4-3x^2=0; x=?

Question

I think I'm doing something wrong...

4-3x^2=0
-3x^2=-4
x^2 = 4/3
x= +/- sqrt(4/3)

I think the correct answer is: +/- (2*sqrt(3))/3

What am I doing wrong?

Answer 1

What you have in your answer is:

sqrt(4/3) and that is the same as sqrt4/sqrt3

But, you should not have a sqrt on the denominator (a stylistic thing).

So what you do to remove the root is multiply the fraction sqrt4/sqrt3 by sqrt3/sqrt3 (essentially you are multiplying by one because the numerator=denominator)

So by multiplying by sqrt3/sqrt3 you get [(sqrt4)(sqrt3)] / [(sqrt3)(sqrt3)]

So simplify the denominator to 2 * sqrt3
Simplify the denominator to 3.

Your answer is (2 * sqrt3)/3

Answer 2

3.x² = 4
x ² = 4 / 3
x = ± 2 / √3
x = ± 2.√3 / 3

Answer 3

You need to rationalize the denominator. The 4 and 3 both have squareroots on them. Squareroot of 4 is 2 because 2 times 2 is for. But you have to get rid of the squareroot of 3...

Answer 4

You are right, you just didn't rationalize the denominator:
√(4/3) = 2√(1/3) = (2√(3))/3

Answer 5

sqrt(4/3) = 2/sqrt(3) = 2 * sqrt(3) / 3

They're equivalent answers, really.

Answer 6

4-3x^2=0
-4 -4

-3x^2=-4
_______
-3

square root (x^2=4/3)

x=1.1547

Hope this helped!

Answer 7

you are not wrong..

the square root of (4/3) is 2/3 root 3....plus or minus of course, which is your answers, just rationalized..

Answer 8

you cant have a square root on the bottom of the fraction

Answer 9

october 26th