View attachment 3356We did this problem in class yesterday evening, Is it incorrect, should the -8 be a -16?
I have difficulty reading your writing, but it appears to me that you have three errors in your solution, but two of them cancel out.
Solve for x: \(\displaystyle 3x^2 - 2x + 1 = 0.\)
You properly identified the quadratic formula as the way to solve and you correctly identified the parameters for the formula
a = 3, b = - 2, and c = 1.
But you entered those parameters into the formula incorrectly.
\(\displaystyle x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 * 3 * 1}}{2 * 3} \ne \dfrac{-(-2) \pm \sqrt{-2^2 - 4 * 3 * 1}}{2 * 3}.\)
\(\displaystyle -2^2 = - (2 * 2) = - 4\ whereas\ (-2)^2 = (-2) * (-2) = 4.\)
So the discriminant should have been \(\displaystyle \sqrt{4 - 12},\ not\ \sqrt{-4 - 12}.\)
But then your luck held: you added - 4 and - 12 and got - 8 instead of the correct - 16. But you should have been adding + 4 and - 12,
which does give - 8.
After this I am not exactly sure what you did because I can't decipher your writing, but I suspect you made a mistake.
\(\displaystyle x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 * 3 * 1}}{2 * 3} = x = \dfrac{2 \pm \sqrt{4 - 12}}{6} = \dfrac{2 \pm \sqrt{- 8}}{6}= \dfrac{2 \pm \sqrt{- 2 * 4}}{6} = \dfrac{2 \pm 2\sqrt{- 2}}{6} = \dfrac{1 \pm i\sqrt{2}}{3}.\)
Now CHECK your answer. Let's use the plus answer to cut down on minus signs.
\(\displaystyle 3\left(\dfrac{1 + i\sqrt{2}}{3}\right)^2 - 2 * \dfrac{1 + i\sqrt{2}}{3} + 1 = 3 * \dfrac{1 + 2i\sqrt{2} + 2(-1)}{9} - \dfrac{2 + 2i\sqrt{2}}{3} + \dfrac{3}{3} =\)
\(\displaystyle \dfrac{1 + 2i\sqrt{2} - 2}{3} - \dfrac{2 + 2i\sqrt{2}}{3} + \dfrac{3}{3} = \)
\(\displaystyle \dfrac{1 + 3 - 2 - 2 + 2i\sqrt{2} - 2i\sqrt{2}}{3} = \dfrac{0}{3} = 0.\)