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
Explanation:
Note that:
#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alpha beta#
So if
#{ (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#
Explanation:
We know that,
Squaring again,
This gives us,