like 6x^2+2x-5=0 I can find solutions but they are generally wrong like (2x-5)(6x+2)=0. I think polynomial long-division might work, come to think of it.
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Here's a kick-off (ax+b)(cx+d) where ac=6 and bd=-5 now expand (ax+b)(cx+d) out, what do you get?
Originally Posted by bournouli like 6x^2+2x-5=0 I can find solutions but they are generally wrong like (2x-5)(6x+2)=0. I think polynomial long-division might work, come to think of it. Since the discriminant is irrational it will not factor with rational coefficients. Here is a way it can be done. Now complete the square to get Now we can write this as the difference of squares
It was really that easy huh? -thanks.
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