3

According to the law of radicals:

$$ \sqrt {an} = \sqrt a \cdot \sqrt n $$

Wouldn't it make sense that: $\sqrt {x^3}= \sqrt {x^2} \cdot \sqrt x = x \sqrt x$

Obviously this doesn't work if you plug in values and compare, but this makes logical sense?

Uys of Spades
  • 223
  • 3
    . . . But it does work for plugging in values. – HDE 226868 Nov 10 '15 at 02:37
  • 8
    Assuming you're plugging in negative values and that's where you're getting issues, then it's because $\sqrt{x^2}=|x|$, not $\sqrt{x^2}=x$. – Hayden Nov 10 '15 at 02:38
  • 1
    What value are you using? It works for positive values $x$ – Thomas Andrews Nov 10 '15 at 02:41
  • Note that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only works for $a, b \ge 0$, and so does $\sqrt{a^2} = a$. So yes, if $a \ge 0$, we have $\sqrt{a^3} = a \sqrt{a}$. And otherwise, if $a < 0$, then the square root is complex. – Joel Cohen Nov 10 '15 at 02:46
  • As everybody said before, the only problem is for negative real numbers, because then, neither the left or the right side make sense since you would have the square root of a negative numbers. If you were working in the complex field, this makes complete sense. – mathstu15 Nov 10 '15 at 02:47
  • This is context-dependent. Would you care to give us some, so we can answer you? – Cameron Buie Nov 10 '15 at 03:33

1 Answers1

2

Your formula is mostly correct, except that it should read $$\sqrt {x^3}= \sqrt {x^2} \cdot \sqrt x = \color{red}{|x| \sqrt x}$$

Brevan Ellefsen
  • 14,793