- #1
chwala
Gold Member
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- Homework Statement
- find the set of values of ##k## for which the equation
##3x^4+4x^3-12x^2+k =0## has 4 real roots
- Relevant Equations
- quartic polynomials
##x^2(3x^2+4x-12) +k=0##
##(3x^2+4x-12)= \frac{-k}{x^2}##
or
##(4x^3-12x^2)=-k-3x^4##
##4(3x^2-x^3)=3x^4+k##
##4x^2(3-x)= 3x^4+k##
or using turning points,
let ##f(x)= 3x^4+4x^3-12x^2+k##
it follows that,
##f'(x)=12x^3+12x^2-24x=0##
##12x(x^2+x-2)=0##
##12x(x-1)(x+2)=0## the turning points for this graph are at,
##x=0##, ## x=1## & ##x=-2##
or considering the first term and the constant, we have
##(ax+k=0)##, where ##a=3##, possible values for ## k = ±1, ±3...##
##(3x^2+4x-12)= \frac{-k}{x^2}##
or
##(4x^3-12x^2)=-k-3x^4##
##4(3x^2-x^3)=3x^4+k##
##4x^2(3-x)= 3x^4+k##
or using turning points,
let ##f(x)= 3x^4+4x^3-12x^2+k##
it follows that,
##f'(x)=12x^3+12x^2-24x=0##
##12x(x^2+x-2)=0##
##12x(x-1)(x+2)=0## the turning points for this graph are at,
##x=0##, ## x=1## & ##x=-2##
or considering the first term and the constant, we have
##(ax+k=0)##, where ##a=3##, possible values for ## k = ±1, ±3...##
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