Question 146049
<pre><font size = 4 color = "indigo"><b>
 {{{sqrt(2x-3)+x=3}}}

Isolate the radical by adding {{{-x}}} to both sides:

 {{{sqrt(2x-3)+x-x=3-x}}}

 {{{sqrt(2x-3)=3-x}}}

Now square both sides:

 {{{(sqrt(2x-3))^2=(3-x)^2}}}

On the left side, to square a square root takes 
away the radical and the exponent. We write the
right side as {{{(3-x)(3-x)}}}

 {{{2x-3=(3-x)(3-x)}}}

Use FOIL on the right side

{{{2x-3=9-3x-3x+x^2}}}

{{{2x-3=9-6x+x^2}}}

See if you can go from there to:

{{{x^2-8x+12=0}}}

Then we factor the left side:

{{{(x-6)(x-2)=0}}}

Use the zero factor principle,
and set each factor = 0:

{{{x-6=0}}} gives {{{x=6}}}

{{{x-2=0}}} gives {{{x=2}}}

But we must check both solutions in the
original equations, because in radical 
equations, there may be extraneous (bogus) 
solutions.

Checking {{{x=6}}} in the original equation:

 {{{sqrt(2x-3)+x=3}}}
 {{{sqrt(2(6)-3)+(6)=3}}}
 {{{sqrt(12-3)+6=3}}}
 {{{sqrt(9)+6=3}}}
 {{{3+6=3}}}
 {{{9=3}}}

That is false, so {{{x=6}}} is not a solution
to the original equation.

Checking {{{x=2}}} in the original equation:

 {{{sqrt(2x-3)+x=3}}}
 {{{sqrt(2(2)-3)+(2)=3}}}
 {{{sqrt(4-3)+2=3}}}
 {{{sqrt(1)+2=3}}}
 {{{1+2=3}}}
 {{{3=3}}}

That is true, so {{{x=2}}} is the only solution
to the original equation.

---------------------

If you look at the graph of the left side of 

{{{sqrt(2x-3)+x=3}}}

that is, plot {{{y=sqrt(2x-3)+x}}}

you get this curve:

{{{drawing(400,400,-2,7,-2,7,

graph(400,400,-2,7,-2,7,sqrt(2x-3)+x) )}}}

and if you plot the right side of

{{{sqrt(2x-3)+x=3}}}


that is, plot {{{y=3}}}

on the same graph you see that the only

solution is the x-coordinate of the point

where they intersect. 

{{{drawing(400,400,-2,7,-2,7,
locate(2,3,"(2,3)"), 
graph(400,400,-2,7,-2,7,sqrt(2x-3)+x,3) )}}}


They intersect only at (2,3). The x-coordinate
of that point is 2, so {{{x=2}}} is the only
solution to the system.  But we wonder why we
got the extraneous solution.

We now notice that the curve of the left
side is really part of a slanted parabola,
and if we were to draw in the rest of the 
parabola, that is, the part in light blue 
below: 

{{{drawing(400,400,-2,7,-2,7,
locate(2,3,"(2,3)"), locate(6,3,"(6,3)"),
graph(400,400,-2,7,-2,7,sqrt(2x-3)+x,3),

graph(400,400,-2,7,-2,7,0,0,0,0,0,-sqrt(2x-3)+x) )}}}

we can see that the other part of that parabola
does indeed intersect the graph of the right side,
(the green line) at the point (6,3).  But we 
discard {{{x=6}}} because the curve of the left side
of the original equation does not include the light
blue part of the parabola.

Edwin</pre>