How do you solve #sqrtx+9=4#?

1 Answer
Aug 11, 2015

Answer:

#x = O/#

Explanation:

As it stands, your equation has no possible solutions among real numbers.

To see why that is the case, isolate the radical term on one side of the equation by adding #-9# to both sides

#sqrt(x) + color(red)(cancel(color(black)(9))) - color(red)(cancel(color(black)(9))) = 4-9#

#sqrt(x) = -5#

For real numbers, you have

#color(blue)(sqrt(x)>=0, (AA)x >=0)#

Simply put, the square root of a positive real number, which is what must be to begin with, is always positive.

In your case, the square root of #x# is equal to a negative number, which cannot happen for real numbers.

Therefore, your equation has no solution for any #x in RR#.