I understand by convention ,if $\sqrt{x^2}=a$ , a is defined to be the positive root of x or the principal square root. but what does this mean for exponential equations- does $x^{0.5}=-5$ have no solutions?

  • $\begingroup$ In any situation where the inverse of a function has, or may have, multiple values you need to specify which one you are choosing. You can, for example, allow complex solutions to your example but then there are two possibilities and if you want to work with a single number you'll have to specify which one you want. Keep in mind that the choice you make for one value in the image may not be readily compatible with the choice you make for another value. $\endgroup$
    – lulu
    Jul 7, 2015 at 12:13
  • $\begingroup$ there is no real solution of the equation $x^{0.5}=-5$ $\endgroup$ Jul 7, 2015 at 12:15
  • $\begingroup$ Is this $x^{5}= -5$ or $x^{0.5} = -5$? $\endgroup$ Jul 7, 2015 at 12:15
  • $\begingroup$ @WarrenHill made an edit to get rid of the confusion. It is $\frac{1}{2}$ (the edit is being peer-reviewed) $\endgroup$
    – alisianoi
    Jul 7, 2015 at 12:16
  • 1
    $\begingroup$ Ah, I misinterpreted the example. Yes...if you define $x^{0.5}$ as positive or non-existant then there are no cases wherein it equals -5. $\endgroup$
    – lulu
    Jul 7, 2015 at 12:23

1 Answer 1


There is no real number $x$ for which $x^{0.5} = -5$, because $x^{0.5} \geq 0$ for all $x$ for which the value is defined.

This means that the equation $x^{0.5}=-5$ has no solution in the real numbers.

  • $\begingroup$ why is x^0.5 greater than or equal to zero, is it because of the principal square root convention $\endgroup$
    – axcvbk
    Jul 7, 2015 at 12:14
  • $\begingroup$ @axcvbk Because of the definition of $a^{0.5}$. $a^{0.5}$ is defined as the positive solution to the equation $x^2=a$. $\endgroup$
    – 5xum
    Jul 7, 2015 at 12:18
  • $\begingroup$ so a^0,5 is defined as positive or non existant? $\endgroup$
    – axcvbk
    Jul 7, 2015 at 12:27
  • $\begingroup$ @axcvbk On the reals, $a^{0.5}$ is defined only for $a\geq 0$. $\endgroup$
    – 5xum
    Jul 7, 2015 at 12:30
  • $\begingroup$ @axcvbk Yes, if you are considering only real numbers, $x^{1/2}$ is either positive or nonexistent, for the reasons/definitions discussed in various responses above. $\endgroup$
    – rogerl
    Jul 7, 2015 at 12:34

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