SOLUTION: 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 sid

Algebra ->  Equations -> SOLUTION: 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 sid      Log On


   

Question 697088: solve the following algebraically:
x-3=+sqrt%28x%5E2%2B3x%29
my work:
%28x-3%29%5E2=+sqrt+%28x%5E2%2B3x%29%5E2
x%5E2-18x%2B9=x%5E2%2B3x

Then i subtracted x%5E2%2B3x from both sides and got:
-21x+9=0 and pulled out a 3 and got 3(7x+3) as my final answer.
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

solve the following algebraically:
x-3=+sqrt%28x%5E2%2B3x%29
my work:
%28x-3%29%5E2=+sqrt+%28x%5E2%2B3x%29%5E2....perfect, you did it right
x%5E2-18x%2B9=x%5E2%2B3x....perfect

Then i subtracted x%5E2%2B3x from both sides and got:
-21x%2B9=0 and pulled out a 3 and got 3%287x%2B3%29=0 ...just finish this and solve for x
3%287x%2B3%29=0 ...this product will be equal to zero if and only if
7x%2B3=0; so,
7x=-3
x=-3%2F7....or as decimal x=-0.43


Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!
solve the following algebraically:
x-3=+sqrt%28x%5E2%2B3x%29
my work:
%28x-3%29%5E2=+sqrt+%28x%5E2%2B3x%29%5E2
x%5E2-18x%2B9=x%5E2%2B3x

Then i subtracted x%5E2%2B3x 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%28x%5E2+%2B+3x%29

%28x+-+3%29%5E2+=+%28sqrt%28x%5E2+%2B+3x%29%29%5E2 ---- Correct up to this point

x%5E2+-+18x+%2B+9+=+x%5E2+%2B+3x ----- Incorrect

s/b: x%5E2++-+6x+%2B+9+=+x%5E2+%2B+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.