Question from my trials (1 Viewer)

[Affinity]

here's a part of a question from my trials today which I believe to be incorrect:

P(x)= x^5 - 5cx + 1
Prove that for c > (1/4)^5, P(x) has 3 distinct real roots.

Any ideas?

 

[underthesun]

I checked it with graph calculator.. and using that minimum c value, it doesn't even touch the x axis, where it's supposed to have a minimum double root there, if the question is right..

 

[underthesun]

Your teacher made a mistake.

4^-1/5 is the root (X_o), if there is a double root with varying c.

The minimum c value is 4^-4/5. Try it on a graph calculator.

The difficult part is the indices.. with minus and fractions.

 

[Affinity]

thanks guys, feel alot more reassured

 

[Richard Lee]

Originally posted by Affinity
here's a part of a question from my trials today which I believe to be incorrect:

P(x)= x^5 - 5cx + 1
Prove that for c > (1/4)^5, P(x) has 3 distinct real roots.
Any ideas?

u r right. there r two turning pts, sub them back to the P(x), it must be:
P(-4th sqrt(c))*P(4th sqrt(c))<0. So, the answer should be: c>(1/4)^(4/5)

 

[Harimau]

I suck... Can someone please post on how they got the value for C? I got the part about the two stationary points by myself, but couldnt go further than that.

 

[ND]

Originally posted by Harimau
I suck... Can someone please post on how they got the value for C? I got the part about the two stationary points by myself, but couldnt go further than that.

When you have the two turning pts (say they're x1 and x2), for the curve to have 3 distinct roots, P(x1)*P(x2)<0. Just sub in and solve for C.