Question 229099
To solve equations with square roots, you must eliminate the square roots first. And to eliminate a square root:<ol><li>Isolate the square root (i.e. get the square root term alone on one side of the equation)</li><li>Square both sides of the equation. (Important: Whenever you square both sides of an equation, we may be introducing what are called extraneous solutions (solutions which fit the squared equation but do not fit the original equation). Because of this, checking your answers is no longer just a nice idea. It is necessary so you can identify and reject extraneous solutions, if any.)</li><li>If there are still square roots (you may have had more than one to start), then repeat this process until the square roots are all gone.</li></ol>
Let's try this on your equation. Since your two square roots are both square roots of x, we will only need to go through the above steps once.
1. Isolate
{{{2sqrt(x) + 20 = 15 + 3sqrt(x)}}}
I like positive coefficients so I'll subtract {{{2sqrt(x)}}} from each side:
{{{20 = 15 + sqrt(x)}}}
Then subtract 15 from each side:
{{{5 = sqrt(x)}}}
2. Square both sides:
{{{(5)^2 = (sqrt(x))^2}}}
{{{25 = x}}}
At this point, not only have we eliminated the square root, but we actually have a solution. (This does not always happen. Often, after the square roots are gone, there is an "regular" equation which still needs to be solved.)<br>
Now that we have a solution, we <b>must</b> check it because, as was explained above, it may be an extraneous solution. Always use the original equation when checking:
{{{2sqrt(x) + 20 = 15 + 3sqrt(x)}}}
Checking x = 25:
{{{2sqrt(25) + 20 = 15 + 3sqrt(25)}}}
Since {{{sqrt(25) = 5}}}
{{{2*(5) + 20 = 15 + 3*(5)}}}
{{{10 + 20 = 15 + 15}}}
{{{30 = 30}}} Check!<br>
So x=25 is the only solution to your equation. (Since x=25 was the only possible solution, if it had actually turned out to be extraneous it would have meant that your equation had no solutions.)