Odd factoring in textbook

hibernaldream · #1 $Old$ October 25th, 2009, 05:43 PM

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.

skeeter · #2 $Old$ October 25th, 2009, 05:53 PM

Quote:

Originally Posted by hibernaldream $View Post$

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.

...

hibernaldream · #3 $Old$ October 25th, 2009, 05:57 PM

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