Question

Step-by-Step How do i solve
4x^(2)-5x+5

Answer 1

it has no real roots

Answer 2

quadratic formula, there is no step by step to show, just plug in
a=4
b=-5
c=5
if you don't know the quadratic formula look it up and memorize it, I don't feel like typing it

Answer 3

is there something to solve here?

Answer 4

i don't see anything to solve

Answer 5

oh yeah, is it
4x^(2)-5x+5=0
?
otherwise myininaya is right
should've noticed that

Answer 6

@joyce
you have to put an equals sign if you want us to do anything other than simplify, which in this case cannot be done

be sure to type the entire problem, it makes all the difference

Answer 7

this parabola opens up and vertex is above x-axis

Answer 8

IF it is 4x^(2)-5x+5=y

Answer 9

well the Q is asking me to factor 4x^(2)-5x+5.

Answer 10

\[x=\frac18(5\pm\sqrt{16y-55})\]

Answer 11

can't be done

Answer 12

not over the integers anyways

Answer 13

Oh, it can be done... just not over the reals.

Answer 14

since the discriminant
\[b ^{2}-4ac = 25 - 4(4)(5) = -55\]

Answer 15

right, of course

Answer 16

so it can be factored over the complex
do you guys want to see?
i can factor this by grouping

Answer 17

yeah, let's see it

Answer 18

\[\frac{1}{16} (-8 i x+\sqrt{55}+5 i) (8 i x+\sqrt{55}-5 i)\]

Answer 19

there it is, eh?

Answer 20

That's the lowest form I can find, at least.

Answer 21

But, I suck at factoring :D

Answer 22

let's see what myininaya gets

Answer 23

\[ax^2+bx+c=4x^2-5x+5=> a=4; b=-5;c=5\]
\[b=(\frac{-5}{2}+z)+(\frac{-5}{2}-z)\]
\[a \cdot c=(\frac{-5}{2}+z)(\frac{-5}{2}-z)\]
We need to find what z
remember that a*c=4(5)=20 so we can solve for z
\[20=\frac{25}{4}-z^2 => 20-\frac{25}{4}=-z^2 => z^2=\frac{25}{4}-20\]
\[z^2=\frac{25}{4}-\frac{20 \cdot 4}{4}=\frac{25-80}{4}=\frac{-55}{4}\]
\[z=\pm \sqrt{\frac{-55}{4}}=\pm i \frac{\sqrt{55}}{2}\]
We will only need \[z=i \frac{\sqrt{55}}{2}\]
so we have
\[bx=(\frac{-5}{2}+i \frac{\sqrt{55}}{2})x+(-\frac{5}{2}-i \frac{\sqrt{55}}{2})x\]
so we have
\[4x^2-5x+5\]
now we will replace -5x as follows
\[4x^2+(\frac{-5}{2}+i \frac{\sqrt{55}}{2})x+(\frac{-5}{2}-i \frac{\sqrt{55}}{2})x+5\]
Now the factor by grouping happens! :)
\[4x(x+\frac{-5+i \sqrt{55}}{8})+\frac{-5 - i \sqrt{55}}{2}(x+5 \cdot \frac{2}{-5 - i \sqrt{55}})\]
Believe it or not we are almost there
Now we will multiply top and bottom of 2/(-5-i sqrt{55}) by the conjugate of bottom
\[4x(x+\frac{-5+i \sqrt{55}}{8})+\frac{-5-i \sqrt{55}}{2}(x+5 \cdot \frac{2(-5+i \sqrt{55})}{25-i^2 \cdot 55})\]
\[4x(x+\frac{-5 + i \sqrt{55}}{8})+\frac{-5 - i \sqrt{55}}{2}(x+5 \cdot \frac{2(-5+i \sqrt{55})}{80})\]
\[4x(x+\frac{-5 + i \sqrt{55}}{8})+\frac{-5- i \sqrt{55}}{2}(x+\frac{-5 + i \sqrt{55}}{8})\]
Notice we have a common factor in each term
\[(x+\frac{-5 + i \sqrt{55}}{8})(4x+\frac{-5 - i \sqrt{55}}{2})\]

Answer 24

No, I just now finished reading both your link and proof and studied it closely, so it took a while to really get each step. I've never seen that before and really thought it was awesome, and saved it to my bookmarks. Thanks for showing me that it I really like to know how to do those tricky algebra things, they come in handy so often. Really epic, thanks! :D

Answer 25

in the link? no way!

