Question

If zeros of the polynomial $f\left( x \right)={{x}^{3}}-3p{{x}^{2}}+qx-r$ are in A.P., then?
A.$2{{p}^{3}}=pq-r$
B.$2{{p}^{3}}=pq+r$
C.${{p}^{2}}=pq-1$
D.None of these

Answer 1

Hint: Suppose zeros of the given cubic equation as $a-d,a,a+d$ . Now, use the relation among the zeros of cubic equation with the coefficient of the polynomial, which is given as
Sum of zeros $=\dfrac{-\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}$
Get the value of ‘a’ and put it to the given polynomial and equate it to 0 as ‘a’ is a zero of the polynomial. Now, get the required answer.

Complete step-by-step answer:
Polynomial $f\left( x \right)$ in the problem is given as
$\Rightarrow f\left( x \right)={{x}^{3}}-3p{{x}^{2}}+qx-r$ ……………………………………….(i)
   As, it is also given that zeros of the above polynomial are in A.P. and hence, we need to find relation among p, q and r.
So, as we know A.P. is a sequence of numbers with same successive difference which is termed as common difference of an A.P.
As the given polynomial is cubic, it will have three zeros. So, let us suppose zeroes of the polynomial are
$\Rightarrow a-d,a,a+d$ …………………………………………(ii)
Where the common difference between successive terms is ‘d’. So, they will be in A.P.
Now, as we know the relation among coefficients of a cubic polynomial and zeros of that polynomial is given as
Sum of zeros $=-\left( \dfrac{\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}} \right)$ …………………………..(iii)
Hence, we can write the equation according to the above identity and using equation (i) and (ii) as
$\Rightarrow \left( a-d+a+a+d \right)=-\left( \dfrac{-3p}{1} \right)$
$\Rightarrow 3a=3p$
$\Rightarrow a=p$ …………………………..(iv)
Now, we get the value of ‘a’ as ‘p’. As ‘a’ is a term of the A.P. given in equation (ii), so, it will be a zero of the polynomial $f\left( x \right)$ given in equation (i). So, ‘a’ will satisfy the equation (i) as well i.e. $f\left( a \right)=0$ .
Hence, as $a=p$ , so, we can write $f\left( p \right)=0$ ……………………………….(v)
Now, we can write the above equation from equation (i) by putting $x=p$ to it as
$\Rightarrow f\left( p \right)={{\left( p \right)}^{3}}-3p{{\left( p \right)}^{2}}+qp-r=0$
\[\Rightarrow {{p}^{3}}-3{{p}^{3}}+qp-r=0\]
\[\Rightarrow -2{{p}^{3}}+qp-r=0\]
\[\Rightarrow qp-r=2{{p}^{3}}\]
\[\Rightarrow 2{{p}^{3}}=qp-r\]
Hence, the relation among p, q, r is given as \[2{{p}^{3}}=qp-r\]
So, option (a) is the correct answer of the problem.

Note:One may suppose the terms of A.P. as \[a,a+d,a+2d\] which is the general representation of terms of an A.P. We can get the same answer with this may as well. Here, we will get value of \[\left( a+d \right)\] by using the relation
Sum of zeros $=\dfrac{-\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}$
Now, put the value of \[\left( a+d \right)\] to the given equation as, \[\left( a+d \right)\] is a root of a given expression.
We need to know the other relations of coefficients of any cubic with it’s zeroes as well. So, let us suppose a cubic equation as
\[\Rightarrow a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
And suppose roots/zeros of equation as \[\alpha ,\beta ,\gamma \] so, relations are given as:
\[\Rightarrow \alpha +\beta +\gamma =\dfrac{-b}{a}\]
\[\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}\]
\[\Rightarrow \alpha \beta \gamma =\dfrac{-d}{a}\]
Another approach to get relation among p, q, r for the problem would be that we can use above three relations altogether as
\[\Rightarrow \left( a-d \right)+a+\left( a+d \right)=3p\]
\[\Rightarrow \left( a-d \right)a+a\left( a+d \right)+\left( a-d \right)\left( a+d \right)=q\]
\[\Rightarrow a\left( a-d \right)\left( a+d \right)=r\]
Solve above equations to get the required answer.