SOLUTION: Solve problem by completing the square, leaving the results in simplest radical form Check. I am confused with this one: 3xsquared - 6x =3 divide by 3 x squared - 3x = 9 x

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Question 231170: Solve problem by completing the square, leaving the results in simplest radical form Check.
I am confused with this one:
3xsquared - 6x =3
divide by 3
x squared - 3x = 9
x squared - 2x = 1 (Add one to both sides)......x squared -2x + 1 = 4
now I am lost....please help me! Thanks

Found 3 solutions by unlockmath, rapaljer, Theo:
Answer by unlockmath(1688)   (Show Source): You can put this solution on YOUR website!
Hello,
Check out when you added 1 to both sides on the last step.
Shouldn't it look like this on the last step you did above?
x(sq) - 2x + 1 = 2(not 4)
Now, let's factor x(sq)-2x + 1 = 2
It'll look like this.
(x - 1)sq = 2
Let's solve by sq rooting both sides.
Therefore: x - 1 = sq rt of 2
Add 1 to both sides gives us
x = 1 +/- sq rt of 2
RJ Toftness
Author "Unlock the Mystery to Math"
www.math-unock.com

Answer by rapaljer(4671)   (Show Source): You can put this solution on YOUR website!
You seem to have divided wrong!

3x^2 -6x=3

x^2-2x=1

Now add +1 to each side:
x^2-2x+1=1+1
(x-1)^2=2

Square root each side:


Add +1 to each side:


Dr. Robert J. Rapalje

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
Your equation is:

3x^2 - 6x = 3

You need to divide both sides of this equation by 3 to get:

x^2 - 2x = 1

You take 1/2 of 2 to get 1.

Your factors are (x-1)^2 = 1 + 1

This becomes (x-1)^2 = 2

Take the square root of this to get:

x-1 = +/- sqrt(2)

Add 1 to both sides to get:

x = 1 +/- sqrt(2)

This means that:

x = 1+sqrt(2)
or:
x = 1-sqrt(2)

To see if this works out ok, you need to substitute in the original equation to see if the equation is true.

Your original equation is 3x^2 - 6x = 3

If x = 1 + sqrt(2), this becomes:

3 * (1+sqrt(2)^2) - 6 * (1+sqrt(2)) = 3

This becomes:

3 + 6*sqrt(2) + 6 - 6 - 6*sqrt(2) = 3

6 and 6*sqrt(2) cancel out leaving 3 = 3

This is true so the value for x is good.

I'll leave you to replace x with 1 - sqrt(2) to see if that's good too.

You might want to check out the following website to review how to complete the square.

Completing the Square (algebrahelp)


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