SOLUTION: I need help on my algebra I have to find the roots of an equation and this problem has me stumped 3x^2=48

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Question 382645: I need help on my algebra I have to find the roots of an equation and this problem has me stumped 3x^2=48
Found 4 solutions by rfer, solver91311, kingme18, mananth:
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
first divide by 3
3x^2=48
x^2=16
sqrt 16
x=4

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Multiply both sides by

Take the square root of both sides. Remember to consider both the positive and negative root.

John

My calculator said it, I believe it, that settles it
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Answer by kingme18(98) About Me  (Show Source):
You can put this solution on YOUR website!
3%2Ax%5E2=48
If this just said 3%2Ax=48, you would divide by 3 on both sides to solve for x. You're going do the same here: when you divide by 3 on both sides, you get x%5E2=16.
To get rid of squaring, you take the square root on both sides. Don't forget that this means the answer could be positive or negative!
x=+%2B-+sqrt%2816%29
x=+%2B-+4
The roots are the answers: x=4 and x=-4

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
+3x%5E2=+48
...
/3
+x%5E2=+48%2F3
...
x%5E2=16
...
take the square root
x= +/- 4
...
m.ananth@hotmail.ca