Asked by Amy
How would you solve polynomial equation for, a to the 8th power minus a to the 2nd power and b to the 6th power?
Answers
Answered by
Steve
you mention no equation, but if it is
a^8 - a^2b^6 = 0
then
a^2(a^6-b^6) = 0
a^2(a^2-b^2)(a^4+a^2b^2+b^4) = 0
so, a=0
or a = ±b
or a^2 = (-b^2 ± √(b^4 - 4a^4b^4))/2
= (-b^2 ± b^2√(1-4a^4))/2
= -b^2/2 (1 ± √(1-4a^4))
a^8 - a^2b^6 = 0
then
a^2(a^6-b^6) = 0
a^2(a^2-b^2)(a^4+a^2b^2+b^4) = 0
so, a=0
or a = ±b
or a^2 = (-b^2 ± √(b^4 - 4a^4b^4))/2
= (-b^2 ± b^2√(1-4a^4))/2
= -b^2/2 (1 ± √(1-4a^4))
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