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anonymous
 5 years ago
does anyone understand this finding the real solution? the 3 and 2 are exponets ???x^3/^2 27=0
anonymous
 5 years ago
does anyone understand this finding the real solution? the 3 and 2 are exponets ???x^3/^2 27=0

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shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Hm. What do you mean? The real solution of each is the appropriate root. e.g.: $$ x^2  27 = 0 $$ $$ x^2 = 27 $$ $$ x = \pm \sqrt{27} $$

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, sorry, misunderstood. You have \(x^{\frac{3}{2}}\) In that case the process is similar, but you have to remember that \(\sqrt{x}\) is the same as \(x^2\), so \(x^{\frac{3}{2}}\) is really the same as \(\sqrt{x^3}\). So, you end up with: $$ \sqrt{x^3} = 27 $$ You can reverse the square root by squaring and the cube by taking the cubed root.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0I meant \(\sqrt{x}\) is the same as \(x^{\frac{1}{2}}\).
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