If #alpha#, #beta# are the roots of the equation #x^2 +px + q = 0#, find the value of #alpha^4 + alpha^4 beta^4 + beta^4#?

2 Answers
Jun 23, 2018

#alpha^4+alpha^4beta^4+beta^4 = p^4-4p^2q+2q^2+q^4#

Explanation:

Note that:

#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alpha beta#

So if #alpha, beta# are the roots of #x^2+px+q = 0# then:

#{ (alpha + beta = -p), (alpha beta = q) :}#

Then we find:

#alpha^2+beta^2 = (alpha+beta)^2-2alpha beta#

#color(white)(alpha^2+beta^2) = (-p)^2-2q#

#color(white)(alpha^2+beta^2) = p^2-2q#

Then:

#alpha^4+beta^4 = (alpha^2+beta^2)^2-2(alpha^2 beta^2)#

#color(white)(alpha^4+beta^4) = (p^2-2q)^2-2q^2#

#color(white)(alpha^4+beta^4) = p^4-4p^2q+4q^2-2q^2#

#color(white)(alpha^4+beta^4) = p^4-4p^2q+2q^2#

So:

#alpha^4+alpha^4beta^4+beta^4 = (alpha^4+beta^4)+(alphabeta)^4#

#color(white)(alpha^4+alpha^4beta^4+beta^4) = p^4-4p^2q+2q^2+q^4#

Jun 23, 2018

# p^4-4p^2q+2q^2+q^4#.

Explanation:

We know that, #alpha, beta#, being the roots of the quadr. eqn.

#x^2+px+q=0#, satisfy the following relation :

#alpha+beta=-p, and, alphabeta=q#.

#:. (alpha+beta)^2=(-p)^2,#

# i.e., alpha^2+2alphabeta+beta^2=p^2, #

# or, alpha^2+2(q)+beta^2=p^2#.

# rArr alpha^2+beta^2=p^2-2q#.

Squaring again, # alpha^4+2alpha^2beta^2+beta^4=(p^2-2q)^2#.

#:. alpha^4+2(q)^2+beta^4=p^4-4p^2q+4q^2#,

# or, alpha^4+beta^4=p^4-4p^2q+2q^2#.

This gives us,

#"The Desired Value"=alpha^4+beta^4+alpha^4beta^4,#

#=p^4-4p^2q+2q^2+q^4#.