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mathmath333
 one year ago
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mathmath333
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you asking who has a question

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0no he is gonna post lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0she is* my bad

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{Let }\ a\in \mathbb{R} \ \normalsize \text{and let }\ f: \mathbb{R}\rightarrow \mathbb{R} \normalsize \text{be given by } \hspace{.33em}\\~\\ & f(x)=x^55x+a \hspace{.33em}\\~\\ & \normalsize \text{then } \hspace{.33em}\\~\\ & A.) \quad f(x)\ \normalsize \text{has three real roots if }\ a>4 \hspace{.33em}\\~\\ & B.) \quad f(x)\ \normalsize\text{has only one real roots if }\ a>4 \hspace{.33em}\\~\\ & C.) \quad f(x)\ \normalsize\text{has only three real roots if }\ a<4 \hspace{.33em}\\~\\ & D.) \quad f(x)\ \normalsize\text{has only three real roots if }\ 4<a<4 \hspace{.33em}\\~\\ \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm what did you try?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This info might be useful: https://en.wikipedia.org/wiki/Quintic_function#Quintics_in_Bring.E2.80.93Jerrard_form

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0honestly this question is beyond my IQ

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0why a=5 you are trying to find the constraints for a

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0for first 2 options it says to check roots for a>4

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0oh i see your line of thinking

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0answer given is both option (B.) and (D.)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm not sure if this might help suppose that has 3 distinct real roots therefore \[f(x)=x^55x+a=(xl)(xh)(xr)(Ax^2+Bx+C)\] where Ax^2+Bx+C has no real roots

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0and how do i factor that

P0sitr0n
 one year ago
Best ResponseYou've already chosen the best response.0Long divide by a factor (xroot)

P0sitr0n
 one year ago
Best ResponseYou've already chosen the best response.0via the Fund. thm of algebra

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i don't even know what thm is but i think this needs some algebra thm application

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i just need a value for \(a\) so that the expression is factorrable

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0for all constraints of \(a\) as given in options

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0you mean you want to go trial and error ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0if it is the shortest soln

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0this is doable by calculus if we look for f" but you didn't take that yet

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@freckles any bright ideas

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0let's google haha

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0can we use descartus rule of signs here

freckles
 one year ago
Best ResponseYou've already chosen the best response.7Hey @mathmath333 you can't do derivatives right?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i believe that rule won't help here

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0well i cannt but if u can use simple calculus then i can understand

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0example sith and giggles used calculus here ,that level i can understand http://openstudy.com/users/mathmath333#/updates/55818ea7e4b0636b8cc4098d

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0from here this confirms what you said B and D

freckles
 one year ago
Best ResponseYou've already chosen the best response.7\[f(x)=x^55x+a \\ f'(x)=5x^45 \\ f'(x)=0 \text{ when } 5x^45=0 \\ 5x^45=0 \\ 5(x^41)=0 \\ (x^21)(x^2+1)=0 \\ x^21=0 \\ x=\pm 1 \\ f(1)=15+a \text{ and } f(1)=1+5+a \\ f(1)=a4 \text{ and } f(1)=a+4 \] I think we should play with the signs of f(1) and f(1)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0why did u took first derivative and equated it to zero

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0that's to find maxima and minima

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0y we need maxima and minima

freckles
 one year ago
Best ResponseYou've already chosen the best response.7at 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 \[f'(x)=5x^45=5(x^41)=5(x^21)(x^2+1) \\ 5(x^21)(x^2+1)>0 \\ x^21>0 \\ x^2>1 \text{ so increasing on } x \in (\infty,1) \cup (1,\infty)\] \[\text{ so decreasing on } x \in (1,1)\] 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
 one year ago
Best ResponseYou've already chosen the best response.7as 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
 one year ago
Best ResponseYou've already chosen the best response.7\[f(1) >0 \text{ and } f(1)>0 \text{ then we have } a4>0 \text{ and } a+4>0 \\ \text{ so } a>4 \text{ and } a>4 \\ \text{ that intersection is when } a>4 \text{ and this is for one solution } \\ \] \[f(1)<0 \text{ and } f(1)<0 \text{ then } a4<0 \text{ and } a+4<0 \\ \text{ so } a<4 \text{ for one solution }\] 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 \[f(1)<0 \text{ and } f(1)>0 \\ a4<0 \text{ and } a+4>0 \\ a<4 \text{ and } a>4 \\ \text{ intersection is } 4<a<4 \text{ and this is for 3 real roots }\] but hmm.. I think I'm missing something because this gives us two answers above

freckles
 one year ago
Best ResponseYou've already chosen the best response.7or can you choose more than one answer?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0two options B and D are correct

freckles
 one year ago
Best ResponseYou've already chosen the best response.7this is why learning calculus is much better than not learning calculus and just using your graphing calculator :) go calculus! :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i will learn that after 2 years

freckles
 one year ago
Best ResponseYou've already chosen the best response.7fun stuff so if you haven't learned calculus than do you think we were suppose to use some other approach?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0may be calculus was necessarily needed here, i thought it could be done with some algebra

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0as other questions

freckles
 one year ago
Best ResponseYou've already chosen the best response.7maybe 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i dont see how B is correct

Here_to_Help15
 one year ago
Best ResponseYou've already chosen the best response.0Wonderful explanation by @freckles :D

freckles
 one year ago
Best ResponseYou've already chosen the best response.7@ganeshie8 why do you say that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Oh sry misread second option, thought it was saying 3 roots for a > 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nice problem, nice discussion
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