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I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.

The equation is this:

$$ \begin{align} 3x^2 - 4x -2 = 0 \\ 3x^2 - 4x = 2 \end{align} $$

Now, divide both sides by three:

$$x^2 - \frac{4}{3}x = \frac{2}{3}$$

Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:

$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$

Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:

$$ \begin{align} x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\ \left(x - \frac{2}{3}\right)^2 = \frac{10}{9} \end{align} $$

Can anyone please explain how they went from the first step to the second step?

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    $\begingroup$ We add $\,a^2\,$ to $\,x^2+2a\,$ to get $\,(x+a)^2,\,$ i.e. to complete the square. Here $\,2a = -4/3\,$ so $\,a = -2/3$ $\ \ \ $ $\endgroup$ – Bill Dubuque Apr 26 at 13:58
  • $\begingroup$ They add the square of half of the coefficient of x: basically, they know that that will complete the square on the LHS. It's not obvious and as some answers point out, it's most easily seen by working "backwards" from the last line, multiplying out the square. Once you' ve done it a few times, it becomes second nature and becomes another arrow in your quiver. $\endgroup$ – NickD Apr 26 at 14:27
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Try going backward; expand the square $(x-\tfrac{2}{3})^2$ to find that $$\left(x-\frac{2}{3}\right)^2=\left(x-\frac{2}{3}\right)\left(x-\frac{2}{3}\right)=x^2-\frac43x+\frac49.$$

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Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=\frac{2}{3}$ gives the last line.

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Note that it is better to multiply by $3$, thus transforming the given equation as follows:

$3x^2 - 4x - 2 = 0$

$9x^2 - 12x - 6 = 0$

$(3x-2)^2 = 10$

To get the last line, you want $(3x+?)^2$ to match the $9x^2$ so you twiddle the $?$ until you match the $9x^2-12x$.

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  • $\begingroup$ See my post the closely-related AC-method for more on the ideas behind this method. $\endgroup$ – Bill Dubuque Apr 26 at 21:01
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Maybe it would be easier with prettier numbers? If you were to solve the equation $$x^2+200x+10000=0,$$ you’d recognize that $$(x+100)^2 = x^2+200x+10000$$ which would simplify things greatly, quickly leading to the solution $x=-100$.

If, however, your equation does not match the $(x+a)^2 = x^2 + 2ax + a^2$ formula perfectly, you'll have to add something to make it match. For example, to solve $$x^2+200x+9999=0,$$ you’ll have to transform it like this: $$x^2 + 200x+9999+1-1=0$$ $$x^2 + 200x+10000-1=0$$ $$(x-100)^2-1=0$$ Note that $x^2+200x$ can only be completed by 10000 and not by any other number.

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Let me do this with some diagrams. You have started with the task of figuring out what x is in the expression three squares of side-length x on the left hand side of an equals sign, one rectangle of height 4 and width x, plus a square of area 2, on the right hand side.

And the first thing we do is divide by 3 to make this simpler; scaling the boxes up gives: only one x-square box remains on the left, the right contains a 4x/3 rectangle with height 4/3 and width still x, plus a square with area 2/3. Completing the square in this case works by subtraction, so we need to invent negative areas, which I will represent in red. The fundamental idea is that we divide this chunk whose side length is the unknown x into two equal parts and collect them next to our $x^2$ term. the same x^2 is joined on the left of the equals sign by two red boxes, one of height 2/3 and width x, one of width 2/3 and height x, representing negative parts of the 4x/3. The right side only has the box of area 2/3. You can maybe see what we're aiming to do by this 90 degree rotation that has happened in the above diagram, we can use these boxes to annihilate some of that area systematically. But after we do this, if we do it carefully, we will “accidentally” leave over a red square:

the left-hand side of the equation now contains one unknown box, (x-2/3) squared, plus one red box, (2/3) squared. The right hand side still contains one box of size 2/3.

The issue is that we used one of our negative rectangles to cut away part of the square into a rectangle that was $x$ high but now $x - 2/3$ long, and then we tried to use the other negative rectangle to cut away the remaining rectangle some more: but it had a little bit of extra red left over because of the first cut. The extra red was of size $(2/3)^2 = 4/9$ as marked above, while the unknown square is of size $\left(x - \frac23\right)^2.$

The rest of the argument is just shifting the red square back to the right side and combining it with the square of area $2/3$ to create a square of area $\frac49 + \frac23 = \frac{10}9.$ What they did was they realized in advance that exactly such a square would be necessary, so they added it from the start in the hopes that you would not notice that it is a red square of negative area since that can be somewhat hard to think about.

This argument is somewhat easier in the opposite situation, where you are adding terms, like if you have $x^2 + 6x = 21.$ Then you split up this term into two added rectangles of size 3-by-$x$, you rotate one of them again by 90 degrees, but then you see that you can almost make a new square of size $(x+3)$-by-$(x+3)$ except you are missing a certain 3-by-3 square in the corner. So you just add that to both sides and discover that $(x+3)^2 = 21 + 3^2 = 30.$ Then you can take the square root to find that $x = \pm\sqrt{30} - 3.$

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We get from $$3x^2-4x-2=0$$ to $$3x^2-4x=2$$ by adding two on both sides. Then dividing by $3$ we obtain $$x^2-\frac{4}{3}x=\frac{2}{3}$$ then we can write $$x^2-2\cdot \frac{2}{3}x+\frac{4}{9}=\frac{2}{3}+\frac{4}{9}$$

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