User:Xyrael/Sandbox

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[edit] From abc to de

What I'm giving you to start with is

4x2 + 3x + 8 = 0

What I'm asking you to do it, using valid mathematical manipulations(*)

is to convert it into the form

x2 + dx + e = 0

where d and e will be numbers you can easily find.

(The reason you are doing this, by the way, is because the next step on the method is easier if the coefficient of the quare of x is 1)


(*)As a list, your valid operations are:

    • Manipulating one side of the equation without changing it's value
    • adding a number to both sides (same as taking a number away from both sides)
    • multiplying (== dividing, provided not dividing by 0) both sides by a number
    • raising both sides to a power (squaring both sides, say)
    • There are other things you can do, such as differentiating both sides and so on, but these are for later.

Surely: \frac {x*y*z} x = y*z

Yes (except in the case that x=0, in which case the division isn't valid). The only thing is that what you are doing is this:

\frac {x*y} x + x*z = y + z

instead of:

\frac {x*y + x*z} x = y + z

i.e. while you *are* dividing both sides by a number, you are not dividing the *whoile* of both sides. Clear as mud?

\frac {4x^2+3x+8} 4 = \frac 0 4 becomes x2 + 3 / 4x + 2 = 0

ax^2+bx+c=0 \Rightarrow x^2+bx/a+c/a=0

[edit] Next part

OK, we now have x2 + dx + e = 0 as our general equation.

Now, for simplicity, we'll move e across to the other side (by subtracting c from both sides) giving x2 + dx = − e --- I imagine this will be the hardest bit for me to explain, so don't hesitate to ask questions:

What we to do is convert out general equation into the form:

(x + p)2 = q

for some p and q. The reason is that you can then square root to get

x + p = \pm \sqrt{q}

giving you the solution:

x = -p \pm \sqrt{q}

Reason for +/- :

Find both (-\sqrt{q})^2

and

(\sqrt{q})^2

(-\sqrt{q})^2 = (-q^{\frac 1 2})^2 = -q^1 = -q and (\sqrt{q})^2 = (q^{\frac 1 2})^2 = q^1 = q

Observe

(-\sqrt{q}) == (-1)*(\sqrt{q})

So,

(-\sqrt{q})^2 = ((-1)*(\sqrt{q}))^2 = (-1)^2*(\sqrt{q})^2 = 1 * q = q

Consider that

(x + p)2

can be expanded out like this:

(x + p)2 = x2 + 2px + p2

Yes?

So,

(x + p)2 = q

can be written as:

x2 + 2px + p2 = q

We'll rearrange it a little to get the following:

x2 + 2px + (p2 − q) = 0

Happy?


Now, compare this to our d-e equation:

x2 + dx + e = 0

Notice the similarity?

We can definately say that 2p == d, right? OTherwise the two equations we want to be identical have different numbers of x's.

so p == (d/2)

So in the p-q equation, replace every p with (d/2) to get:

(x + (d / 2))2 = q

Right?

So all we need to do is find what q is...


x2 + dx + e = 0 contrasted with: x2 + 2px + (p2 − q) = 0

So, this time observing the constant terms must be equal, we get: e == (p^2 - q ). But we know p == (d/2) .

So:

e = = ((d / 2)2 − q)

which (check it yourself) simplifies to

q == (\frac{d^2}{4}) - e

So the p-q equation becomes:

(x + \frac{d}{2})^2 = (\frac{d^2}{4}) - e

Which, as you've seen, is simple to solve (much easier with real numbers!)

Now you've seen how to derive it in quite a circituitous manner, I'll try and find a way to type in the simple resoning behind it.

Try to solve the following using this method:

x^2 + 4x + 2 = 0

2x^2 + 2x - 24 = 0 (remember we need a = 1)

x^2 + 1 = 0 (you'll get an odd answer here!)

(1) x^2 + 4x + 2 = 0

x2 + 4x = − 2

(x + 2)2 = 6

x^2 + 4x + 4 \ne 6 <-- help :\

I don't think I really understand the method at the moment.

OK, let's assume we are starting with a monic polynomial (leading coefficient = 1)

x2 + dx + e = 0

And we want it in the form

(x + p)2 + q = 0

since you know how to solve that really easily.

OK, got that? For now, forget the rest of the above, it is confusing and backwards.

The first step is to expand out the equation we WANT:

x2 + 2px + p2 + q = 0

Compare it back with the d-e equation. Look how in both equations there is only one x term. They must be equal, right (if doubtful, ask)

So

dx = 2px \Rightarrow d = 2p \Rightarrow p = d/2

So, write this int our p-q equation:

(x + \frac{d}{2})^2 + q = 0

Thinking about this in terms of words, when we expand out (x + p)^2, the x term is twice p - so the p that we put a number on must be half of the d shown in our unsimplified equation, to cancel this out.

Next, we need to get a nice way of representing it in words. Clearly, it will mostly be made up of the e in the unsimplified equation, but remeber that the (x + p)^2 creates a p^2 term, which we want to remove, because it is connected to the d, not the e. So q would logically seem to be e - p^2 = e - (d/2)^2, right?

Let's look at the algebra:

(x + \frac{d}{2})^2 + q = 0

expands out to:

x^2 + 2*\frac{d}{2} + (\frac{d}{2})^2 + q = 0

Which simplies to:

x^2 + dx + (\frac{d}{2})^2 + q = 0

Notice how the (d/2) term comes out to make the nice dx we are looking for.

Comparing the constant terms, we get:

e = (\frac{d}{2})^2 + q \Rightarrow q = e - (\frac{d}{2})^2

converting the nasty d-e equation into the nice p-q form:

(x + \frac{d}{2})^2 + e - (\frac{d}{2})^2 = 0

Now, you could simply memorize this formula (indeed, if you follow through both the step used to get rid of the a term, the step used to get it into p-q form, and the step used to solve the p-q form, you would get the quadratic equation exactly), but as far as I'm concerned it is both easier to remember how to do it from first principles and less prone to errors (once I had it practised).

The key ideas are:

  • Get rid of the a^2 coefficient, by dividing *all through* by a.
  • Work out the p term (is is simply half of a, since when you expand out the (x + p)^2 bracket, the p goes to 2px)
  • Work out the q term - it is simply e take away that p^2 (= (d/2)^2 )the brackts spit out by accident.
  • Square root both sides (not forgetting the \pm and move the p across.
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