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Given a quadratic equation:  x^2 - sqrt(5a^2 - 26a - 8) * x + (a^2 - 6a + 12) = 0

It has 2 integer roots. If a is an integer as well, what is the value of a?

Jan 7, 2021

#1
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I think you mean $$x^2-x\sqrt{5a^2-26a-8}+a^2-6a+12=0$$

Jan 7, 2021
edited by Guest  Jan 7, 2021
edited by Guest  Jan 7, 2021
#2
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Uhhhh......yah......that is what was posted ......

Guest Jan 7, 2021