How do you solve #\sqrt { 2x - 5} - \sqrt { 4x - 9} = 0#?

1 Answer
Cesareo R.
Dec 7, 2016

No real solution

Explanation:

The feasible solutions are for #2x-5 ge 0 nn 4x-9 ge0# so

#x ge 5/2#

Now squaring term to term the equivalent equation

#\sqrt { 2x - 5} = \sqrt { 4x - 9} # we have

#2x-5=4x-9# so #2x=4# and #x = 2# but #2 < 5/2# so no real solution for the equation

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