Question 162354
{{{sqrt(x+2) + sqrt(3*x+4) = 2}}}
{{{sqrt(x+2) = 2 - sqrt(3*x+4)}}}
multiply both sides by {{{sqrt(x+2)}}} to get
{{{x+2 = sqrt(x+2)*(2-sqrt(3*x+4))}}}
since {{{sqrt(x+2) = 2 - sqrt(3*x+4)}}}
this equation becomes
{{{x+2 = (2 - sqrt(3*x+4))*(2-sqrt(3*x+4))}}}
which is the same as
{{{x+2 = (2-sqrt(3*x+4))^2}}}
multiplying out on the right hand side of the equation and it becomes
{{{x+2 = 4 - 4*sqrt(3*x+4) + (3*x + 4)}}}
this becomes
{{{x+2 = 4 - 4*sqrt(3*x+4) + 3*x + 4}}}
which becomes
{{{x+2 = 8 + 3*x - 4*sqrt(3*x+4)}}}
subtracting 8 and 3*x from both sides of the equation and it becomes
{{{-2*x - 6 = -4*sqrt(3*x+4)}}}
multiplying both sides of the eqution by (-1) and it becomes
{{{2*x+6=4*sqrt(3*x+4)}}}
squaring both sides of the equation and it becomes
{{{4*x^2 + 24*x + 36 = 16*(3*x+4)}}}
which becomes
{{{4*x^2 - 24*x + 36 = 48*x + 64}}}
subtracting 48*x and 64 from both sides of the equation and it becomes
{{{4*x^2 - 24*x - 28 = 0}}}
dividing both sides of the equation by 4 and it becomes
{{{x^2 - 6*x - 7 = 0}}}
this factors out into
{{{(x-7)*(x+1) = 0}}}
x is either 7 or -1.
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substituting in original equation, x = 7 doesn't work, but x = -1 does.
answer is x = -1.