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- mathmath333

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- schrodinger

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- anonymous

are you asking who has a question

- xapproachesinfinity

no he is gonna post lol

- xapproachesinfinity

she is* my bad

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## More answers

- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{Let }\ a\in \mathbb{R} \ \normalsize \text{and let }\ f: \mathbb{R}\rightarrow \mathbb{R} \normalsize \text{be given by } \hspace{.33em}\\~\\
& f(x)=x^5-5x+a \hspace{.33em}\\~\\
& \normalsize \text{then } \hspace{.33em}\\~\\
& A.) \quad f(x)\ \normalsize \text{has three real roots if }\ a>4 \hspace{.33em}\\~\\
& B.) \quad f(x)\ \normalsize\text{has only one real roots if }\ a>4 \hspace{.33em}\\~\\
& C.) \quad f(x)\ \normalsize\text{has only three real roots if }\ a<-4 \hspace{.33em}\\~\\
& D.) \quad f(x)\ \normalsize\text{has only three real roots if }\ -4

- xapproachesinfinity

hmm what did you try?

- anonymous

This info might be useful: https://en.wikipedia.org/wiki/Quintic_function#Quintics_in_Bring.E2.80.93Jerrard_form

- mathmath333

putting a=5

- mathmath333

honestly this question is beyond my IQ

- xapproachesinfinity

why a=5
you are trying to find the constraints for a

- mathmath333

for first 2 options it says to check roots for a>4

- xapproachesinfinity

oh i see your line of thinking

- mathmath333

answer given is both option (B.) and (D.)

- xapproachesinfinity

hmm not sure if this might help
suppose that has 3 distinct real roots
therefore
\[f(x)=x^5-5x+a=(x-l)(x-h)(x-r)(Ax^2+Bx+C)\]
where Ax^2+Bx+C has no real roots

- mathmath333

and how do i factor that

- P0sitr0n

Long divide by a factor (x-root)

- P0sitr0n

via the Fund. thm of algebra

- xapproachesinfinity

i don't even know what thm is
but i think this needs some algebra thm application

- mathmath333

i just need a value for \(a\) so that the expression is factorrable

- mathmath333

for all constraints of \(a\) as given in options

- xapproachesinfinity

you mean you want to go trial and error ?

- mathmath333

yes

- mathmath333

if it is the shortest soln

- xapproachesinfinity

this is doable by calculus if we look for f"
but you didn't take that yet

- xapproachesinfinity

@freckles any bright ideas

- xapproachesinfinity

let's google haha

- mathmath333

haha lol

- mathmath333

can we use descartus rule of signs here

- freckles

Hey @mathmath333 you can't do derivatives right?

- freckles

or calculus stuff?

- xapproachesinfinity

i believe that rule won't help here

- freckles

not yet at least

- mathmath333

well i cannt but if u can use simple calculus then i can understand

- mathmath333

example sith and giggles used calculus here ,that level i can understand
http://openstudy.com/users/mathmath333#/updates/55818ea7e4b0636b8cc4098d

- xapproachesinfinity

from here this confirms what you said B and D

- mathmath333

https://www.desmos.com/calculator/dsmahabxqt

- freckles

\[f(x)=x^5-5x+a \\ f'(x)=5x^4-5 \\ f'(x)=0 \text{ when } 5x^4-5=0 \\ 5x^4-5=0 \\ 5(x^4-1)=0 \\ (x^2-1)(x^2+1)=0 \\ x^2-1=0 \\ x=\pm 1 \\ f(1)=1-5+a \text{ and } f(-1)=-1+5+a \\ f(1)=a-4 \text{ and } f(-1)=a+4 \]
I think we should play with the signs of f(-1) and f(1)

- mathmath333

why did u took first derivative and equated it to zero

- xapproachesinfinity

that's to find maxima and minima

- mathmath333

y we need maxima and minima

- freckles

at first I was looking at cases for the above and deleted all of that because it was totally unnecessary after playing with the cases and a little calculus on the cases
to get a better sense of the graph)
and where the function is increasing or decreasing if at all
\[f'(x)=5x^4-5=5(x^4-1)=5(x^2-1)(x^2+1) \\ 5(x^2-1)(x^2+1)>0 \\ x^2-1>0 \\ x^2>1 \text{ so increasing on } x \in (-\infty,1) \cup (1,\infty)\]
\[\text{ so decreasing on } x \in (-1,1)\]
anyways if both f(1) and f(-1) are positive then your graph looks like this:
|dw:1434569492292:dw|
and if we have both f(1) and f(-1) are negative then your graph looks like:
|dw:1434569524643:dw|
now if f(1) is negative and f(-1) is positive your graph looks like this:
|dw:1434569608801:dw|

- freckles

as we see when both f(1) and f(-1) are same sign we have one real root
but if f(-1) and f(1) have difference sign we have 3 real root

- freckles

\[f(1) >0 \text{ and } f(-1)>0 \text{ then we have } a-4>0 \text{ and } a+4>0 \\ \text{ so } a>4 \text{ and } a>-4 \\ \text{ that intersection is when } a>4 \text{ and this is for one solution } \\ \]
\[f(1)<0 \text{ and } f(-1)<0 \text{ then } a-4<0 \text{ and } a+4<0 \\ \text{ so } a<-4 \text{ for one solution }\]
now if they have difference signs and we need f(1) to be negative on this case since between -1 and 1 the function is decreasing
\[f(1)<0 \text{ and } f(-1)>0 \\ a-4<0 \text{ and } a+4>0 \\ a<4 \text{ and } a>-4 \\ \text{ intersection is } -4

- freckles

or can you choose more than one answer?

- mathmath333

two options B and D are correct

- freckles

this is why learning calculus is much better than not learning calculus and just using your graphing calculator
:)
go calculus! :)

- mathmath333

i will learn that after 2 years

- freckles

fun stuff
so if you haven't learned calculus than do you think we were suppose to use some other approach?

- mathmath333

may be calculus was necessarily needed here, i thought it could be done with some algebra

- mathmath333

as other questions

- freckles

maybe
not sure
I can't see algebra along being used here
though that doesn't mean it can't be
it could just mean I'm blind

- freckles

alone *

- ganeshie8

i dont see how B is correct

- Here_to_Help15

Wonderful explanation by @freckles :D

- freckles

@ganeshie8 why do you say that?

- ganeshie8

Oh sry misread second option, thought it was saying 3 roots for a > 4

- anonymous

nice problem, nice discussion

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