How do you solve #sqrt(4x+1)+3=0#?

2 Answers
Mar 16, 2018

Answer:

You can't.

Explanation:

#sqrt(4x+1)+3=0#

Subtract #-3# from both sides,

#sqrt(4x+1)=-3#

Since the square root of any number cannot be negative, there are no values of #x# that satisfies the equation.

Mar 25, 2018

Answer:

Actually, there is no solution.

Explanation:

I know everyone else said that the solution is #x=2#, but if you plug it in, it doesn't hold up:

#color(white)=>sqrt(4x+1)+3=0#

#=>sqrt(4(2)+1)+3=0#

#color(white)=>sqrt(8+1)+3=0#

#color(white)=>sqrt(9)+3=0#

#color(white)=>3+3=0#

#color(white)=>6!=0#

You can also look at the graph of #sqrt(4x+1)+3# and see that there are no zeroes:

graph{sqrt(4x+1)+3 [-10.24, 15.07, -3.37, 9.29]}

Here's the flaw in trying to solve it:

#sqrt(4x+1)+3=0#

#sqrt(4x+1)=-3#
#" "color(red)uarr#
Here's the mistake. A square root cannot equal a negative number, so the process needs to stop here. Unfortunately, you can't square both sides.

Hope this helped!