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Mathematics OpenStudy (anonymous):

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

8 years ago OpenStudy (shadowfiend):

Hm. What do you mean? The real solution of each is the appropriate root. e.g.: $$x^2 - 27 = 0$$ $$x^2 = 27$$ $$x = \pm \sqrt{27}$$

8 years ago OpenStudy (shadowfiend):

Ah, 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.

8 years ago OpenStudy (shadowfiend):

I meant $$\sqrt{x}$$ is the same as $$x^{\frac{1}{2}}$$.

8 years ago OpenStudy (anonymous):

Thanks so much:)

8 years ago
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