Question

if one root of the equation x²+p x+q=0 be the square of the other , then p³+q²+q=3pq​

Answer 1

Answer:

If one root of the equation x2+px+q=0 is the square of the other, then how do I show that p3−q(3p−1)+q2=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

Answer 2

Question :-

if one root of the equation x²+p x+q=0 be the square of the other , then p³+q²+q=3pq

Solution :-

$\small{ \mathrm{For \: quadratic \: equation \: {ax}^{2} +bx+c=0 :}}$

• sum of roots =−b/a
• product of roots =c/a

If one root of the equation

$\small{ \mathrm{ {x}^{2} +px+q=0 \: is \: the \: square \: of \: the \: other, \: then \: roots \: are \: ω \: and \: {ω}^{2} }}$

$\small{ \mathrm{sum \: of \: roots \: = \: −p \: ⟹ \: p= \: − {ω}^{2} \: −ω \: product \: of \: roots \: }}$

$\small{ \mathrm{ = \: q \: ⟹ \: q \: = \: {ω}^{2} \: ⋅ \: ω \: = \: {ω}^{3} }}$

$\small{ \mathrm{ {p}^{3} −q(3p−1)+ {q}^{2} {p}^{3} −q(3p−1)+ {q}^{2} }}$

$\small{ \mathrm{=(− {ω}^{2} −ω)^3− {ω}^{3} (3(−ω2−ω)−1)+( {ω}^{3} )^2}}$

$\small{ \mathrm{=(−ω(ω+1))^3− {ω}^{3} (−3 {ω}^{2} −3ω−1)+ {ω}^{6} </p><p>}}$

$\small{ \mathrm{=−ω^3(ω^3+3ω^2+3ω+1)+3ω^5+3ω^4+ω^3+ω^6}}$

$\small{ \mathrm{=−ω^6−3ω^5−3ω^4−ω^3+3ω^5+3ω^4+ω^3+ω^6}}$

$\small{ \mathrm{= \: 0}}$