#### Anthonyk2013

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- Thread starter Anthonyk2013
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No, it should not be -16 because the -4 in the prior step is wrong; it should be positive 4: (-2)^2 = 4.View attachment 3356We did this problem in class yesterday evening, Is it incorrect, should the -8 be a -16?

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.

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

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.

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

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

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

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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).

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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Because as you yourself said, b = - 2, and b^2 = b * b = (-2) * (-2).−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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So it's b*b not b squared ?Because as you yourself said, b = - 2, and b^2 = b * b = (-2) * (-2).

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\(\displaystyle b\cdot b=b^2\) and \(\displaystyle -2\cdot -2=4=(-2)^2\)So it's b*b not b squared ?

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

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That is very complicated question. To understand the answer one must understand the logic of programming languages.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.

However, in mathematics it is true that \(\displaystyle -2^2=-4\).

But if \(\displaystyle b=-2\) then \(\displaystyle b^2=4\).

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 \(\displaystyle (b)^2\).