^2 + kx + 8 = 0]


to have exactly one root


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x^2 + kx + 8]


must be a perfect square trinomial, so knowing that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (ax + b)^2 = (ax)^2 + 2abx + b^2]


you can clearly see that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k = 2ab]


where *[tex \LARGE a^2 = 10] and *[tex \LARGE b^2 = 8]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k = \left(\sqrt{10}\right)\left(\sqrt{8}\right) = 8\sqrt{5}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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