## mathmath333 one year ago question

1. anonymous

are you asking who has a question

2. xapproachesinfinity

no he is gonna post lol

3. xapproachesinfinity

4. 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<a<4 \hspace{.33em}\\~\\ \end{align}}

5. xapproachesinfinity

hmm what did you try?

6. anonymous

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

7. mathmath333

putting a=5

8. mathmath333

honestly this question is beyond my IQ

9. xapproachesinfinity

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

10. mathmath333

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

11. xapproachesinfinity

oh i see your line of thinking

12. mathmath333

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

13. 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

14. mathmath333

and how do i factor that

15. P0sitr0n

Long divide by a factor (x-root)

16. P0sitr0n

via the Fund. thm of algebra

17. xapproachesinfinity

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

18. mathmath333

i just need a value for $$a$$ so that the expression is factorrable

19. mathmath333

for all constraints of $$a$$ as given in options

20. xapproachesinfinity

you mean you want to go trial and error ?

21. mathmath333

yes

22. mathmath333

if it is the shortest soln

23. xapproachesinfinity

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

24. xapproachesinfinity

@freckles any bright ideas

25. xapproachesinfinity

26. mathmath333

haha lol

27. mathmath333

can we use descartus rule of signs here

28. freckles

Hey @mathmath333 you can't do derivatives right?

29. freckles

or calculus stuff?

30. xapproachesinfinity

i believe that rule won't help here

31. freckles

not yet at least

32. mathmath333

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

33. mathmath333

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

34. xapproachesinfinity

from here this confirms what you said B and D

35. mathmath333
36. 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)

37. mathmath333

why did u took first derivative and equated it to zero

38. xapproachesinfinity

that's to find maxima and minima

39. mathmath333

y we need maxima and minima

40. 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|

41. 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

42. 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 $f(1)<0 \text{ and } f(-1)>0 \\ a-4<0 \text{ and } a+4>0 \\ a<4 \text{ and } a>-4 \\ \text{ intersection is } -4<a<4 \text{ and this is for 3 real roots }$ but hmm.. I think I'm missing something because this gives us two answers above

43. freckles

or can you choose more than one answer?

44. mathmath333

two options B and D are correct

45. freckles

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

46. mathmath333

i will learn that after 2 years

47. freckles

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

48. mathmath333

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

49. mathmath333

as other questions

50. 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

51. freckles

alone *

52. ganeshie8

i dont see how B is correct

53. Here_to_Help15

Wonderful explanation by @freckles :D

54. freckles

@ganeshie8 why do you say that?

55. ganeshie8

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

56. anonymous

nice problem, nice discussion

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