Gerry decided to spend $450 on fireworks. If he buys regular

The part that you are unsure of has to do with quadratic equations (I think that's the right name, I've been out of the "loop" for awhile. Someone in-the-know correct me if i'm wrong).

This one is not the best for you to learn on. It's much easier on simpler equations.

Have you ever heard of F.O.I.L.? First Outside Inside Last?

(x + 3)(x - 2)

First Outisde = x * -2 = -2x
then you take x * x and +3 * -2
Inside Last = +3 * x

So all of this becomes

\(-2x + x^2 - 6 + 3x\) (now combine where possible) => -2x + 3x = x (this is all we can combine)

We are left with \(x^2 + x - 6\) Now it is easiest to go in reverse so you can see how the process works this way too.

When you see a pattern like this, you know you'e working with quadratic equations or binomials (x+3) is a "nomial" and there are 2 hence the "bi" part.

so if you have \(x^2\), you know you need x * x so start with (x )(x )

Now, then plus or minus signs tell you a lot. If the last number, the integer, is a +, then you know that both signs in the ( ) are going to be + because a positive * positive = positive. If the second sign is a negative, like -6 here, then we know we need a + and a - because a positive * negative = negative.

So the way this works is what factors of -6, when added together, give us a +1 because + x is an implied + 1x.

We have 3 and 2, but one has to be positive and one has to be negative to give us -6, so since we need a +1x, the larger number must be positive because +3 - 2 = +1.

Another example my help out.

\(^2 + 5x + 6\) => (x +/- ?)(x +/- ?) because it's a +6 we know both signs will be +

(x + ?)(x + ?)

We're dealing with 6 so we know we can only choose from 1 * 6 or 2 * 3. The middle is 5, so we know we could actually get 5 with either one. 6-1 is 5 and 2 + 3 is 5 BUT, we know both signs have to be positive so 6-1 would not work because we'd have to have a - sign in there. we know the result MUST be 2 and 3.

\((x + 2)(x + 3) = x^2 + 5x + 6\)

One other important variation you are VERY likely to see is the perfect square.

\(x^2 + 4 => (x + 2)^2\) or \((x - 2)^2\) Keep this in mind because neg * neg = pos and pos*pos = pos, so it could be either one.

\(x^2 - 4 => (x + 2)(x - 2)\)

i hope this helps.

PS. - This is also how you get an irrational number out of the denominator in a fraction.

Example:

\(\frac{5}{3x - sqrt{2}} = 12\)

Multiple by \(\frac{3x + sqrt{2}}{3x + sqrt{2}}\)

We do this because \((3x + sqrt{2})(3x - sqrt{2}) = 9x^2 + 3sqrt{2}x - 3sqrt{2}x - 2\)

so we then have \(\frac{5(3x + sqrt{2})}{9x^2 - 2}=12\)

A square root * square root gets rid of the square root and leaves the number inside.

I have no idea if \(x\) actually comes out to any interesting number, these numbers were chosen just for explanation purposes. Hope this helps you.

450/x-3 -450/x= 40
plugging is very easy here.. solving yes if you r good in equations

450(x)-450(x-3) = 40X^2-120X
1350 = 40X^2-120X
4X^2-12X-135 =0
factors of 135*4 (540) are 2^2*3^3*5
thus
4X^2-30X+18X-135 =0
2x(2x-15)+9(2x-15)=0
x cant be negative; 2x-15 =0 ; x= 7.5