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- Jan 26, 2012

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Well, that's what completing the squareIn your opinion what is the best way to introduce completing the square, in fact I do not like the way sub and add the squared half of x coefficient, after making x^2 coefficient 1

saying it is complicated...

Thanks

$$x^{2}+2fx+f^{2}+g=x^{2}+4x+7.$$

Equating coefficients of like powers gives you immediately that $2f=4$, or $f=2$. Plugging that in to the rest of it says that $4+g=7$, so $g=3$. Evidently, then,

$$(x+2)^{2}+3=x^{2}+4x+7.$$

Then you can work up to the rest of it.

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- Jan 26, 2012

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Certainly, adding ~~half~~ the square of HALF the linear coefficient is *the result*, but I would take a slightly different approach...

We start with squaring binomials of the form $(x + d)$ for integers $d$. I prefer to skip $d = 1,2$, as they don't necessarily lend themselves to an educated guess as to*what is going on*.

So we proceed. $$(x+3)^2 = x^2 + 6x + 9$$

$$(x + 4)^2 = x^2 + 8x + 16$$

$$(x + 5)^2 = x^2 + 10x + 25$$

etc.

Then students should start to see the pattern (and indeed, the general formula for $(x + y)^2$) that $(x + d) = x^2 + (2d)x + d^2$

But this is going*from left to right* (i.e. we are beginning with the square completed already). I'll try to recreate with Latex what I usually write on paper/whiteboard. Let's see...

"Solve by completing the square: $x^2 + 6x -3 = 0$"

Solution: $x^2 + 6x + $ (###) $ = 3 +$ (###)

Then under this line, I would write $(x + $@$)^2$

"So what is @? It's half of 6, so let's write that in..." Then it looks like:

$x^2 + 6x + $ (###) $ = 3 +$ (###)

$(x + 3)^2$

"So what is ### going to be when we 'FOIL it out'? Right..."

$x^2 + 6x + 9 = 3 + 9$

$(x + 3)^2 = 12 \cdots$

-------------

I always make my students do it this way... "begin with the end in mind", where "the end" is a binomial raised to the second power. After a few iterations, the somewhat opaque (to them) idea of $(\dfrac{b}{2}) ^2$ becomes clear.

Hope that helps!

We start with squaring binomials of the form $(x + d)$ for integers $d$. I prefer to skip $d = 1,2$, as they don't necessarily lend themselves to an educated guess as to

So we proceed. $$(x+3)^2 = x^2 + 6x + 9$$

$$(x + 4)^2 = x^2 + 8x + 16$$

$$(x + 5)^2 = x^2 + 10x + 25$$

etc.

Then students should start to see the pattern (and indeed, the general formula for $(x + y)^2$) that $(x + d) = x^2 + (2d)x + d^2$

But this is going

"Solve by completing the square: $x^2 + 6x -3 = 0$"

Solution: $x^2 + 6x + $ (###) $ = 3 +$ (###)

Then under this line, I would write $(x + $@$)^2$

"So what is @? It's half of 6, so let's write that in..." Then it looks like:

$x^2 + 6x + $ (###) $ = 3 +$ (###)

$(x + 3)^2$

"So what is ### going to be when we 'FOIL it out'? Right..."

$x^2 + 6x + 9 = 3 + 9$

$(x + 3)^2 = 12 \cdots$

-------------

I always make my students do it this way... "begin with the end in mind", where "the end" is a binomial raised to the second power. After a few iterations, the somewhat opaque (to them) idea of $(\dfrac{b}{2}) ^2$ becomes clear.

Hope that helps!

Last edited:

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- #5

- Jan 26, 2012

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I think you mean the square of half the linear coefficient, right?Certainly, adding half the square of the linear coefficient isthe result,

[EDIT]: Corrected.

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- Jan 26, 2012

- 24

Doh! I'll have to fix it the next time I'm at a proper computer

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Then I say it would be much easier if it was \(\displaystyle \displaystyle x^2 + 6x + 9\), because then it factorises easily to \(\displaystyle (x + 3)^2\).

Of course, we can't just turn \(\displaystyle \displaystyle x^2 + 6x + 1\) into \(\displaystyle \displaystyle x^2 + 6x + 9\) because they're not equal. But we CAN write \(\displaystyle x^2 + 6x + 1 = x^2 + 6x + 9 - 8 = (x + 3)^2 - 8\).

I then remind them of the difference of two squares rule, and point out that the first lot of "stuff" is clearly a square, and that the final term can be written as a square if written as \(\displaystyle \displaystyle \left( \sqrt{8} \right) ^2\), and so can be factorised with DOTS.

Then I leave as a task for THEM to figure out how to find the "missing" term to create the first square, i.e. to complete the square. They nearly all get it for themselves.

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Creative way thanks, Prove

Then I say it would be much easier if it was \(\displaystyle \displaystyle x^2 + 6x + 9\), because then it factorises easily to \(\displaystyle (x + 3)^2\).

Of course, we can't just turn \(\displaystyle \displaystyle x^2 + 6x + 1\) into \(\displaystyle \displaystyle x^2 + 6x + 9\) because they're not equal. But we CAN write \(\displaystyle x^2 + 6x + 1 = x^2 + 6x + 9 - 8 = (x + 3)^2 - 8\).

I then remind them of the difference of two squares rule, and point out that the first lot of "stuff" is clearly a square, and that the final term can be written as a square if written as \(\displaystyle \displaystyle \left( \sqrt{8} \right) ^2\), and so can be factorised with DOTS.

Then I leave as a task for THEM to figure out how to find the "missing" term to create the first square, i.e. to complete the square. They nearly all get it for themselves.

- Mar 4, 2013

- 188

Consider the equation \(\displaystyle x^2+ax+b\)

.This is a reduced form of the general quadratic equation.

We cannot solve this We don't know how to undo the square operation and multiplication of a operation on x simulatneously.

So we look for a simpler case.y^2+C=0

We know how to solve this.

But we should be able to convert all quadratic equations to this form.For that we need to get the ay term deleted.The only way to get the ay term deleted is to include -ay in x^2.

This can be seen by looking at the identity (y+d)^2=y^2+2dy+d^2

We will find that surprisingly enough all equations of the form \(\displaystyle x^2+ax+b\) are convertible to y^2+C=0

We define x=y+d

And we have the 2dy term in it which can be made into -ay by altering the d.

We make d = -a/2

Now substituting x=y-a/2

gives an equation of the form y^2+C=0

We have completed the square.

This method is also used for finding the roots of a general cubic.

It is too important to remember that in completing the square we are actually making a coefficient zero.

Otherwise we would go looking for completing the cube to solve a cubic.But all cubics cannot be completed into a cube.