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OpenStudy (anonymous):
sqrt3x+16=x+2 What is the solution set?
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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
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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.
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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?
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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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