Is this possible? "If 3x^2 - 2x + 7 = 0, then find value of (x - 1/3)^2."

RVic

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Nov 21, 2017
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I will be taking the Accuplacer test at my school in December. I was given a practice test at my school's testing center and this is one of the questions:

If 3x^2 - 2x + 7 = 0, then (x - 1/3)^2 =

A. 20/9
B. 7/9
C. -7/9
D. -8/9
E. -20/9

By using the quadratic formula I get A. 20/9. However, in the answer sheets it lists that the answer is E. -20/9. Not only do I think it's wrong, but I also think it's impossible to get a negative answer out of (x - 1/3)^2 since I understand it to be the case that any exponent that's a multiple of 2 will always give you a positive answer.

Am I off base here, guys? I appreciate any comments and people shipping in with what they think of the problem and answer situation.
 

mmm4444bot

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If 3x^2 - 2x + 7 = 0, then (x - 1/3)^2 =

… in the answer sheets it lists that the answer is -20/9. Not only do I think it's wrong, but I also think it's impossible to get a negative answer out of (x - 1/3)^2 since I understand it to be the case that any exponent that's a multiple of 2 will always give you a positive answer.
I get -20/9.

It is possible for even powers to produce negative products when imaginary numbers are involved.

You did not show your work, so I cannot comment on what you did wrongly.

Did you expand the expression (x - 1/3)^2 first?

What are your solutions for the quadratic equation?
 

tkhunny

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How did you pick which solution for x that you would use? It doesn't matter. I just wondered.

Please apply "Completing the Square" to the expression on the LeftHandSide. Start by dividing by 3.
 

RVic

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How did you pick which solution for x that you would use? It doesn't matter. I just wondered.

Please apply "Completing the Square" to the expression on the LeftHandSide. Start by dividing by 3.
You guys are right. I avoided using the negative solution because I kept getting 'DOMAIN ERROR' on my Scientific calculator. So I would just use the positive one and matching the decimal it gave me with its next approximate which was choice A (20/9). I really need to study imaginary numbers it seems.
 

RVic

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I get -20/9.

It is possible for even powers to produce negative products when imaginary numbers are involved.

You did not show your work, so I cannot comment on what you did wrongly.

Did you expand the expression (x - 1/3)^2 first?

What are your solutions for the quadratic equation?

You are right, I'm sorry.
I will now show you my most recent work:

First, I replace a, b and c with the numbers given in the quadratic expression to get:


x = 2 + or - √-2^2 - 4(3)(7)/2(3)


Then, I do order of operation to simplify to:

x = 2 + or - √-80/6

But since I kept getting 'DOMAIN ERROR' on the negative value, I acted the opposite of mathematically and must have changed some things around.

I still don't know where to go from there.
 

JeffM

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Sep 14, 2012
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I will be taking the Accuplacer test at my school in December. I was given a practice test at my school's testing center and this is one of the questions:

If 3x^2 - 2x + 7 = 0, then (x - 1/3)^2 =

A. 20/9
B. 7/9
C. -7/9
D. -8/9
E. -20/9

By using the quadratic formula I get A. 20/9. However, in the answer sheets it lists that the answer is E. -20/9. Not only do I think it's wrong, but I also think it's impossible to get a negative answer out of (x - 1/3)^2 since I understand it to be the case that any exponent that's a multiple of 2 will always give you a positive answer.

Am I off base here, guys? I appreciate any comments and people shipping in with what they think of the problem and answer situation.
You don't tell us what answer you got from the quadratic formula or how you got it. So it is impossible to tell where your train flew off the tracks.

\(\displaystyle 3x^2 - 2x + 7 = 0 \implies x = \dfrac{-\ (-\ 2) \pm d}{2 * 3} = \dfrac{2 \pm d}{6} \text {, where}\)

\(\displaystyle d = \sqrt{(-\ 2)^2 - 4 * 3 * 7} = \sqrt{4 - 84} =\sqrt{-\ 80} = \sqrt{-\ 1} * \sqrt{16} * \sqrt{5} = 4i\sqrt{5}.\)

\(\displaystyle \therefore x = \dfrac{2 \pm 4i\sqrt{5}}{6} = \dfrac{1 \pm 2i\sqrt{5}}{3}.\)

Now finish it up.
 

mmm4444bot

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Make sure that you're using a scientific calculator programmed to handle the square root of negative one (i.e., the imaginary unit). You may need to switch modes, too, in order to enable complex numbers. :cool:
 
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