# How to obtain Roots from indicial equation!?

Discussion in 'Homework Help' started by mcdara, Jan 2, 2013.

1. ### mcdara Thread Starter New Member

Feb 23, 2011
11
0
Hi

I am stuck on a Frobunius equation. The part is where I have to obtain roots from a indical equation as follows

4r^2 -4r + 2r = 0

=> 2r(2r-1) = 0

Roots = r1 = 0 r2 = 1/2

If anyone could tell me how these roots were obtained it would be great, when I google help the level of sums are in great detail and over my head
Any step by step method would be great, im sure its simple but I cant get my head around it.

2. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
Wolfram Alpha

it's nothing more than a quadratic formula:
$\frac{-b+-\sqrt{b^{2} - 4ac}}{2a}$

4r^2 -4r + 2r = 0 => 4r^2 - 2r = 0

a = 4
b = -2
c = 0

$\frac{-(-2)+-\sqrt{(-2)^{2} - 4(4)(0)}}{2(4)}$

=

$\frac{2+-2}{8}$

r = 0 & 1/2

3. ### mcdara Thread Starter New Member

Feb 23, 2011
11
0
I had a feeling it was something simple!!

Thank you!

4. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Which part of the process don't you understand?

I have no idea what a Frobunius equation or indical equation are, but you don't need the quadratic equation to find the roots of the equation you've given.

Is getting from 4r^2 -4r + 2r = 0 to 2r(2r-1) = 0 the stumbling point?

If so, the distributive property says you can factor out the same thing from each term in a group of terms. All of your terms are divisible by 2r, so factoring out 2r gives you your second equation.

4r^2 -4r + 2r = 0
2r(2r) - 2r(2) + 2r(1) = 0
2r(2r - 2 + 1) = 0
2r(2r - 1) = 0

Note that we could have combined the last two terms earlier since

-4r + 2r = -2r

Making the equation

4r^2 - 2r

And, after factoring out 2r, we again get

2r(2r - 1)=0

Is the problem that you don't understand what a "root" of an equation is?

If so, a root is nothing more than a value of the variable that results in the equation being equal to zero. Well, if I have xy=0, then the equation overall is equal to zero if either x is zero or y is zero. So I can set each factor equal to zero and solve for the roots that way. In your equation you have two factors (technically three), namely (2r) and (2r-1). I say "technically three" because the three factors are really (2), (r), and (2r-1) and, if you recognize this, then things become even a bit easier. But let's say we didn't recognize this. So our root are:

The values of r that make
2r = 0
or
2r-1 = 0

2r = 0 => r=0
2r-1 = 0 => r = 1/2

mcdara likes this.
5. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
I just did a quick look and Frobunius method is a means of finding an infitite power series solution to an second order ordinary differential equation. And its "indicial", not "indical" (easy typo to make).

Is this the level of math you are supposed ot be working at, namely solving differential equaitons?

6. ### justtrying Active Member

Mar 9, 2011
330
836
knowing how to solve quadratic equations is essential for survival. The basic approaches are factoring, completing the square, and using the formula. I suggest you practice all of them.

mcdara likes this.
7. ### Papabravo Expert

Feb 24, 2006
11,779
2,495
It seems odd to me that you would be playing with ininite series and differential equations, having slept through Algebra I, which is practically dedicated to the quadratic equation, factoring, synthetic division, and Newton's method.

8. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Let's not jump too quickly to that conclusion. It's possible that it is a much lower level course and these terms are just being thrown about in some effort to convince the students that what they are studying is relevant and will be useful down the road. But it certainly would be useful to know what level the OP is really at, because the indications thus far paint a contradictory picture, making it hard to choose a suitable way to describe things.

9. ### Papabravo Expert

Feb 24, 2006
11,779
2,495
I agree that the evidence does not paint a clear picture and that multiple interpretations are possible.

10. ### mcdara Thread Starter New Member

Feb 23, 2011
11
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Thanks!! it was the best class for a nap!

11. ### mcdara Thread Starter New Member

Feb 23, 2011
11
0
I'm a 4th year electrical engineering student. It was a very long 12 hour day and I was stressed over the sum and I came to the site for help. After a sleep I was able to realize I should have use that simple equation

12. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
That's not helping to clear up the picture any!

13. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Okay, now that IS helping clear things up. Lag in the server meant I didn't see this when making my last (off the cuff and not-to-be-taken-too-seriously) remark.