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are you asking who has a question

no he is gonna post lol

she is* my bad

\(\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?

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.)

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

from here this confirms what you said B and D

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

why did u took first derivative and equated it to zero

that's to find maxima and minima

y we need maxima and minima

\[f(1) >0 \text{ and } f(-1)>0 \text{ then we have } a-4>0 \text{ and } a+4>0 \\ \text{ so } a>4 \text{ and } a>-4 \\ \text{ that intersection is when } a>4 \text{ and this is for one solution } \\ \]
\[f(1)<0 \text{ and } f(-1)<0 \text{ then } a-4<0 \text{ and } a+4<0 \\ \text{ so } a<-4 \text{ for one solution }\]
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

i will learn that after 2 years

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

as other questions

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