Question #ae3bc

1 Answer
Jan 3, 2018

Compare with a general second degree polynomial (a quadratic equation), then look at the discriminant of the quadratic formula and find the condition that p24q.

Explanation:

Ah, that happens to be a quadratic equation! Compare:

ax2+bx+c=0

With what we have:

x2px+q=0

It seems to be a special case where a=1, then b is now p and c is renamed to q. Anyways, there are (at least?) two ways to go about this:

  1. Treat it like a quadratic equation and directly use the quadratic formula etc.
  2. Play around with it like it was any other polynomial

Hmm, quite a difficult decision... I'll do both, firstly doing the first option for readers with a limited amount of time, then (also) the second option for leisure readers with plenty of time.

1. Treating It As A Special Quadratic Equation

Well, we need to solve for x in either case. However, mathematicians have long known an equation (along with techniques to derive it, one of which is shown in the second part) that will return x-intercepts of a second degree polynomial given its coefficients:

x=b±b24ac2a

For any ax2+bx+c=0.

Our polynomial is a special case of this where a=1, then b is p and c is q. Let's adjust:

x=(p)±(p)24(1)(q)2(1)

x=p±p24q2

Currently, we are asked about which conditions where the "zeros" of this polynomial exist. That means what the x-intercepts are. That's what this formula does for us!

However, we need to make sure we aren't getting imaginary numbers so that it appears in the graph, since it should be for real numbers. How do we do that? Well, let's look inside the metaphorical possible breeding spot for imaginary numbers: the radical!

p24q

We need to make sure that whatever number that shows up in there isn't negative!

p24q0

So, that's the condition! You could rearrange, if needed:

p24q

2. Playing Around With It Like Any Other Polynomial

Well, we have

x2px+q=0

And we first want to solve for x. Let's start by subtracting q:

x2px+qq=0q

x2px=q

Now, there are two xs, how can we "combine" them into one? Maybe we could somehow use the following:

(a+b)2=a2+2ab+b2

Hmm... let's try to match it with what we have! We could turn p into 1p, into 212p:

x2212px=q

Rearrange this a bit:

x2+2x(12)p=q

Then... add (12p)2 to both sides?

x2+2x(12p)+(12p)2=q+(12p)2

Now we could match it! We could have a=x and b=12p, so:

a2+2ab+b2=(a+b)2

x2+2x(12p)+(12p)2=(x+(12p))2

So let's substitute this back:

(x+(12p))2=q+(12p)2

Nice! We only have one x, making it easier! Expand:

(x12p)2=q+(12p)2

(x12p)(x12p)=q+(12p)(12p)

x212px12px+14p2=q+14p2

x2212px+14p2=q+14p2

x2px+14p2=q+14p2

Subtract 14p2 from both sides:

x2px=q

Finally, add back q:

x2px+q=0

Ah, that gets us back to where we started! Good indicator that we're in the right direction, but we're not really getting anywhere, so... how about if we don't expand the parenthesis containing x this time?

(x+(12p))2=q+(12p)2

(x12p)2=q+(12p)(12p)

(x12p)2=q+14p2

(x12p)2=14p2q

Now, to really "take the x out", we're going to have to... take the square root! What a birthplace for imaginary numbers! Anyways, we also have to be careful: 32 and (3)2 both equal 9, so the square root of 9 should be both 3 and 3, for example, so we need the ± sign!

(x12p)2=14p2q

x12p=±14p2q

And add 12p to both sides:

x12p+12p=12p±14p2q

x=12p±14p2q

Eh, maybe there's something we can do with what's inside the radical... Maybe make q also a fraction?

x=12p±p244q4

What I just did here, is multiply q by 4, and also divide by 4. They should cancel out, but here we're reversing the process! This is so that we could subtract the fractions:

x=12p±p24q4

Then we can "split" the square root:

x=12p±p24q4

x=12p±p24q2

x=p±p24q2

Whew! Now we focus on the contents of the radical:

p24q

Which shouldn't be a negtive number, otherwise we'd have imaginary numbers, and such x-intercepts wouldn't appear in the graph:

p24q0

Rearranging, by adding 4q to both sides:

p24q+4q0+4q

p24q

Ah, so that's our condition!

By the way, what I just did was a "special case" for a general method used to derive the quadratic formula, which can be seen here: