Root help?!

If #c = d^2#, does #d = sqrt c#?

1 Answer
Jun 8, 2017

Answer:

Yes, but that's only half the story.

Explanation:

The thing to remember here is that every positive real number has two square roots

  • a positive square root called the principal square root
  • a negative square root

That is the case because the square root of a positive real number #c#, let's say #d# to use the variables you have in your example, is defined as the number that, if multiplied by itself, gives you #d#.

In other words, if you have

#d xx d = d^2 = c#

then you can say that

#d = sqrt(c)#

is the square root of #c#.

However, notice what happens if we multiply #-d# by itself

#(-d) xx (-d) = (d xx d) = d^2 = c#

This time, you can say that

#d = -sqrt(c)#

is the square root of #c#.

Therefore, for every positive real number #c#, you have two possible square roots denoted using a plus-minus sign

#d = +- sqrt(c)#

You can thus say that if

#c = d^2#

then

#d = +- sqrt(c)#

You can check that this is the case because if you square both side, you will end up with

#d^2 = (+sqrt(c))^2" "# and #" "d^2 = (-sqrt(c))^2#

which is

#d^2 = sqrt(c) * sqrt(c)" "# and #" " d^2 = (-sqrt(c)) * (-sqrt(c))#

#d^2 = sqrt(c) * sqrt(c)" "# and #" " d^2 = sqrt(c) * sqrt(c)#

#d^2 = c" "# and # " "d^2 = c#

So, for example, you can say that the square roots of #25# are

#sqrt(25) = +-5#

The principal square root of #25# is equal to #5#, which is why we always say that

#sqrt(25) = 5#

but do not forget that #-5# is also a square root for #25#, since

#(-5) * (-5) = 5 * 5 = 5^2 = 25#