if one root of ax²+bx+c=0 be square of another, then b³+a²c+ac²=kabc.find k=?
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given equation ax^2+ bx+c=0.
let its roots be £ and £^2.
£+£^2=-b\a
£^3=c\a
we have to find b^3+a^2c+ac^2=kabc then k =?
divide the whole by a^3.
b^3+a^2c+ac^2willbe equal to b^3\a^3+c\a+c^\a^2.
kabc will become kbc\a^2.
b\a= -(£^2+£).
c\a=£^3.
c^2\a^2=(£^3)^2
{-(£^+£)}^3+£^3+(£^3)^2
= -[£^6+£^3+3(£^3){£+£^2}]+£^3+£^6
= -£^6 -£^3-3(£^3){£+£^2}+£^3+£^6
=-3*c\a*(-b\a)
=3bc\a^2
3bc\a^2=kbc\a^2
so we get k=3
let its roots be £ and £^2.
£+£^2=-b\a
£^3=c\a
we have to find b^3+a^2c+ac^2=kabc then k =?
divide the whole by a^3.
b^3+a^2c+ac^2willbe equal to b^3\a^3+c\a+c^\a^2.
kabc will become kbc\a^2.
b\a= -(£^2+£).
c\a=£^3.
c^2\a^2=(£^3)^2
{-(£^+£)}^3+£^3+(£^3)^2
= -[£^6+£^3+3(£^3){£+£^2}]+£^3+£^6
= -£^6 -£^3-3(£^3){£+£^2}+£^3+£^6
=-3*c\a*(-b\a)
=3bc\a^2
3bc\a^2=kbc\a^2
so we get k=3
yashasvi47:
nice done
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