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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: $ -\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. $\endgroup$ – Michael Burr Aug 16 '15 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. $\endgroup$ – Ethan Bolker Aug 16 '15 at 19:50
  • $\begingroup$ 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. $\endgroup$ – Martin Sleziak Aug 17 '15 at 9:11
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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. $\endgroup$ – Rolazaro Azeveires Aug 16 '15 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 $\endgroup$ – Kevin Brown Aug 17 '15 at 2:05
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    $\begingroup$ Yes, you are right $\endgroup$ – Harish Chandra Rajpoot Aug 17 '15 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 $\endgroup$ – Rolazaro Azeveires Aug 18 '15 at 2:08
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    $\begingroup$ (+1) But if roots were assumed real, this problem made no sense. $\endgroup$ – Amad27 Oct 20 '15 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" $\endgroup$ – Rolazaro Azeveires Aug 16 '15 at 22:43
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    $\begingroup$ A pretty inefficient approach, and what if the OP doesn't know about complex numbers ? $\endgroup$ – Yves Daoust Sep 21 '16 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. $\endgroup$ – Adrian Buzea Oct 21 '16 at 10:10
  • $\begingroup$ @YvesDaoust, if $x$ is supposed to be real, then $42$ is another valid answer. $\endgroup$ – Carsten S Jan 10 '17 at 18:44
  • $\begingroup$ @CarstenS: of course. This exercise is about "completing the square" and can be solved without caring about $x$. $\endgroup$ – Yves Daoust Jan 10 '17 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 $\endgroup$ – Bhaskara-III Sep 11 '16 at 12:10
  • $\begingroup$ i have edited a little $\endgroup$ – Bhaskara-III Sep 21 '16 at 12:53
  • $\begingroup$ Obviously, the OP is precisely stuck because he is hasn't learnt how to complete the square. $\endgroup$ – Yves Daoust Sep 21 '16 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 '16 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. $\endgroup$ – Michael Burr Aug 16 '15 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. $\endgroup$ – barak manos Aug 16 '15 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. $\endgroup$ – Daniel Fischer Jan 17 '17 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 '16 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. $\endgroup$ – Yves Daoust Sep 21 '16 at 13:32
  • $\begingroup$ I did indeed miss that, sorry. You might want to use another arrow for your purpose? $\endgroup$ – Hirshy Sep 21 '16 at 13:35
  • $\begingroup$ @Hirshy: I didn't find one that pleased me. $\endgroup$ – Yves Daoust Sep 21 '16 at 13:54

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