completeing the sqaure

nikkiNEEDShelp

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Jan 9, 2006
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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.
 
nikkiNEEDShelp said:
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?
 
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
 
nikkiNEEDShelp said:
okay so you have the x^2-2x+1=5/3
No: You have x<sup>2</sup> - 2x + 1 = 5/3 + 1.

nikkiNEEDShelp said:
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.

nikkiNEEDShelp said:
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.
 
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