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Simplify: -8+- SR/4 SQ/43 over -6
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\[-8\pm \sqrt{4}\sqrt{43}\]
over -6
\[\frac{-8+-SR}{4} \cdot \frac{SQ}{\frac{43}{-6}} ?\] lol i couldn't understand at first so we have \[\frac{-8 \pm 2 \sqrt{43}}{6}=\frac{-8}{6} \pm \frac{2 \sqrt{43}}{6}=\frac{-4}{3} \pm \frac{\sqrt{43}}{3}\]
Sorry, not sure how to type some of the functions
But your answer is the same as what the homework gave as correct answer. Just can't figure out where the -8 and -6 turned to 4 and three?
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i reduced the fraction
\[\frac{-8}{6} =\frac{-8 \div 2}{6 \div 2}=\frac{-4}{3}\]
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