Question 1150463
<br>
{{{x^(1/2)+x^(1/4) = 30}}}<br>
Introduce a new variable to make it easier to see how to solve the problem.<br>
Let y = x^(1/4)
Then y^2 = x^(1/2)<br>
Then the equation is<br>
{{{y^2+y = 30}}}
{{{y^2+y-30 = 0}}}
{{{y+6)(y-5) = 0
{{{y = -6}}} or y = {{{y = 5}}}<br>
y = -6 gives us x^(1/4) = -6, which is not possible.<br>
y = 5 gives us x^(1/4) = 5, which makes x = 5^4 = 625.<br>
ANSWER: The problem asks for the sum of the digits of x, which is 6+2+5 = 13.<br>
Of course, if a formal algebraic solution is not required, the recognition that x^(1/4) is the square of x^(1/2) means we are looking for two integers, one the square of the other, whose sum is 30.  That obviously is 5 and 25, which means x is 5^4 = 35^2 = 625.<br>