How do you solve #3sqrtx + x = 10#?

1 Answer
Jun 24, 2016

#x=4#

Explanation:

'Transpose' #x# to the other side of the equation and square both sides.
Expand the binomial on the right side.
'Transpose #9x# to the right side of the equation.
Use the quadratic formula.

It should look like this:

#3sqrtx=10-x#

#9x=(10-x)^2#

#9x=100-20x+x^2#

#0=100-29x+x^2#

#x=-25#

and

#x=4#

However, we will not accept #x=-25# as an answer because it will yield an imaginary number in the original equation. The squaring step (step1b) created this extra root.