Question

Hey!_____100 points_____!
Find the condition that the roots of the equation
${x}^{3} - p {x}^{2} + qx - r = 0$
may be in A. P. ​

Answer 1

Answer:

$2 {p}^{3} - 9pq + 27r = 0$

Solution :

Step-by-step explanation:

Let the roots of the given equation be a-d, a, a+d

Then:

(a-d) +a+(a+d)=

$\frac{ - ( - p)}{1}$

$= > \: a = \frac{p}{3}$

Since a is a root of the given equation Therefore :

${a}^{3} - p {a}^{2} + qa - r = 0$

$= > \: \frac{ {p}^{3} }{27} - \frac{ {p}^{3} }{9} + \frac{qp}{3} - r = 0$

Taking LCM of the LHS (i. e. 3, 9, 27) it will be 27 itself.

Now we will cross multiply and the LCM (i. e..27) obtained will be going to RHS according to cross multiplication rule and therefore RHS will be (27*0=0) 0.

$= > 2 {p}^{3} - 9pq + 27r = 0$

This is the required equation.

HOPE IT HELPS YOU