How do you solve #sqrt(4x - 3) = 2 + sqrt(2x - 5)#?

Question

1259 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-05T15:54:17

x=7

First let's fix the domain:

#4x-3>=0# and #2x-5>=0#

#x>=3/4# and #x>=5/2#

that's #x>=5/2#

Then let's square both parts:

#(sqrt(4x-3))^2# and #(2+sqrt(2x-5))^2#

#4x-3=4+4sqrt(2x-5)+2x-5#

let's put the irrational term on the right and the remaining terms on the left:

#4x-3-4-2x+5=4sqrt(2x-5)#

#2x-2=4sqrt(2x-5)#

let's divide all terms by 2:

#x-1=2sqrt(2x-5)#

let's again square both parts:

#(x-1)^2=(2sqrt(2x-5))^2#

#x^2-2x+1=4(2x-5)#

#x^2-2x+1=8x-20#

#x^2-10x+21=0#

whose solutions are x=2 and x=7

but only the second one belongs to the domain