How do you solve #sqrt(5x^2-9)=2x# and check your solution?

1 Answer
Alan P.
Sep 4, 2017

#x=+3#
(see below for solution and check)

Explanation:

Given
#color(white)("XXX")sqrt(5x^2-9)=2x#

Squaring both sides (this may introduce extraneous solutions; we will need to check later)
#color(white)("XXX")5x^2-9=4x^2#

Subtract #4x^2# from both sides
#color(white)("XXX")x^2-9=0#

Factor the left side
#color(white)("XXX")(x+3)(x-3)=0#

which implies
#color(white)("XXX")x=-3color(white)("xxx") or color(white)("xxx")x=+3#

Checking for extraneous solutions:
#{: ("with "x=-3,,color(white)("xxx"),"with "x=+3,), (sqrt(5 * x^2-9)=+6,2x=-6,,sqrt(5^2-9)=+6,2x=6), ("extraneous",,,"valid",) :}#

The only valid solution is #x=+3#

Related questions
See all questions in Radical Equations
Impact of this question
1584 views around the world
You can reuse this answer
Creative Commons License