Solving $3x^2 - 4x -2 = 0$ by completing the square
$begingroup$
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?
algebra-precalculus quadratics self-learning
edited Apr 26 at 20:51
user21820
41k545166
asked Apr 26 at 13:55
neuronneuron
30217
$endgroup$
add a comment |
$begingroup$
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?
algebra-precalculus quadratics self-learning
edited Apr 26 at 20:51
user21820
41k545166
asked Apr 26 at 13:55
neuronneuron
30217
$endgroup$
6
$begingroup$
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
add a comment |
$begingroup$
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?
algebra-precalculus quadratics self-learning
edited Apr 26 at 20:51
user21820
41k545166
asked Apr 26 at 13:55
neuronneuron
30217
$endgroup$
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?
algebra-precalculus quadratics self-learning
algebra-precalculus quadratics self-learning
edited Apr 26 at 20:51
user21820
41k545166
asked Apr 26 at 13:55
neuronneuron
30217
edited Apr 26 at 20:51
user21820
41k545166
asked Apr 26 at 13:55
neuronneuron
30217
edited Apr 26 at 20:51
user21820
41k545166
edited Apr 26 at 20:51
user21820
41k545166
edited Apr 26 at 20:51
user21820
41k545166
41k545166
asked Apr 26 at 13:55
neuronneuron
30217
asked Apr 26 at 13:55
neuronneuron
30217
asked Apr 26 at 13:55
neuronneuron
30217
30217
6
$begingroup$
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
add a comment |
6
$begingroup$
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
6
6
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Try going backward; expand the square $(x-tfrac{2}{3})^2$ to find that
$$left(x-frac{2}{3}right)^2=left(x-frac{2}{3}right)left(x-frac{2}{3}right)=x^2-frac43x+frac49.$$
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
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add a comment |
$begingroup$
Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=frac{2}{3}$ gives the last line.
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
$endgroup$
add a comment |
$begingroup$
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 $9x^2-12x$.
answered Apr 26 at 20:56
user21820user21820
41k545166
$endgroup$
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
add a comment |
$begingroup$
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.
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
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add a comment |
$begingroup$
Let me do this with some diagrams. You have started with the task of figuring out what x is in the expression
And the first thing we do is divide by 3 to make this simpler; scaling the boxes up
gives: 
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.
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 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 $left(x - frac23right)^2.$
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 $x = pmsqrt{30} - 3.$
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
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add a comment |
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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 $$x^2-2cdot frac{2}{3}x+frac{4}{9}=frac{2}{3}+frac{4}{9}$$
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try going backward; expand the square $(x-tfrac{2}{3})^2$ to find that
$$left(x-frac{2}{3}right)^2=left(x-frac{2}{3}right)left(x-frac{2}{3}right)=x^2-frac43x+frac49.$$
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
$endgroup$
add a comment |
$begingroup$
Try going backward; expand the square $(x-tfrac{2}{3})^2$ to find that
$$left(x-frac{2}{3}right)^2=left(x-frac{2}{3}right)left(x-frac{2}{3}right)=x^2-frac43x+frac49.$$
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
$endgroup$
add a comment |
$begingroup$
Try going backward; expand the square $(x-tfrac{2}{3})^2$ to find that
$$left(x-frac{2}{3}right)^2=left(x-frac{2}{3}right)left(x-frac{2}{3}right)=x^2-frac43x+frac49.$$
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
$endgroup$
Try going backward; expand the square $(x-tfrac{2}{3})^2$ to find that
$$left(x-frac{2}{3}right)^2=left(x-frac{2}{3}right)left(x-frac{2}{3}right)=x^2-frac43x+frac49.$$
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
answered Apr 26 at 13:59
ServaesServaes
33.2k444103
33.2k444103
add a comment |
add a comment |
$begingroup$
Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=frac{2}{3}$ gives the last line.
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
$endgroup$
add a comment |
$begingroup$
Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=frac{2}{3}$ gives the last line.
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
$endgroup$
add a comment |
$begingroup$
Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=frac{2}{3}$ gives the last line.
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
$endgroup$
Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=frac{2}{3}$ gives the last line.
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
answered Apr 26 at 13:59
WuestenfuxWuestenfux
6,8221513
6,8221513
add a comment |
add a comment |
$begingroup$
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 $9x^2-12x$.
answered Apr 26 at 20:56
user21820user21820
41k545166
$endgroup$
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
add a comment |
$begingroup$
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 $9x^2-12x$.
answered Apr 26 at 20:56
user21820user21820
41k545166
$endgroup$
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
add a comment |
$begingroup$
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 $9x^2-12x$.
answered Apr 26 at 20:56
user21820user21820
41k545166
$endgroup$
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 $9x^2-12x$.
answered Apr 26 at 20:56
user21820user21820
41k545166
answered Apr 26 at 20:56
user21820user21820
41k545166
answered Apr 26 at 20:56
user21820user21820
41k545166
answered Apr 26 at 20:56
user21820user21820
41k545166
41k545166
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
add a comment |
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
$begingroup$
See my post the closely-related AC-method for more on the ideas behind this method.
$endgroup$
– Bill Dubuque
Apr 26 at 21:01
add a comment |
$begingroup$
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.
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
$endgroup$
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.
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
answered Apr 26 at 19:10
Roman OdaiskyRoman Odaisky
25116
25116
add a comment |
add a comment |
$begingroup$
Let me do this with some diagrams. You have started with the task of figuring out what x is in the expression
And the first thing we do is divide by 3 to make this simpler; scaling the boxes up
gives:
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.
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 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 $left(x - frac23right)^2.$
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 $x = pmsqrt{30} - 3.$
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
$endgroup$
add a comment |
$begingroup$
Let me do this with some diagrams. You have started with the task of figuring out what x is in the expression
And the first thing we do is divide by 3 to make this simpler; scaling the boxes up
gives:
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.
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 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 $left(x - frac23right)^2.$
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 $x = pmsqrt{30} - 3.$
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
$endgroup$
add a comment |
$begingroup$
Let me do this with some diagrams. You have started with the task of figuring out what x is in the expression
And the first thing we do is divide by 3 to make this simpler; scaling the boxes up
gives:
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.
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 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 $left(x - frac23right)^2.$
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 $x = pmsqrt{30} - 3.$
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
$endgroup$
Let me do this with some diagrams. You have started with the task of figuring out what x is in the expression
And the first thing we do is divide by 3 to make this simpler; scaling the boxes up
gives:
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.
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 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 $left(x - frac23right)^2.$
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 $x = pmsqrt{30} - 3.$
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
answered Apr 26 at 21:20
CR DrostCR Drost
1,919811
1,919811
add a comment |
add a comment |
$begingroup$
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 $$x^2-2cdot frac{2}{3}x+frac{4}{9}=frac{2}{3}+frac{4}{9}$$
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
$endgroup$
add a comment |
$begingroup$
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 $$x^2-2cdot frac{2}{3}x+frac{4}{9}=frac{2}{3}+frac{4}{9}$$
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
$endgroup$
add a comment |
$begingroup$
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 $$x^2-2cdot frac{2}{3}x+frac{4}{9}=frac{2}{3}+frac{4}{9}$$
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
$endgroup$
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 $$x^2-2cdot frac{2}{3}x+frac{4}{9}=frac{2}{3}+frac{4}{9}$$
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
answered Apr 26 at 13:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.6k42867
81.6k42867
add a comment |
add a comment |
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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