Completing the Square

[szcoup]

If someone could point out where I've gone wrong, I would appreciate it.

9x^2 - 18x + 3 = 0
x^2 - 2x + 1/3 = 0
x^2 - 2x = -1/3
x^2 - 2x + 1 = 1 - 1/3
(x-1)^2 = 2/3
x - 1 = square root 2/3
x = 1 plus/minus square root 2/3

The answer should be 1 plus/minus the square root 6 over 3.
Where did I go wrong? Thanks.

 

[mmm4444bot]

Your result is also correct, but you did not rationalize the denominator.

Multiply the radical by sqrt(3)/sqrt(3).

Everything else looks good. Cheers :cool:

 

[szcoup]

Thanks!

 

[mmm4444bot]

Here's a third form of the solution:

x = [3 ± sqrt(6)]/3

Of course, students should turn in an answer that meets their instructor's expectation. :cool:

 

[lookagain]

If someone could point out where I've gone wrong, I would appreciate it.

9x^2 - 18x + 3 = 0
x^2 - 2x + 1/3 = 0
x^2 - 2x = -1/3
x^2 - 2x + 1 = 1 - 1/3
(x-1)^2 = 2/3


x - 1 = square root 2/3 No, this must be plus/minus square root (2/3) *


x = 1 plus/minus square root 2/3

The answer should be 1 plus/minus the square root 6 over 3.
Where did I go wrong? Thanks.

** The intended point is that you must have the plus-or-minus symbol at this stage also.

\(\displaystyle x - 1 \ = \ \pm\sqrt{\dfrac{2}{3}}\)

 

[mmm4444bot]

Ah -- good catch, lookagain.

Readers, note also the grouping symbols around the radicand.

 

[szcoup]

Great! It makes complete sense now. Thanks!