How do you solve and check for extraneous solutions in #sqrt(5x) - x = 0#?

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Answer 1 · 2015-08-01T17:56:31

#color(red)(x=0)# and #color(red)(x=5)# are solutions.
There are #color(red)("no")# extraneous solutions.

SOLVE:

#sqrt(5x)-x = 0#

Add #x# to each side.

#sqrt(5x) = x#

Square each side.

#5x=x^2#

Subtract #5x#from each side.

#0 = x^2-5x#

Factor.

#0 = x(x-5)#

#x=0# and #x-5=0#

#x=0# and #x=5#

CHECK FOR EXTRANEOUS SOLUTIONS

#sqrt(5x)-x = 0#

If #x=0#,

#sqrt(5(0)) -0 = 0#

#sqrt(0)-0 = 0#

#0-0=0#

#0=0#

#x=0# is a solution.

If #x=5#

#sqrt(5(5)) -5 = 0#

#sqrt25 -5 = 0#

#5-5=0#

#0=0#

#x=5# is a solution.

There are no extraneous solutions.