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are you asking who has a question
no he is gonna post lol
she is* my bad

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\(\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
hmm what did you try?
This info might be useful: https://en.wikipedia.org/wiki/Quintic_function#Quintics_in_Bring.E2.80.93Jerrard_form
putting a=5
honestly this question is beyond my IQ
why a=5 you are trying to find the constraints for a
for first 2 options it says to check roots for a>4
oh i see your line of thinking
answer given is both option (B.) and (D.)
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
and how do i factor that
Long divide by a factor (x-root)
via the Fund. thm of algebra
i don't even know what thm is but i think this needs some algebra thm application
i just need a value for \(a\) so that the expression is factorrable
for all constraints of \(a\) as given in options
you mean you want to go trial and error ?
yes
if it is the shortest soln
this is doable by calculus if we look for f" but you didn't take that yet
@freckles any bright ideas
let's google haha
haha lol
can we use descartus rule of signs here
Hey @mathmath333 you can't do derivatives right?
or calculus stuff?
i believe that rule won't help here
not yet at least
well i cannt but if u can use simple calculus then i can understand
example sith and giggles used calculus here ,that level i can understand http://openstudy.com/users/mathmath333#/updates/55818ea7e4b0636b8cc4098d
from here this confirms what you said B and D
https://www.desmos.com/calculator/dsmahabxqt
\[f(x)=x^5-5x+a \\ f'(x)=5x^4-5 \\ f'(x)=0 \text{ when } 5x^4-5=0 \\ 5x^4-5=0 \\ 5(x^4-1)=0 \\ (x^2-1)(x^2+1)=0 \\ x^2-1=0 \\ x=\pm 1 \\ f(1)=1-5+a \text{ and } f(-1)=-1+5+a \\ f(1)=a-4 \text{ and } f(-1)=a+4 \] I think we should play with the signs of f(-1) and f(1)
why did u took first derivative and equated it to zero
that's to find maxima and minima
y we need maxima and minima
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|
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
\[f(1) >0 \text{ and } f(-1)>0 \text{ then we have } a-4>0 \text{ and } a+4>0 \\ \text{ so } a>4 \text{ and } a>-4 \\ \text{ that intersection is when } a>4 \text{ and this is for one solution } \\ \] \[f(1)<0 \text{ and } f(-1)<0 \text{ then } a-4<0 \text{ and } a+4<0 \\ \text{ so } a<-4 \text{ for one solution }\] 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
or can you choose more than one answer?
two options B and D are correct
this is why learning calculus is much better than not learning calculus and just using your graphing calculator :) go calculus! :)
i will learn that after 2 years
fun stuff so if you haven't learned calculus than do you think we were suppose to use some other approach?
may be calculus was necessarily needed here, i thought it could be done with some algebra
as other questions
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
alone *
i dont see how B is correct
Wonderful explanation by @freckles :D
@ganeshie8 why do you say that?
Oh sry misread second option, thought it was saying 3 roots for a > 4
nice problem, nice discussion

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