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

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Answer 1 · 2015-08-11T18:41:58

#x = O/#

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#.