Compare with a general second degree polynomial (a quadratic equation), then look at the discriminant of the quadratic formula and find the condition that #p^2 >= 4q#.
Explanation:
Ah, that happens to be a quadratic equation! Compare:
#ax^2 + bx + c = 0#
With what we have:
#x^2 - px + 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:
Treat it like a quadratic equation and directly use the quadratic formula etc.
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 pm sqrt(b^2 - 4ac))/(2a)#
For any #ax^2 + 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) pm sqrt((-p)^2 - 4(1)(q)))/(2(1))#
#x = (p pm sqrt(p^2 - 4q))/(2)#
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!
#sqrt(p^2 - 4q)#
We need to make sure that whatever number that shows up in there isn't negative!
#p^2 - 4q >= 0#
So, that's the condition! You could rearrange, if needed:
#p^2 >= 4q#
#"2. Playing Around With It Like Any Other Polynomial"#
Well, we have
#x^2 - px + q = 0#
And we first want to solve for #x#. Let's start by subtracting #q#:
#x^2 - px + q - q = 0 - q#
#x^2 - px = -q#
Now, there are two #x#s, how can we "combine" them into one? Maybe we could somehow use the following:
#(a + b)^2 = a^2 + 2ab + b^2#
Hmm... let's try to match it with what we have! We could turn #p# into #1 * p#, into #2 * 1/2 * p#:
Nice! We only have one #x#, making it easier! Expand:
#(x - 1/2 p)^2 = -q + (-1/2 p)^2#
#(x - 1/2 p)(x - 1/2 p) = -q + (-1/2 p)(- 1/2 p)#
#x^2 - 1/2 p x - 1/2 p x + 1/4 p^2 = -q + 1/4 p^2#
#x^2 - 2 * 1/2 p x+ 1/4 p^2 = -q + 1/4 p^2#
#x^2 - px+ 1/4 p^2 = -q + 1/4 p^2#
Subtract #1/4 p^2# from both sides:
#x^2 - px = -q#
Finally, add back #q#:
#x^2 - px + 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 + (-1/2 p))^2 = -q + (-1/2 p)^2#
#(x - 1/2 p)^2 = -q + (-1/2 p)(-1/2 p)#
#(x -1/2 p)^2 = -q + 1/4 p^2#
#(x -1/2 p)^2 = 1/4 p^2 - q#
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: #3^2# and #(-3)^2# both equal #9#, so the square root of #9# should be both #3# and #-3#, for example, so we need the #pm# sign!
#sqrt((x -1/2 p)^2) = sqrt(1/4 p^2 - q)#
#x -1/2 p = pm sqrt(1/4 p^2 - q)#
And add #1/2 p# to both sides:
#x -1/2 p + 1/2 p = 1/2 p pm sqrt(1/4 p^2 - q)#
#x = 1/2 p pm sqrt(1/4 p^2 - q)#
Eh, maybe there's something we can do with what's inside the radical... Maybe make #q# also a fraction?
#x = 1/2 p pm sqrt((p^2)/4 - (4q)/4)#
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 = 1/2 p pm sqrt((p^2 - 4q)/4)#
Then we can "split" the square root:
#x = 1/2 p pm sqrt(p^2 - 4q)/sqrt(4)#
#x = 1/2 p pm sqrt(p^2 - 4q)/2#
#x = (p pm sqrt(p^2 - 4q))/2#
Whew! Now we focus on the contents of the radical:
#sqrt(p^2 - 4q)#
Which shouldn't be a negtive number, otherwise we'd have imaginary numbers, and such #x#-intercepts wouldn't appear in the graph:
#p^2 - 4q ≥ 0#
Rearranging, by adding #4q# to both sides:
#p^2 - 4q + 4q ≥ 0 + 4q#
#p^2 ≥ 4q#
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:
