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okay, so ive been doing better with my algebra II, but i am still having tons of trouble. these questions in particular i have no idea about. the book/commentary didnt even explain double roots, and i can't call the teacher for help with a test. any help?
https://i391.photobucket.com/albums/...lgebrahelp.png
According to my memory and this page
A double root is when the parabola just touches the x axis instead of crossing it. If the equation were to cross the x axis at -6 and 5 then those would be your answers. If the equation touches the x axis at 3 but doesn't cross it then the answer would be double root at 3.
Hopefully that helps :sweatdrop:
.......................sorry my brain just broke for a couple minutes.........
uuuuuhhhhhhhhhhh............................what?
double root is omega
riiiiiiiiiiiightttt.................................................???????
If you were to graph the equations, typically roots are the part where the parabola (curvy line) crosses the x-axis (horizontal straight line) a double root simply means that it is the only place where the curvy line (parabola) touches the horizontal straight line (x-axis).
It's called a double root because usually there are two roots. But because there is one, it's double. Just to keep all the math stuff uniform.
How do you extract the roots? The "abc formula"? Whenever b^2-4ac = 0, you've got a double root, where the quadratic equation is on the form ax^2 + bx + c.
E.g. x^2 + 2x + 1 gives the double root -1.
My calculus books calls them single roots; though since I am no mathematician I shan't enter an argument..
Basically all x^a functions have a roots. If you have less solutions than roots, then some roots are double or larger roots.
x^87=0 have 87 roots but 1 solution (namely 0), so that's one 87 root. Understand the concept, not my words (as they're probably described differently in the books).
They appear when a term is repeated (x+4)^3*(x-3)=0 have 4 roots and 2 solutions (-4, 3). Because the first term is repeated three times, -4 becomes a trippel root.
(aka full answer is -4, -4, -4, 3)
As I take it this is part of a test, I'll refrain from giving the correct answers.
okay, guys, try to explain this like you're explaining it to a 10 year old. i simply have no idea even when you are trying to help. the book makes no sense of it either.
I put it as simply as I could dude...Quote:
Originally Posted by Prussian Iron
What method do you use to find the roots? Don't you have any?Quote:
Originally Posted by Prussian 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.
Roots are where the line crosses the x-axis.Quote:
Originally Posted by Prussian Iron
okay. now how fo i find the roots?
Algebra 2 makes my brain hurt, but I'll see if I can help.
This is a parabola:
https://i225.photobucket.com/albums/.../parabola1.gif
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.Quote:
Originally Posted by Zradha Pahlavan
Understand now?
Roots are the value where the equation can be solved and the answer is 0.Quote:
Originally Posted by Prussian Iron
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.
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.Quote:
Originally Posted by Prussian Iron
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.
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?
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.
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.
I think I got a nice way to explain a double root here.Quote:
Originally Posted by Tellos Athenaios
(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.
You mean you don't like to do iterations by hand to solve partial differential equations? :whip:Quote:
Originally Posted by Owen Glyndwr
Spoiler Alert, click show to read:
What...do you mean that you had to solve those equations numerically by hand? :inquisitive:Quote:
Originally Posted by Ironside
so in that example the double root would be....Quote:
Originally Posted by Ironside
1,???
and
3,???
correct?
A double root is a root that occurs twice. In the first example, the root '2' occurs twice. In the second, no root occurs twice.Quote:
Originally Posted by Prussian Iron
A root is the x value for which the function is zero, which is also a place where it crosses the x-axis. So, if you plot the funtcion y = (x-2)(x-3), the function will cross the x-axis at x = 2 and x = 3, because
y(2) = (2-2)*(2-3)
= 0*(-1)
= 0
Same goes for x = 3.
What you should do, though, is apparently to plot the function and find out where it crosses the x-axis.
Let's say you want to find the roots for
x^2 + 3x = - 2
you have to rearrange the equation such that the right hand side of it equals zero. This can be done by adding 2 on both sides:
x^2 + 3x + 2 = - 2 + 2
x^2 + 3x + 2 = 0
now you plot y = x^2 + 3x + 2, because this is the function you want to equal zero.
Alternatively, you may plot y = x^2 + 3x and find out where this equals -2.
so would number 10 be C? or A? I'm guessing A right?
Try to insert e.g. x = 3 in the equation and see what you get. Remember that you are looking for the solutions of an equation.
so C then!
Does the left hand side of the equation equal the right hand side if x is the number three?
If you have the equation
(x-2)(x-1) = 2
and set x = 2
you get...
(2-2)(2-1) = 2
0*(2-1) = 2
0 = 2
which is impossible. Hence x = 2 cannot be a root.
I think you should rearrange the equation and plot it, as I suggested; otherwise you are meant to do it otherwise. How does the example in the book find the root(s)?
there is no example in the book; if there is i have no idea where it is. the teacher likes to use future material in tests for no reason.
These are the answers I got:
9. A
10. B
11. A
12. D
If you want me to show you how I solved them just ask.
tahts exactly what i want; a detailed walkthrough of each problem. if i can see how it is solved, it makes it much easier to do in the future.
Question 9:
Factor X^2 out of the equation:
X^2(X^2=36)
That X^2 is now equal to zero. You take the square root of both sides which gives you + and - zero because a negative squared comes out as the same as a positive squared. However in this case there is no such thing as positive or negative zero, there is just zero twice. So zero is a root twice, or a double root.
Next you have the X^2=36. Again take the square root of each side and you get + and - 6. So 6 and -6 are roots as well. Altogether you have 0 twice, 6, -6 as your answers. Which means A is your answer.
Question 10:
Multiply the factors (X-3)(X+3) together to get X^2-9=40.
Add nine to each side to balance it out to X^2=49. Take the square root of each to get your + and - 7.
Your answer is B.
Question 11:
Multiply the factors (X+2)(X+2) together (because its squared or in other words multiplied by itself once) to get (X^2+4X+4)+(7(X+2))=18.
Then multiply that 7(X+2) out to get (7X+14). Add this to the first polynomial (or equation if you don't know what that means) to get X^2+11X+18=18.
You then subtract each side by 18 to get it all balanced which as it happens makes the whole thing equal to zero.
So now you have X^2+11X=0. Take X out like this X(X+11)=0. Now that X is simply equal to zero. Which means zero is a root (and there is only one of it).
With X+11=0 you can easily see that the last root is X= -11. So your two roots are 0 and -11. The answer is A.
Question 12:
This one is easy if you know what you are doing. You won't need to multiply them all together since it is equal to zero. What you see there with that cubed root means that written out long it looks like ((X^2)-4)((X^2)-4)((X^2)-4)=0.
Just see that when you balance each side by adding 4 you will have X^2=4 and when you take the square root of each side after that the roots are + and - 2. And then since there is two other factors exactly the same your roots will be -2,2 -2,2 and -2,2 or -2 three times or a triple root and 2 three times or a triple root.
So basically the answer is -2 (triple root) and 2 (triple root). The answer is D.
Is that helpful?