Question

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

Answer 1

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

$hope \: \: it \: \: helps .....$

Answer 2

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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......