If two roots of the equations $x^{3} - p x^{2} + q x - r = 0$ are equal in magnitude but opposite in sign, for

pr = q

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qr = p

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pq = r

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p^{2} q^{2} = r

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Solution

The correct option is D pq = r

let those are m, -m
now sum of three roots = p
hence third root will be p
now 
m*(-m) + m*p + (-m)*p = q
hence  –m² = q
now m*( –m) * p = r
 –m² p  = r
put value of  –m² = q
hence  pq = r

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If two roots of the equations x3−px2+qx−r=0 are equal in magnitude but opposite in sign, for

The correct option is D pq = rlet those are m, -mnow sum of three roots = phence third root will be pnow m*(-m) + m*p + (-m)*p = qhence –m2 = qnow m*( –m) * p = r –m2 p = rput value of –m2 = qhence pq = r

If two roots of the equations $x^{3} - p x^{2} + q x - r = 0$ are equal in magnitude but opposite in sign, for

The correct option is D pq = r
let those are m, -m
now sum of three roots = p
hence third root will be p
now
m(-m) + mp + (-m)p = q
hence –m² = q
now m( –m) * p = r
–m² p = r
put value of –m² = q
hence pq = r