SOLUTION: sqrt(x+2) + sqrt(3x+4) = 2

Algebra ->  Radicals -> SOLUTION: sqrt(x+2) + sqrt(3x+4) = 2      Log On


   



Question 162354: sqrt(x+2) + sqrt(3x+4) = 2
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B2%29+%2B+sqrt%283%2Ax%2B4%29+=+2
sqrt%28x%2B2%29+=+2+-+sqrt%283%2Ax%2B4%29
multiply both sides by sqrt%28x%2B2%29 to get
x%2B2+=+sqrt%28x%2B2%29%2A%282-sqrt%283%2Ax%2B4%29%29
since sqrt%28x%2B2%29+=+2+-+sqrt%283%2Ax%2B4%29
this equation becomes
x%2B2+=+%282+-+sqrt%283%2Ax%2B4%29%29%2A%282-sqrt%283%2Ax%2B4%29%29
which is the same as
x%2B2+=+%282-sqrt%283%2Ax%2B4%29%29%5E2
multiplying out on the right hand side of the equation and it becomes
x%2B2+=+4+-+4%2Asqrt%283%2Ax%2B4%29+%2B+%283%2Ax+%2B+4%29
this becomes
x%2B2+=+4+-+4%2Asqrt%283%2Ax%2B4%29+%2B+3%2Ax+%2B+4
which becomes
x%2B2+=+8+%2B+3%2Ax+-+4%2Asqrt%283%2Ax%2B4%29
subtracting 8 and 3*x from both sides of the equation and it becomes
-2%2Ax+-+6+=+-4%2Asqrt%283%2Ax%2B4%29
multiplying both sides of the eqution by (-1) and it becomes
2%2Ax%2B6=4%2Asqrt%283%2Ax%2B4%29
squaring both sides of the equation and it becomes
4%2Ax%5E2+%2B+24%2Ax+%2B+36+=+16%2A%283%2Ax%2B4%29
which becomes
4%2Ax%5E2+-+24%2Ax+%2B+36+=+48%2Ax+%2B+64
subtracting 48*x and 64 from both sides of the equation and it becomes
4%2Ax%5E2+-+24%2Ax+-+28+=+0
dividing both sides of the equation by 4 and it becomes
x%5E2+-+6%2Ax+-+7+=+0
this factors out into
%28x-7%29%2A%28x%2B1%29+=+0
x is either 7 or -1.
-----
substituting in original equation, x = 7 doesn't work, but x = -1 does.
answer is x = -1.