problem from evening class, I think its incorrect

Anthonyk2013

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maths 001.jpgWe did this problem in class yesterday evening, Is it incorrect, should the -8 be a -16?
 

wjm11

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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

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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

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−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).
 
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JeffM

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−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

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pka

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Anthonyk2013

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\(\displaystyle b\cdot b=b^2\) 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

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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 \(\displaystyle (b)^2\).
 
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