Question 260291
Here is the original problem:
{{{sqrt(3x) + sqrt(12) = (x+5) / sqrt(3)}}}
--
I don't know if the first x is in or outside of the square root. Lets first assume in and see where it takes us.
step 1 - multiply evreything by sqrt(3) to get
{{{sqrt(3)*sqrt(3x) + sqrt(12)*sqrt(3) = x+5}}}
step 2 - simplify the square roots to get
{{{3sqrt(x) + 6 = x+5}}}
step 3 - subtract 6 from both sides to get
{{{3sqrt(x) = x-1}}}
step 4 - divide by 3 to get
{{{sqrt(x) = (1/3)x - (1/3)}}}
step 5 - square both sides to get
{{{x = (1/9)x^2 -(2/9)x + 1/9}}}
step 6 - multiply everything by 9 to get
{{{9x = x^2 -2x + 1}}}
step 7 - set = 0 and factor
{{{0 = x^2 -11x + 1}}}
by quadratic, we get
{{{x = (11 +- sqrt( 121-4*1*1 ))/(2) }}}
then
{{{x = (11 +- sqrt( 117 ))/(2) }}}
and then
{{{x = (11 +- 3sqrt( 13 ))/(2) }}}
--
Now, suppose the x is outside the square root. We get
{{{x*sqrt(3) + sqrt(12) = (x+5) / sqrt(3)}}}
step 1 - multiply evreything by sqrt(3) to get
{{{x*sqrt(3)*sqrt(3) + sqrt(12)*sqrt(3) = x+5}}}
step 2 - simplify the square roots to get
{{{3x + 6 = x+5}}}
step 3 - subtract x from both sides to get
{{{2x + 6 = 5}}}
step 4 - subtract 6 from both sides to get
{{{2x = -1}}}
divide by 2 to get
{{{x = -1/2}}}