Solve using the quadratic formula: 3x^2 = 6x - 2

3x^2=6x-2
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First rearrange to get 0 on one side:

3x^2 – 6x + 2 = 0

Identify a, b, and c:

a = 3
b = -6
c = 2

Can you take it from there?
 
Here is the formula: \(\displaystyle \L
\,x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}\)
 
Yes I have went that far, let me show you what I have done.

3x^2=6x-2
3x^2-6x+2=0
A=3 B=-6 C=2

-(-6)+or-square root of -6^2-4,3,2 divided by 6=0

6+or- square root of 36-24 divided by 6=0

6+or- square root of 12 divided by 6=0

6+or-2 square root of 3 divided by 6=0

final answer is

1+or- 1/3 times the square root of 3

but my book says the answer is

3+or- square root of 3 divided by 3
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Good work, Mikeh,

(Excet that all those "...divided by 6=0" should read "...divided by 6=x". Also, please show all your work as Replies to the original question so that others can assist/benefit from them, rather than using a PM.)

You got: 1 +/- (sqrt3)/3

Your book said: [3 +/- (sqrt3)]/3

Notice any similarities? (They're the same.)

Note: For future reference, you can also show a square root as a number raised the the 1/2 power:

sqrt3 = 3^(1/2) = 3^(.5)
 
I got: 1+/- (1/3) (sqrt 3)

I don't understand that it is the same????
 
Mikeh said:
I got: 1+/- (1/3) (sqrt 3)

I don't understand that it is the same????
Click to expand...

What is 3 ÷ 3?

What is sqrt[3] ÷ 3?

Eliz.
 
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