Question 179343
Can you please help me solve this radical equaqtion? {{{sqrt(x-1)+x=3}}}
This problem did not come from a text book, it came from a worksheet.

I began by isolating the radical:  {{{sqrt(x-1)=-x+3}}}
Then I squared each side to eliminate the radical:  {{{sqrt(x-1)^2=(-x+3)^2}}}
Which gave me:  {{{ x-1=-x^2+9 }}}
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There's your error!

{{{(-x+3)^2}}} IS NOT {{{-x^2+9}}}!
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When there are two terms added to be squared, you must
write it in parentheses TWICE and multiply it by itself, 
using FOIL. Also you must remember that when you square 
a negative, such as {{{(-x)}}} you get a positive {{{""+x^2}}},
not {{{-x^2}}}:

Therefore when you got to

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

the next step should have been

{{{x-1=(-x+3)(-x+3)}}}

Then you must use "FOIL":

{{{x-1=(-x)(-x) + (-x)(3) + (3)(-x) + (3)(3)}}}

{{{matrix(1,3,x-1, "=",""+x^2 - 3x - 3x + 9)}}}

You don't need to write that {{{""+""}}}. I just
put it there to remind you that when you square
a negative you get a positive.

{{{matrix(1,3,x-1,"=",x^2-3x-3x+9)}}}

{{{x-1=x^2-6x+9}}}

Then move everything right and you get:

{{{0=x^2-7x+10}}}

or if you like the 0 on the right, just swap sides:

{{{x^2-7x+10=0}}}

Then it factors nicely as

{{{(x-5)(x-2)=0}}}

Then you get the solutions {{{x=5}}} and {{{x=2}}}

But you must check these, because when you square
both sides or multiply both sides by a variable,
you may get extraneous solutions:

Checking {{{x=5}}}
{{{sqrt(x-1)+x=3}}}
{{{sqrt(5-1)+5=3}}}
{{{sqrt(4)+5=3}}}
{{{2+5=3}}}
{{{7=3}}}

That doesn't check, so {{{x=5}}} is not a 
genuine solution, but is called "extraneous".

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

That checks. So {{{x=2}}} is the only solution

Edwin</pre>