completeing the sqaure

[nikkiNEEDShelp]

3x^2-6x=5

I need help with completing the square...I think that you have to bring the 5 over to the left side, but after that I'm clueless.

 

[Denis]

http://www.purplemath.com/modules/sqrquad.htm

 

[tkhunny]

3x^2-6x=5

I need help with completing the square...I think that you have to bring the 5 over to the left side, but after that I'm clueless.

You can move the 5 if you wish, but is that your goal?

3x^2-6x=5

Perhaps divide by 3, first.

x^2 - 2x = 5/3

Now, the meat of the issue

-2/2 = -1
(-1)^2 = 1

x^2 - 2x + 1 = 5/3 + 1

Now what?

 

[nikkiNEEDShelp]

okay so you have the x^2-2x+1=5/3

now is this what you would do?

sqrt (x-1)^2=5/3+1
x-1=sqrt8/3
x=1 +or- sqrt 8/3

Is that the final answer, I also need to get the same answer using the quadratic formula which I know but I haven't been able to get the same answer

 

[stapel]

okay so you have the x^2-2x+1=5/3

No: You have x² - 2x + 1 = 5/3 + 1.

now is this what you would do?
sqrt (x-1)^2=5/3+1

No: You can't take the square root of only the one side of the equation.

Note: When you solve "by square roots", you have to put "plus/minus" on one side of the equation. Also, you cannot leave a radical in the denominator. To get a properly-formatting answer, you need to rationalize the denominator. Then convert the "1" and the radical-containing term to common-denominator fractions, and combine. In this way, you can put your "completing the square" answer in "Quadratic Formula" format.

I also need to get the same answer using the quadratic formula which I know

What answer have you obtained by using the Formula?

Thank you.

Eliz.