Question 697088
solve the following algebraically:

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

my work:

{{{(x-3)^2= sqrt (x^2+3x)^2}}}
{{{x^2-18x+9=x^2+3x}}}
 
Then i subtracted {{{x^2+3x}}} from both sides and got:

-21x+9=0 and pulled out a 3 and got 3(7x+3) as my final answer. 


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


{{{(x - 3)^2 = (sqrt(x^2 + 3x))^2}}} ---- Correct up to this point


{{{x^2 - 18x + 9 = x^2 + 3x}}} ----- Incorrect


s/b: {{{x^2  - 6x + 9 = x^2 + 3x}}}


After solving this, you should get: x = 1, but this makes the left-side negative, and, as the square root of an expression CANNOT equal a negative value, an EXTRANEOUS solution exists. Therefore, there is NO SOLUTION.