problem from evening class, I think its incorrect

[Anthonyk2013]

We did this problem in class yesterday evening, Is it incorrect, should the -8 be a -16?

 

[wjm11]

View attachment 3356We did this problem in class yesterday evening, Is it incorrect, should the -8 be a -16?

No, it should not be -16 because the -4 in the prior step is wrong; it should be positive 4: (-2)^2 = 4.

Also, I would suggest that you not mix fractions and decimals in your answers (the 1/3 and the .471). I'd suggest leaving everything in fractions (which are exact), rather than decimals that are approximations -- unless otherwise instructed.

 

[JeffM]

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.

$\displaystyle 3\left(\dfrac{1 + i\sqrt{2}}{3}\right)^2 - 2 * \dfrac{1 + i\sqrt{2}}{3} + 1 = 3 * \dfrac{1 + 2i\sqrt{2} + 2(-1)}{9} - \dfrac{2 + 2i\sqrt{2}}{3} + \dfrac{3}{3} =$

$\displaystyle \dfrac{1 + 2i\sqrt{2} - 2}{3} - \dfrac{2 + 2i\sqrt{2}}{3} + \dfrac{3}{3} =$

$\displaystyle \dfrac{1 + 3 - 2 - 2 + 2i\sqrt{2} - 2i\sqrt{2}}{3} = \dfrac{0}{3} = 0.$

 

[Anthonyk2013]

−22=−(2∗2)=−4 whereas (−2)2=(−2)∗(−2)=4.


So the discriminant should have been 4−12−−−−−√, not −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,



Where I'm confused is the -(2*2), why is it (-2)*(-2) and not -(2*2).

 

[JeffM]

−22=−(2∗2)=−4 whereas (−2)2=(−2)∗(−2)=4.


So the discriminant should have been 4−12−−−−−√, not −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,



Where I'm confused is the -(2*2), why is it (-2)*(-2) and not -(2*2).

Because as you yourself said, b = - 2, and b^2 = b * b = (-2) * (-2).

 

[Anthonyk2013]

Because as you yourself said, b = - 2, and b^2 = b * b = (-2) * (-2).

So it's b*b not b squared ?

 

[pka]

So it's b*b not b squared ?

$\displaystyle b\cdot b=b^2$ and $\displaystyle -2\cdot -2=4=(-2)^2$

 

[Anthonyk2013]

$\displaystyle b\cdot b=b^2$ and $\displaystyle -2\cdot -2=4=(-2)^2$

Why is my calculator giving me a value of -4 ? I understand two minuses give a positive sign but -2 squared on my calculator is -4.

 

[pka]

Why is my calculator giving me a value of -4 ? I understand two minuses give a positive sign but -2 squared on my calculator is -4.

That is very complicated question. To understand the answer one must understand the logic of programming languages.
However, in mathematics it is true that $\displaystyle -2^2=-4$.
But if $\displaystyle b=-2$ then $\displaystyle b^2=4$. That is confusing is it not?

The first is the result of the order of operations.
On your calculator enter $\displaystyle (-2)^2$ and see what you get.
On your calculator enter $\displaystyle -2^2$ and see what you get.

So if you have $\displaystyle b=-3$ and you need $\displaystyle b^2$ then enter $\displaystyle (b)^2$.