Question

find the condition that the zeroes of the polynomial f(x)=x3+3px2+3qx+r may be in A.P.

Answer 1

this is the answer to your question, follow the attatchment

Answer 2

The condition that the zeroes of the polynomial are in AP is 3pq = 2p³+r.

• Let the zeroes of the polynomial in AP be a-d , a , a+d .

• Now the equation of polynomial is  f(x)=x³+3px²+3qx+r

• The standard form of the equation is x³-(a+b+c)x²+(ab+bc+ca)x-abc = 0.

• Comparing the equation with the standard equation, we get

Sum of roots = -3p

(a-d+a+a+d) = -3p

3a = -3p

p = -a                        (Equation 1)

• Now, product of roots = -r

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

r = ad²-a³

r = a(d²-a²)

r = -p(d²-p²)                                    (From equation 1)

d² = $\frac{p^3-r}{p}$       (Equation 2)

• Now, sum of product of two roots =

a(a-d) + a(a+d) + (a-d)(a+d) = 3q

3q = a²- ad + a² +ad + a² - d²

3q = 3a² - d²                  

Substituting the values of a and d from equation 1 and 2,

3q = 3p² - $\frac{p^3-r}{p}$

3pq = 3p³ - p³+r

3pq = 2p³+r

• 3pq = 2p³ + r is the required condition.