Answer 26

Wow, very impressive. May I ask what you do or where you study?

Answer 27

yeah, truly awesome :)

Answer 28

Well, I knew you were pretty good but...
wow, thanks.

Answer 29

:)

Answer 30

I have to assume you're a math major. Post Grad?

Answer 31

Great stuff, I enjoyed it.

Answer 32

haha... yeah, I figured you were working with thesis-level material here.

Answer 33

No, I just mean you're level was obvious. I'm sure you're not going to post your thesis on open study anytime soon.

Answer 34

your*

Answer 35

There's always a lot to learn.

Answer 36

Myiniaya I think I've seen something similar to your methodology you posted... it's kind of like the shortcut for complex factorization, only you used the quadratic as your mode... I'll look around my stuff and see if I can pull up the one that's similar... might take a bit, I think I learnt it in complex analysis as an undergrad >.<

Answer 37

I'm just a first year physics major who has studied as much as I can online, so I can't relate to much high-level material. I do like differential equations though, and would consider looking more into PDF theory.

Answer 38

PDE*
partial differential equations
typo

Answer 39

Most people have to take at least one physics course for their undergrad years

Answer 40

It seems most have taken at least up to electromagnetism.

Answer 41

you're physics teacher thought it was unfair for him to take math classes?

Answer 42

I mean THE physics teacher, sorry

Answer 43

but yeah I would hope so

Answer 44

The only paper I found from my notebook I thought it would be in was a proof that the roots of a polynomial can be multiplied to give the factorization.

Which was how I did this problem... completed the square for roots and then multiplied them... more or less, the coeffs could be factored back in before the multiplication, so I did that, too.

I'll keep looking around, I coulda sworn I saw something very similar >.<

Answer 45

well that's partly why I tutor people here in math more than in physics.

Answer 46

lol I was exposed to vectors via math years before I started taking physics courses... made that bit much easier ;)

Answer 47

just the opposite for me.

Answer 48

I learned vectors in a physics class

Answer 49

I had to deal with vectors back in high school algebra O_o

Answer 50

how's that possible

Answer 51

yeah, that's all I learned in high school. Now I can do basic vector calculus and a fair amount of linear algebra .

Answer 52

that's about all I know of vectors

Answer 53

I had AP though

Answer 54

Yeah, we only did basic stuff... but it's things I see people learning in physics 1. Of course, by the time I was a senior we were doing basic vector calculus and in my non-euclidian geometry class we dealt with the problems that come up with vectors in those spaces.

Answer 55

...ever since that class i've been in love with projected geometry...

Also, I live in SW Missouri, so you arkansas people were always being made fun of :P

Answer 56

lol, so funny you seem to embrace your stereotype.

Answer 57

Yeah, so where are they all smart? You're gonna say Boston, right?

Answer 58

That's why I live in Mexico!

Answer 59

Projective Geometry is fun, I promise: Just look at how spheres are handled (about 1/2 way down the page)

http://www.nct.anth.org.uk/basics.htm

Answer 60

yup University of Gudalajara

Answer 61

I've been thinking about taking some classes at Universidad Autonoma de Yucatan (Merida)... fell in love with Merida when I first went there... god over a decade ago :X

Answer 62

I'm trying to get to UNAM eventually, but I just got here and first need to get my papers in order... and a few credits under my belt too.

Answer 63

UNAM has an awesome library.

Answer 64

http://bnm.unam.mx/
and such.

Answer 65

That I didn't know; yet another reason to go.

Answer 66

thanks