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If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$

I am so confused. It is a self taught algebra book.

The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.

Please explain.

Thanks for everyone who commented! I understand it now.

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    $\begingroup$ Hint: Multiply out $(x-1/3)^2$ and compare it to the original equality. $\endgroup$ Aug 16, 2015 at 17:52
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    $\begingroup$ You can learn from the two different kinds of solutions below. One strategy is to complete the square and hope to be lucky. That would occur to you if you had some experience knowing when and how to complete the square. The other is to start by multiplying out the sought for square, hoping to find something useful. That's a good strategy if nothing else comes to mind. $\endgroup$ Aug 16, 2015 at 19:50
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    $\begingroup$ Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ Aug 17, 2015 at 9:11

8 Answers 8

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Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+\frac{1}{3}+7-\frac{1}{3}=0$$ $$3\left(x^2-\frac{2x}{3}+\frac{1}{9}\right)+7-\frac{1}{3}=0$$ $$3\left(x-\frac{1}{3}\right)^2+\frac{21-1}{3}=0$$ $$3\left(x-\frac{1}{3}\right)^2=-\frac{20}{3}$$ $$\left(x-\frac{1}{3}\right)^2=-\frac{20}{9}$$ Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\left(x-\frac{1}{3}\right)^2=\color{blue}{-\frac{20}{9}}}}$$

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    $\begingroup$ As is, the first step is not math, it is "magic". There is a "why" missing there. $\endgroup$ Aug 16, 2015 at 22:40
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    $\begingroup$ @Rol looks like my edit went through that added the missing step so it's not as magical now $\endgroup$ Aug 17, 2015 at 2:05
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    $\begingroup$ Yes, you are right $\endgroup$ Aug 17, 2015 at 2:14
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    $\begingroup$ @Harish yes. But I still think Burr's answer is clearer because it it is clearer that we are aiming to get to (x-1/3)^2 $\endgroup$ Aug 18, 2015 at 2:08
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    $\begingroup$ (+1) But if roots were assumed real, this problem made no sense. $\endgroup$
    – Amad27
    Oct 20, 2015 at 12:52
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Observe $$\left(x-\frac{1}{3}\right)^2=x^2-\frac{2}{3}x+\frac{1}{9}$$$$=\frac{1}{3}\left(3x^2-2x\right)+\frac{1}{9}.$$ This is almost the original expression, we're just missing a $7$. Then, $$\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}.$$ Now, use the original equality to simplify. Then, we get $$ \left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}$$ $$=-\frac{7}{3}+\frac{1}{9} $$ $$=-\frac{20}{9} $$

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$$3x^2-2x+7=0\Longleftrightarrow$$ $$x=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot 3 \cdot 7}}{2\cdot 3}\Longleftrightarrow$$ $$x=\frac{2\pm\sqrt{4-4\cdot 3 \cdot 7}}{6}\Longleftrightarrow$$ $$x=\frac{2\pm\sqrt{4-84}}{6}\Longleftrightarrow$$ $$x=\frac{2\pm\sqrt{-80}}{6}\Longleftrightarrow$$ $$x=\frac{2\pm i\sqrt{80}}{6}\Longleftrightarrow$$ $$x=\frac{2\pm 4i\sqrt{5}}{6}\Longleftrightarrow$$ $$x=\frac{2 + 4i\sqrt{5}}{6} \vee x=\frac{2 - 4i\sqrt{5}}{6}\Longleftrightarrow$$ $$x=\frac{2 + 4i\sqrt{5}}{6} \vee x=\frac{2 - 4i\sqrt{5}}{6}$$


$$\left(\left(\frac{2 + 4i\sqrt{5}}{6}\right)-\frac{1}{3}\right)^2 =\left(\frac{2i\sqrt{5}}{3}\right)^2 =\frac{4i^2\cdot 5}{9}=-\frac{20}{9}$$

$$\left(\left(\frac{2 - 4i\sqrt{5}}{6}\right)-\frac{1}{3}\right)^2 =\left(\frac{-2i\sqrt{5}}{3}\right)^2=\frac{4i^2\cdot 5}{9}=-\frac{20}{9}$$

