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

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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

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