# Thread: Odd factoring in textbook

1. ## Odd factoring in textbook

Though this is for a calculus problem, I've gotten all the calc worked out. I just don't understand how the book came to this conclusion after factoring!

$3x^2 - 12x + 11 = 0$
The answer the book comes to is

$(6 + sqrt(3)) / 3$
$(6 - sqrt(3)) / 3$

The way I did it in the first place was:
$3x^2 - 12x = -11$
$3x(x - 4) = -11$
Therefore
$x = -11 / 3$
$x = -7$

What were they doing in my textbook? Unfortunately they give no clues and I've been staring at it for 15 minutes now.

2. Originally Posted by hibernaldream
Though this is for a calculus problem, I've gotten all the calc worked out. I just don't understand how the book came to this conclusion after factoring!

$3x^2 - 12x + 11 = 0$
The answer the book comes to is

$(6 + sqrt(3)) / 3$
$(6 - sqrt(3)) / 3$

The way I did it in the first place was: this is not the correct method to solve a quadratic equation.
$3x^2 - 12x = -11$
$3x(x - 4) = -11$
Therefore
$x = -11 / 3$
$x = -7$

What were they doing in my textbook?
they used the quadratic formula.
...

3. Ah. I forgot that existed. I took algebra five years ago, hence the confusion. Thanks.