So as we see the answer is $\color{red}{-\frac{20}{9}}$

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    $\begingroup$ I actually like the brute force approach for this one. Sometimes we need to get our hands "dirty" $\endgroup$ Aug 16, 2015 at 22:43
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    $\begingroup$ A pretty inefficient approach, and what if the OP doesn't know about complex numbers ? $\endgroup$
    – user65203
    Sep 21, 2016 at 13:07
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    $\begingroup$ I also like the brute force. Of course, not as elegant as other solutions, but very very straightforward. And you can only find elegant solutions with experience. $\endgroup$ Oct 21, 2016 at 10:10
  • $\begingroup$ @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer. $\endgroup$
    – Carsten S
    Jan 10, 2017 at 18:44
  • $\begingroup$ @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$. $\endgroup$
    – user65203
    Jan 10, 2017 at 20:12
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HINT: complete the square in $$3x^2-2x+7$$

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  • $\begingroup$ barak manos has given good hint $\endgroup$ Sep 11, 2016 at 12:10
  • $\begingroup$ i have edited a little $\endgroup$ Sep 21, 2016 at 12:53
  • $\begingroup$ Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square. $\endgroup$
    – user65203
    Sep 21, 2016 at 12:55
  • $\begingroup$ @YvesDaoust why is that so obvious? If this is from a book, why couldn't the chapter this question is taken from be about completing the square? It might be that the OP just didn't know how to proceed while still being able to complete the square; he could have just not seen the advantage of this manipulation. $\endgroup$
    – Hirshy
    Sep 21, 2016 at 13:11
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Starting from

$$3x^2-2x+7=0\\$$ $$x^2-\frac{2}{3}x+\frac{7}{3}=0\\$$ $$x^2-2\cdot\frac{1}{3}\cdot x+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{7}{3}=0\\$$ $$\left(x-\frac{1}{3}\right)^2+\frac{21-1}{9}=0\\$$ $$ \left(x-\frac{1}{3}\right)^2=\frac{-20}{9}$$

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HINT:

$$3x^2-2x+7=3\left(x-\frac13\right)^2+6+\frac23$$

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    $\begingroup$ In general, I find the notation $6\frac{2}{3}$ to be a source of confusion because it is unclear if one means $6+\frac{2}{3}$ or $6\cdot\frac{2}{3}$. I strongly suggest either improper fractions or explicitly writing out the operation instead of mixed fractions. $\endgroup$ Aug 16, 2015 at 20:11
  • $\begingroup$ @MichaelBurr: I thought about it while writing the answer, but figured it could hardly be seen as $6\cdot\frac23$ without explicitly using the dot. In any case, I followed your suggestion and added a plus in order to avoid any possible confusion. Thanks. $\endgroup$ Aug 16, 2015 at 20:25
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Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-\frac{2x}{3}+\frac{7}{3}=0$$ Completing the square: $$\left(x-\frac 13\right)^2-\frac{1}{9}+\frac{7}{3}=0$$ $$\left(x-\frac 13\right)^2=\frac{1}{9}-\frac{7}{3}$$ $$\left(x-\frac 13\right)^2=\frac{1-21}{9}$$ $$\left(x-\frac 13\right)^2=-\frac{20}{9}$$

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  • $\begingroup$ I don't think further editing here is needed. $\endgroup$ Jan 17, 2017 at 10:40
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Expand the binomial and compare it to the trinomial.

$$3x^2-2x+7\iff\left(x-\frac13\right)^2=x^2-\frac23x+\frac19.$$

If you divide the polynomial by $3$, you get closer, with two identical terms

$$\frac{3x^2-2x+7}3=x^2-\frac23x+\frac73.$$

To get a perfect identity, it now suffices to add a well-chosen constant

$$\frac{3x^2-2x+7}3-\frac{20}9=x^2-\frac23x+\frac19.$$

Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.

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  • $\begingroup$ As you removed my edit: $3x^2-2x+7$ can't be equivalent to anything, you are missing the $\ldots =0$. $\endgroup$
    – Hirshy
    Sep 21, 2016 at 13:28
  • $\begingroup$ @Hirshy: no, you missed the intent. The double arrow relates the two expressions that are to be compared. Appending $=0$ turns the left one in an equation, while the right one is an identity, and this would result in a false proposition. $\endgroup$
    – user65203
    Sep 21, 2016 at 13:32
  • $\begingroup$ I did indeed miss that, sorry. You might want to use another arrow for your purpose? $\endgroup$
    – Hirshy
    Sep 21, 2016 at 13:35
  • $\begingroup$ @Hirshy: I didn't find one that pleased me. $\endgroup$
    – user65203
    Sep 21, 2016 at 13:54

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