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Mathematics 10 Online
OpenStudy (anonymous):

sqrt3x+16=x+2 What is the solution set?

OpenStudy (saifoo.khan):

\[\sqrt{3x+16} = x+ 2\]If im not wrong..

OpenStudy (saifoo.khan):

x = 3

Directrix (directrix):

sqrt(3x+16)=x+2 OR sqrt(3x)+16=x+2 ? Cuddlepony -- which one?

OpenStudy (anonymous):

(2)^(3x + 16)^(1/2) = (x +2)^(2) = 3x + 16 = x^(2) + 4x + 4 = 0 = x^(2) + x - 12 = (x + 4)(x-3) so you are wrong thus -4 and 3

OpenStudy (anonymous):

sorry I messed up notation but yeah I just squared both sides of the equation

OpenStudy (anonymous):

to elimanate the radical

OpenStudy (anonymous):

How should I know I wouldn't ask a question like this

OpenStudy (anonymous):

@Directrix

OpenStudy (anonymous):

or you are right and I mis interpreted the question meh

OpenStudy (anonymous):

only 3x+16 is the sqrt.

OpenStudy (saifoo.khan):

Then the answer's 3.

OpenStudy (anonymous):

x=-4 is an extraneous solution. It doesn't work. (Try plugging it in.)

OpenStudy (anonymous):

Oh sorry I didnt complete the question by taking the domain

OpenStudy (anonymous):

You can't take the square root of [0, -infinity)

OpenStudy (anonymous):

did you have to use the quadratic equation?

OpenStudy (anonymous):

After you square both sides to clear the radical, you will almost always have a quadratic equation on these kinds of problems.

OpenStudy (anonymous):

ok, thanks so much!

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