Is the simplified form of 2 √3 - 2√3 rational? I'm going with no.
wat do u get after simplification ?
i would think 0
yup ! Isn't 0 rational ?
oh thank you very much! Can i have you help with another?
shoot
btw, 0 IS a rational number !
I have to simplify this:
simplify the last two terms first
\(\large \sqrt{45} = \sqrt{9 \times 5} = ?\)
sqrt of 45 is 6.70?
dont use calculator :/
whats the square root of 9 ?
\(\large 3^2 = 9\), so \(\large \sqrt{9} = 3\)
use that to simplify^
\(\large \sqrt{9 \times 5} = 3\sqrt{5}\)
i don't know how to do sqrt on calculator anyway. Sorry something came up. I'm back
\(\large 3\sqrt{5} - 2\sqrt{7} + \sqrt{45} - \sqrt{28}\) simplify last two terms : \(\large 3\sqrt{5} - 2\sqrt{7} + \sqrt{9 \times 5} - \sqrt{4 \times 7}\) \(\large 3\sqrt{5} - 2\sqrt{7} + 3\sqrt{5} - 2\sqrt{7}\)
next, combine like terms
\(\large 3\sqrt{5} - 2\sqrt{7} + \sqrt{45} - \sqrt{28}\) simplify last two terms : \(\large 3\sqrt{5} - 2\sqrt{7} + \sqrt{9 \times 5} - \sqrt{4 \times 7}\) \(\large 3\sqrt{5} - 2\sqrt{7} + 3\sqrt{5} - 2\sqrt{7}\) combine like terms : \(\large 6\sqrt{5} - 4\sqrt{7} \)
^thats the final simplified form
can you be more specific? I can't do sqrt or cube roots at all
that's the one i was going for too!! Thanks.
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