If one root of x²+px+q=0 is also the root of the equation x²+mx+n=0, then the other root of the first equation will satisfy the equation
a) nx²+mqx+q²=0
b)mx²+nqx+q²=0
c)mx²-mqx+q²=0
d) none
No spamming . answer needed with explanation
Answers
For quadratic equation ax2+bx+c=0 :
sum of roots =−b/a
product of roots =c/a
If one root of the equation
x2+px+q=0
is the square of the other, then roots are ω and ω2
sum of roots =−p⟹p=−ω2−ω
product of roots =q⟹q=ω2⋅ω=ω3
p3−q(3p−1)+q2
=(−ω2−ω)3−ω3(3(−ω2−ω)−1)+(ω3)2
=(−ω(ω+1))3−ω3(−3ω2−3ω−1)+ω6
=−ω3(ω3+3ω2+3ω+1)+3ω5+3ω4+ω3+ω6
=−ω6−3ω5−3ω4−ω3+3ω5+3ω4+ω3+ω6
=0
For quadratic equation ax2+bx+c=0 :
sum of roots =−b/a
product of roots =c/a
If one root of the equation
x2+px+q=0
is the square of the other, then roots are ω and ω2
sum of roots =−p⟹p=−ω2−ω
product of roots =q⟹q=ω2⋅ω=ω3
p3−q(3p−1)+q2
=(−ω2−ω)3−ω3(3(−ω2−ω)−1)+(ω3)2
=(−ω(ω+1))3−ω3(−3ω2−3ω−1)+ω6
=−ω3(ω3+3ω2+3ω+1)+3ω5+3ω4+ω3+ω6
=−ω6−3ω5−3ω4−ω3+3ω5+3ω4+ω3+ω6
=0
HOPE SO IT WILL HELP......