Take the square of both sides:
(Don't forget to determine the 'existence conditions' because has to be or 0)
Yes, that's correct, but you have understand me wrong, I didn't say that:
You just can wright:
I mean, that after you've calculated the solutions, you have to assure that (conditon 1) and (condition 2) summarized , because you've to assure you don't get a negative number under the square root.
I find it easier to take as being positive since IMO it makes factorising it easier. To do this you can move everything over to the other side (and not forgetting to change the sign) :
. This quadratic does factor but there is nothing wrong with the formula if you prefer it.
Because you squared both sides of the equation you may have introduced extraneous solutions - solutions which work in the squared equation but not in the original one. For example we're given the equation .
If we square it we arrive at which has both 4 and -4 as "solutions" but we can see that -4 clearly doesn't work in the original equation.
You must also consider the domain of the original equation which is what Siron pointed out by saying that .
In this example one of your solutions to the quadratic will be extraneous - you can find out which by subbing the values into the original equation