Question 374196
I assume you meant
{{{sqrt(a+b) = sqrt(a) + sqrt(b)}}}
(If you really meant
{{{(a+b)^2 = sqrt(a) + sqrt(b)}}}
then report your question.<br>
To solve this we will square both sides:
{{{(sqrt(a+b))^2 = (sqrt(a) + sqrt(b))^2}}}
On the right side, since <b>exponents do not distribute</b>, we must use FOIL or the {{{(p + q)^2 = p^2 + 2pq + q^2}}} pattern to square it properly. I like using the pattern:
{{{a + b = (sqrt(a))^2 + 2*(sqrt(a))*(sqrt(b)) + (sqrt(b))^2}}}
which simplifies to:
{{{a + b = a + 2*sqrt(a)*sqrt(b) + b}}}
Subtracting a and b from each side we get:
{{{0 = 2*sqrt(a)*sqrt(b)}}}
We now have a product that is equal to zero. From the Zero Product Property we know that this (or any) product is zero <i>only</i> if one one (or more) of the factors is zero. So
2 = 0 or {{{sqrt(a) = 0}}} or {{{sqrt(b) = 0}}}
This first equation is impossible so we get no solution from it. Solving the other two equations (by squaring both sides) we get:
a = 0 or b = 0<br>
Whenever you square both sides of an equation, which we have done several times so far, it is important (not just a good idea) to check you answer(s). Squaring both sides of an equation can introduce extraneous solutions. Extraneous solutions are solutions which fit the squared equation <i>but do not fit the original equation!</i> This can happen even if no mistakes are made! So we have to check our answers and make sure they actually work. And we use the original equation to check:
{{{sqrt(a+b) = sqrt(a) + sqrt(b)}}}
Checking a = 0:
{{{sqrt((0)+b) = sqrt((0)) + sqrt(b)}}}
{{{sqrt(b) = 0 + sqrt(b)}}}
{{{sqrt(b) = sqrt(b)}}} Check!<br>
Checking b = 0:
{{{sqrt(a + (0)) = sqrt(a) + sqrt((0))}}}
{{{sqrt(a) = sqrt(a) + 0}}}
{{{sqrt(a) = sqrt(a)}}} Check!<br>