Odd factoring in textbook

• Oct 25th 2009, 03:43 PM
hibernaldream
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!

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

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

The way I did it in the first place was:
\$\displaystyle 3x^2 - 12x = -11\$
\$\displaystyle 3x(x - 4) = -11\$
Therefore
\$\displaystyle x = -11 / 3\$
\$\displaystyle 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.
• Oct 25th 2009, 03:53 PM
skeeter
Quote:

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!

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

\$\displaystyle (6 + sqrt(3)) / 3 \$
\$\displaystyle (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.
\$\displaystyle 3x^2 - 12x = -11\$
\$\displaystyle 3x(x - 4) = -11\$
Therefore
\$\displaystyle x = -11 / 3\$
\$\displaystyle x = -7\$

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

...
• Oct 25th 2009, 03:57 PM
hibernaldream
Ah. I forgot that existed. I took algebra five years ago, hence the confusion. Thanks.