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$$\sqrt[4]{(-x)^4}=x$$

This equation was provided to me, and I had to find out if this condition was sometimes, always, or never true.

At first my initial thought was that the condition will ALWAYS be true, since $\sqrt[4]{(-x)^4}$ evaluates to $\sqrt[4]{x^4}$, which evaluates to $x$. Since $x=x$, the condition is always true.

However, as I was about to put Always as my solution, I realized that SOMETIMES might be the correct answer. My process was that $\sqrt[4]{(-x)^4}$ can be restated as $(-x)^{4^{\frac{1}{4}}}$, and after multiplying the exponents, this is simply $(-x)^1$ or $-x$.

$-x=x$ is only true for the value $x=0$, so I put SOMETIMES as my solution.

Is sometimes, or always the solution to this question?

If you find the solution to this, remember, you are answering this important question:

In which order do you simplify the expression, do you carry out the exponent to the base first, or do you simplify exponents first?

Yash Jain
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2 Answers2

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It is always true that $$\sqrt[4]{(-x)^4}=|x|$$

Indeed by definition the square root of $a\geq0$ $$\sqrt[4] a$$ is the non negative value $b$ such that $$b^4=a$$

user
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  • Why is the absolute value of one side being taken? – Yash Jain Jan 19 '18 at 23:03
  • @YashJain This is the definition of the principle square root. – JobHunter69 Jan 19 '18 at 23:04
  • However, our math textbook says that this is sometimes true. Should we assume that the book is wrong then? – Yash Jain Jan 19 '18 at 23:06
  • @YashJain it is sometimes true that $\sqrt[4]{(-x)^4}=x$. It is always true that $\sqrt[4]{(-x)^4}=|x|$. These are different statements and both of them are correct. – JMoravitz Jan 19 '18 at 23:07
  • because by definition of square root, the square root of $a\geq0$ $\sqrt[4] a$ is the non negative value $b$ such that $b^4=a$ – user Jan 19 '18 at 23:07
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It should be sometimes since there are two cases to consider:

CASE 1 ($x \ge 0$): In this case, we have $\sqrt[4]{(-x)^4}=\sqrt[4]{x^4} = x$

CASE 2 ($x < 0$): In this case, we have $\sqrt[4]{(-x)^4} = \sqrt[4]{x^4} = -x$ because $x$ is already a negative number and the result should be positive.

Here, notice that we always do the operation $(-x)^4 = x^4$ first, then the $\sqrt[4]{x^4} = x$ or $-x$ according to the value of $x$.

And we can combine these results as $\sqrt[4]{(-x)^{4}} = |x|$ as you can see from the above cases.

ArsenBerk
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  • Is there any information online which tells me we apply the exponent to the base first? – Yash Jain Jan 19 '18 at 23:12
  • I don't know any online source but I can give you an example where your approach is valid. If you have an expression such as $(\sqrt[4]{-1})^4$ then you can write it as $((-1)^{\frac{1}{4}})^4 = (-1)^1 = -1$. – ArsenBerk Jan 19 '18 at 23:18