-1

The original problem is $\sqrt{(-a)^{3}}\sqrt{(-a)}$

I attempted to solve this as the following way:

$\sqrt{(-a)^{3}}\sqrt{(-a)} = (-a)^{\frac{3}{2}}(-a)^{\frac{1}{2}}=(-a)^{2}=a^{2}$

However, I also saw a different way of solving it:

$\sqrt{(-a)^{3}}\sqrt{(-a)} = \sqrt{(-1)^{3}a^{3}}\sqrt{(-1)a} = \sqrt{(-1)a^{3}}\sqrt{(-1)a} = (-1)^{\frac{1}{2}}a^{\frac{3}{2}}(-1)^{\frac{1}{2}}a^{\frac{1}{2}} = (-1)^{\frac{1}{2}+\frac{1}{2}}a^{\frac{3}{2}+\frac{1}{2}} = -a^{2}$

According to Wolfram Alpha, the second solution is correct. I tried to find out what I'd done wrong, and narrowed it down the following:

  1. $\sqrt{(-a)^3} = ia^{\frac{3}{2}}$,
  2. $((-a)^{3})^{\frac{1}{2}} = ia^{\frac{3}{2}}$, link: https://www.wolframalpha.com/input/?i=((-a)%5E3)%5E(1%2F2)
  3. $(-a)^{3*\frac{1}{2}} = -ia^{\frac{3}{2}}$, link: https://www.wolframalpha.com/input/?i=(-a)%5E(3*1%2F2)
  4. $(-a)^{\frac{3}{2}} = -ia^{\frac{3}{2}}$,

Without the required reputation 10, I am not able to post more than 2 links, which is why I can't link the first and fourth case.

Anyway, assuming Wolfram Alpha is to be trusted, $((-a)^{3})^{\frac{1}{2}} \neq (-a)^{3*\frac{1}{2}}$. I thought this was a general rule, but apparently there is some conditions I was not aware of. This may be connected to complex numbers, which I am less than proficient with. I thought the problem was supposed to be solvable without using complex numbers.

3SAT
  • 7,512
Marius
  • 3

2 Answers2

1

Every non zero complex number has two square roots. That means that the square root of a complex number is not uniquely defined.

There are special cases.

When $x$ is a positive real number:

  • $x^{1/2}$ and $\sqrt x$ are defined to be the positive square root of $x$. So, even though $9$ has two square roots, $3$ and $-3$, we define $9^{1/2} = \sqrt 9 = 3$.

  • $(-x)^{1/2}$ and $\sqrt{-x}$ are defined to be $i \sqrt x$. So, even though $-9$ has two square roots, $3i$ and $-3i$, we define $(-9)^{1/2} = \sqrt{-9} = 3i$.

This is probably done because it provides a quick and easy way to use things like the quadratic formula.

In general, you should try to reduce your problem to an equivalent problem with positive base numbers, then apply the rules of exponentiation.

Here is the most unambiguous way to simplify $\sqrt{(-a)^{3}}\sqrt{(-a)}$

If a is a negative number, then, -a is a positive number. \begin{align} \sqrt{(-a)^{3}}\sqrt{(-a)} &= (-a)^{3/2} (-a)^{1/2} \\ &= (-a)^2 \\ &= a^2 \\ \end{align}

If $a$ is a positive number, then $a^3$ is also a positive number. \begin{align} \sqrt{(-a)^3}\sqrt{(-a)} &= \sqrt{-a^3}\sqrt{-a} \\ &= (i\sqrt{a^3})(i \sqrt a) \\ &= i^2 \sqrt{a^4} \\ &= -a^2 \end{align}

Steven Alexis Gregory
  • 26,769
0

$x=\sqrt{(-a^3)(-a)}=a^2$ but when we will seprate both numbers as different roots then $x=\sqrt{(-a^3)}\sqrt{-a}=(i\times a^{\frac{3}{2}})\times (i\times a^{\frac{1}{2}})=-a^2$.

Hope this will be helpful !

Vineet Mangal
  • 943