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Help please, I have like 30 pages of homework and the ones I'm posting are the ones I don't understand.
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If \[ax^2+bx+c=0 \]then\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \]And the radicand is \[b^2-4ac \] =-21 So \[x=\frac{-b \pm \sqrt{-21}}{2a}=\frac{-b \pm \sqrt{-1}\sqrt{21}}{2a}==\frac{-b \pm i\sqrt{21}}{2a} \] We write the sqaure root of -1 as i by convention. Now, a real number is DEFINED by having nothing i-ey in it (that it, it can be written as x=a+ib WHERE B IS =0), and herre x DOES have something i-ey in it, so there are no real solutions.
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