SOLUTION: Thank you in advance for your assistance. This problem was presented on my pretest. This is how I solved and it is wrong. Will you please correct it for me? Solve for x sqrt5x

Algebra ->  Radicals -> SOLUTION: Thank you in advance for your assistance. This problem was presented on my pretest. This is how I solved and it is wrong. Will you please correct it for me? Solve for x sqrt5x      Log On


   



Question 710177: Thank you in advance for your assistance.
This problem was presented on my pretest. This is how I solved and it is wrong. Will you please correct it for me?
Solve for x
sqrt5x=-5
(sqrt5x)^2=-(5)^2
5x=25
x=5
Again, thank you.
Peter

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%285x%29+=+-5
sqrt%285x%29 refers to the positive number/expression that, when squared, results in 5x. But the equation says that sqrt%285x%29 is a negative number, -5. A positive square root cannot be equal to a negative number. IOW: There is no solution to this equation!

If you don't notice this and proceed to solve the equation, as you did, then you can still find out that there is no solution. Any time you square both sides of an equation, as you did, you must check your answers! Let's check your solution, x = 5, using the original equation:
sqrt%285x%29+=+-5
Checking x = 5:
sqrt%285%285%29%29+=+-5
sqrt%2825%29+=+-5
5+=+-5
Since this is false, the check fails!! x=5 is not a solution. And since we must reject the only "solution" you found, there are no solutions to your equation.