can't for the life of me undestand this

[Prussian to the Iron]

i dont even know what a root is. it doesnt even say, it just tell us to find the roots, gives us an example, than leaves us to screw ourselves over.

[Zain]

i dont even know what a root is. it doesnt even say, it just tell us to find the roots, gives us an example, than leaves us to screw ourselves over.

Roots are where the line crosses the x-axis.

[Prussian to the Iron]

okay. now how fo i find the roots?

[Zradha Pahlavan]

Algebra 2 makes my brain hurt, but I'll see if I can help.

This is a parabola:


Where the curvy line crosses or touches the straight horizontal line (it can be anywhere on that line) is where the root is. I think.

[Zain]

Algebra 2 makes my brain hurt, but I'll see if I can help.

This is a parabola:


Where the curvy line crosses or touches the straight horizontal line (it can be anywhere on that line) is where the root is. I think.

In this example, the root is 0 because that is where the red line (parabola) crosses the horizontal x-axis.

Understand now?

[Ironside]

i dont even know what a root is. it doesnt even say, it just tell us to find the roots, gives us an example, than leaves us to screw ourselves over.

Roots are the value where the equation can be solved and the answer is 0.

Take (x-2)^2=0 for example, the roots are then (2, 2) a double root. (x-2)^2=16 on the other hand have the solutions (6,-2) as you can see if you put those values in.

Now by a general rule, the prefered method is to write it in this form f(x)=y where y=0, so (x-2)^2=16 becomes (x-2)^2-16=0. It has the same solutions, but it is in this form the roots are where the curve touches or crosses zero. So the roots are (6,-2) same as above, so you don't have to make the equation in the f(x)=0 form to solve it, but you'll need to understand the concept.

Viking has given one formula to calculate the roots, but these are easy enough to solve by asking yourself "what values of x make this equation 0?" (that is f(x)=0). The double roots are already explained properly I think.

Can't explain properly why this is done, but it has to do with making more advanced math easier. For you, it would melt your brain atm.

[Prussian to the Iron]

so how would i be able to see how i should graph the parabola so i can see the answer?

[Ironside]

so how would i be able to see how i should graph the parabola so i can see the answer?

Graphs are only really useful if you got a graph drawing calculator. You normally solve them the same way you solve normal equations, with the exception of that they have multiple answers or a single answer with a double root (if it's a x^2 function that is). All examples given are possible to solve by head counting, even if 11 is severely helped by the possible answers.

It will get slightly more complicated soon I suspect, but it is a speciffic formula for those functions.

Anyway, the more literal answer on your question (drawing a graph by hand) would be to put in some x values in the function, get the y values and then plot them together in a x/y system.

[Prussian to the Iron]

I have a graphing calculator, but it only uses Y= when graphing. how can i get to it? could you walk me through one of the questions so i can check if I'm doing it right?

[Tellos Athenaios]

I don't really understand ‘double roots’ as proper terminology either; but if you need to mark the points in which a function f touches upon a certain line this is done not by solving for roots, but by finding the derivative function f' and then solving for f' = g' and f = g; where g' is the derivative of a function g describing said line. E.g. f(x) = x^2; g(x) = 0 => f'(x) = 2x; g' = 0 => f' = g' and f = g if and only if x = 0.

That test is fairly trivial to answer: since it's multiple choice, plug in one of the values see if it works out. E.g.: x^4 = 36x^2 => 36 = x^2 => set of solutions for x = {-6, 6}. Or: x^4 = 36x^2 = 0 => x = 0. So: solutions for x = {-6, 0, 6}. But you could get that easily via process of elimination and a careful review of your answer; at least if you are good at calculation inside your head this'll be faaar faster than properly deriving answers to questions you struggle to understand.

For bonus points:
Compared to the abc formula; for a formula of ax^2 + bx + c; with a, b and c not 0: factorization into forms such as (dx + e)(fx + g) is a more efficient approach to deriving solutions for equations since you want to solve for x that evaluates to 0, and multiplication like this gives you a trivial & painless way to evaluate just that.
@Ironside: (x-2)^2 = 0 => (x-2) = sqrt(0) => (x-2) = 0 => x = 2.

[Owen Glyndwr]

This kind of stuff is why I'm really glad I tested out of math and don't have to take any math classes in University!!

Good luck my friend.

[Ironside]

@Ironside: (x-2)^2 = 0 => (x-2) = sqrt(0) => (x-2) = 0 => x = 2.

I think I got a nice way to explain a double root here.

(x-2)^2=(x-2)(x-2)=0 . Gives one 2 for the first paranthesis (root 1) and a second 2 for the second (root 2) paranthesis. They just happen to be the same number.

Compare to (x-1)(x-3)=0 => (1 (root 1), 3 (root 2))

Same principles, just a special case. Same as elips => circle, rectangle => square etc. Not that much simplification bonus in this case though.

This kind of stuff is why I'm really glad I tested out of math and don't have to take any math classes in University!!

Good luck my friend.

You mean you don't like to do iterations by hand to solve partial differential equations?

Spoiler Alert, click show to read: 

Those are mean, but computers loves them. Until you make them complicated enough to cause work overload...