polynomials + integration q (1 Viewer)

[bally24]

1. The equation x³ + qx + r has roots alpha (a), beta (b) and gamma (c). Find expressions for
a) a² + b² + c²
b) a⁴ + b⁴ + c⁴
c) a⁵ + b⁵ + c⁵
<o></o>
for part a) I just did (a+b+c)² – 2(ab + bc + cd) using the sum of roots, product of roots etc.
however I’m really stuck on parts b) and c) – ive tried multiplying various things together and doing (a + b + c)⁴ then factorising but nothing seems to work to get it in terms of the roots taken one at a time, two at a time, three at a time. so i really have no idea....

2. Integrate (tanx)⁴ dx
<o></o>
I tried changing it to (sinx)⁴/(cosx)⁴ but that doesn’t help at all, I couldn’t make it into a log without adding in pronumerals which you can’t do. I tried expressing it as ((secx)² – 1 )² but that just gets really tricky. Is there a faster way?

 

[hyparzero]

f(x) = x³ + qx + r has roots a, b, c ; hence the roots must satisfy f(x).

Hence
a³ = - (qa + r)
b³ = - (qb + r)
c³ = - (qc + r)

thus;
a⁴ = - a(qa + r)
b⁴ = - b(qb + r)
c⁴ = - c(qc + r)

Hence a⁴ + b⁴ + c⁴ = - [ a(qa + r) + b(qb + r) + c(qc + r)]

Do the same thing for degree 5 roots

 

[bally24]

oh right...that seems so simple now, thanks for your help with that one!

 

[onebytwo]

for the integral try changing it into part sec^2(x) - 1 and part tan^2(x) then try and use a substitution

 

[Riviet]

tan⁴x=tan²x(sec²x-1)
=tan²x.sec²x - tan²x
=tan²x.sec²x - (sec²x - 1)

Split integral into two parts and use substitution u=tanx to integrate tan²x.sec²x with respect to x. The other part is easy.

Hope that helps.

 

[bally24]

ok thanks, i forgot about using substitutions.