If roots of #x^2-2x+3=0# are #alpha# and #beta#, find #alpha^4+beta^4#?

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Answer 1 · 2017-12-05T06:38:49

#alpha^4+beta^4=-14#

In a quadratic equation of the type #x^2-px+q=0#, sum of the roots is #p# and product of roots is #r#.

As roots of #x^2-2x+3=0# are #alpha# and #beta#

while #alpha+beta=2#, #alphabeta=3#

Therefore #alpha^2+beta^2#

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

= #2^2-2xx3=-2#

and #alpha^4+beta^4#

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

= #(-2)^2-2xx3^2#

= #4-18#

= #-14#