Question

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

Answer 1

Hm. What do you mean? The real solution of each is the appropriate root. e.g.:

$$ x^2 - 27 = 0 $$
$$ x^2 = 27 $$
$$ x = \pm \sqrt{27} $$

Answer 2

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.

Answer 3

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

Answer 4

Thanks so much:)