How do you find the number of complex, real and rational roots of $3x^2+11x-10=0$ ?

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Answer 1 · 2016-11-21T18:02:47

The two roots are real, but not rational given by x= $(-11 +- sqrt 241)/6$ . No complex roots.

This is a quadratic expression. Quadratic formula can be used to find the roots

x= $(-11+- sqrt (11^2 -4(3)(-10))) /(2(3)$

= $(-11 +- sqrt 241)/6$

as can be seen now, both the roots are real, they are not rational. There are no complex roots.

How do you find the number of complex, real and rational roots of 3x^2+11x-10=03x^2+11x-10=0?