Quick quadratic functions question

Tragedy

I've just started introduction to Maths & Stats, doing easy Quadratic Functions at the moment and already ran into a problem.

The question is
solve:
x(x-9)(x+11)=0

But it doesn't appear to me to be a quadratic, if it's multiplied out you end up with x3, x2 and x.

Could anyone point me in the right direction?

equivariant

The left hand side is a product (of three factors). A product is equal to 0 if and only if one of the factors is 0.........

Tragedy

This is the question. The rest of the topic is fine, but I've gone through notes/the textbook(maths for economics and business) and really just don't have a clue.

As far as I can see, the leading x needs to be a number(or not be there) for this to form a quadratic?

Fremen

It's not a quadratic. It's a "cubic" because the highest power on the left hand side is three. You also get quartics (power of four) quintics (power of five) and higher.

The question isn't about solving quadratics. They're trying to demonstrate the relationship between roots and factors.

Remember that a function f(x) is a rule. You plug in a number x, and the function spits out another number. A root of the function is a number that you plug in to make the function spit out the number 0.

Solving f(x) = 0 means "find all the roots of f(x)".

In the example you gave, there's a very nice structure that you can exploit. Remember that 0 times anything is 0. Your function is a product of three things:

f(x) = x(x-9)(x+11)

so, say,

f(1) = 1(1-9)(1+11) = 1(8)(12) = 96

but what if we plugged in a number that made one of the three terms equal to 0?

f(9) = 9(9-9)(9+11) = 9(0)(18) = 0

so we've found a root.

Just by looking at the equation, you can see what other numbers you could plug in to find a value of x such that f(x) = 0. Any idea what they might be?

bnt

Tragedy wrote: »

Could anyone point me in the right direction?

I don't know how much I can say without basically giving you the answer! If you multiply it out, you do indeed get a 3rd-order equation, as you say; and you'd have to solve for its roots, which is too much to ask at this stage.

If you really get stuck, have a look at this:

DON'T multiply it out. You'd only have to factorise it again, but here you've been given the problem in factored form already, so all you're doing is the last step in that process. Try solving x^2-4x=0 and try to apply the last step in that process to this problem.

Tragedy

Fremen wrote: »

Just by looking at the equation, you can see what other numbers you could plug in to find a value of x such that f(x) = 0. Any idea what they might be?

0, 9, -11?

bnt wrote: »

DON'T multiply it out. You'd only have to factorise it again, but here you've been given the problem in factored form already, so all you're doing is the last step in that process. Try solving x^2-4x=0 and try to apply the last step in that process to this problem.

x^2-4x=0 => x(x-4)=0 => x=0 or 4

We haven't covered anything like this yet, hence the confusion. We didn't even really do much factoring, just using the quadratic formula to throw out results.

Thanks for the help!

Fremen

Tragedy wrote: »

0, 9, -11?
!

Bingo! problem solved.

How about (x-2)(x+17)(x - 1/3)(2x - 5) = 0?

Tragedy

2, -17, 1/3, 2.5?

I'm a mature(25year old) mature student who last did(pass!) maths 7-8 years ago. I'm not finding any of it difficult so far - once it's explained well enough/examples given, especially as it's fairly progressive in it's structure.

Thanks again

bnt

Tragedy wrote: »

x^2-4x=0 => x(x-4)=0 => x=0 or 4

Cool!

PS: if you know the quadratic form ax²+bx+c, then you'll note that that function has no c term, so c=0. Which means that its y-intercept =0: if you draw the function on a chart, it will cross the y-axis (x=0) at y=0. Whenever the constant term (c in this case) in a polynomial is zero, one of the roots must be zero.