How do you solve `sqrt(3x-2)=1+sqrt(2x-3)`?

Given:`" "sqrt(3x-2)=1+sqrt(2x-3)`

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Square both sides

`3x-2" "=" "1+2sqrt(2x-3)+2x-3`

The `-2` on both sides cancel out

`3xcancel(-2)" "=" "2xcancel(-2)+2sqrt(2x-3)`

`3x=2x+2sqrt(2x-3)`

Subtract `2x` from both sides

`x=2sqrt(2x-3)`

Divide both sides by 2

`x/2=sqrt(2x-3)`

Square both sides

`x^2/4 =2x-3`

giving:`" "x^2/4-2x+3=0`

Using the standardised formula `y=ax^2+bx+c`

`a=1/4"; "b=-2"; "c=3`

where `x=(-b+-sqrt(b^2-4ac))/(2a)`

`=> x= (+2+-sqrt((-2)^2-4(1/4)(3)))/(2(1/4))`

`=> x=4 +-(sqrt(4-12/4))/(1/2)`

`x=4+- 2`

`color(brown)(x=2" and "x=6`

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`color(blue)("However:")`

Consider `sqrt(2x-3)`

For `x in RR: 2x>=3` otherwise `2x-3` is negative

`color(brown)(=>x>=3/2)`

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Consider `sqrt(3x-2)`

For `x in RR: 3x>=2` otherwise `3x-2` is negative

`color(brown)(=> x>=2/3)`
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`color(blue)("Putting it all together")`

`color(brown)("If "x<3/2" then at least one of the roots is negative")`

`color(brown)("As both "x=2" and "x=6" are greater than "3/2)``color(brown)("both are feasible solutions")`

`" "color(green)(bar(ul(|color(white)(2/2)x=2" and "x=6" "|))`
Tony B

Given
`color(white)("XXX")sqrt(3x-2)=1+sqrt(2x-3)`

Square both sides
`color(white)("XXX")3x-2=1+2sqrt(2x-3)+2x-3`

Simplify
`color(white)("XXX")3x-2=2sqrt(2x-3)+2x-2`

Isolate the term with the root on one side
`color(white)("XXX")x= 2sqrt(2x-3)`

Square both sides
`color(white)("XXX")x^2 = 8x-12`

Convert to standard form
`color(white)("XXX")x^2-8x+12=0`

Factor to get
`color(white)("XXX")(x-2)(x-6)=0`

`rArr x=2 or x=6`