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# If the equation ${{x}^{4}}-\left( 3m+2 \right){{x}^{2}}+{{m}^{2}}=0\left( m>0 \right)$ has four real roots which are in A.P then the value of ‘m’ is

Last updated date: 23rd Jun 2024
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Hint: We solve this problem using the standard representation of four terms of an A.P as
$\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)$
Then we use the sum and product of terms of a polynomial of degree 4 that is
If $p,q,r,s$are the roots of equation $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$ then, the sum of roots is
$\Rightarrow p+q+r+s=\dfrac{-b}{a}$
The sum of product of roots taken two at a time is
$\Rightarrow pq+qr+rs+sp+pr+sq=\dfrac{c}{a}$
The product of roots
$\Rightarrow pqrs=\dfrac{e}{a}$
By using the above formulas to given equation we find the value of $'m'$

Complete step by step answer:
We are given that the polynomial equation of degree 4 as
$\Rightarrow {{x}^{4}}-\left( 3m+2 \right){{x}^{2}}+{{m}^{2}}=0$
Let us rewrite the above equation by placing the missing terms as
$\Rightarrow {{x}^{4}}+0{{x}^{3}}-\left( 3m+2 \right){{x}^{2}}+0x+{{m}^{2}}=0$
We are given that the roots of above equation are in A.P
We know that the standard representation of four terms of an A.P as
$\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)$
Let us assume that the roots of given equation as
$\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)$
We know that the sum of roots and product of roots of equation of degree 4 that is
If $p,q,r,s$are the roots of equation $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$ then, the sum of roots is
$\Rightarrow p+q+r+s=\dfrac{-b}{a}$
The sum of product of roots taken two at a time is
$\Rightarrow pq+qr+rs+sp+pr+sq=\dfrac{c}{a}$
The product of roots
$\Rightarrow pqrs=\dfrac{e}{a}$
By using the above formulas to given equation we find the value of $'m'$
Now, by using the sum of roots to given equation we get
\begin{align} & \Rightarrow a-3d+a-d+a+d+a+3d=\dfrac{-0}{1}=0 \\ & \Rightarrow 4a=0 \\ & \Rightarrow a=0 \\ \end{align}
Now the roots of given equation will be
$-3d,-d,d,3d$

Now by using the sum of product of roots taken two at a time we get
\begin{align} & \Rightarrow \left( -3d\times -d \right)+\left( -d\times d \right)+\left( d\times 3d \right)+\left( 3d\times -3d \right)+\left( -3d\times d \right)+\left( -d\times 3d \right)=\dfrac{-\left( 3m+2 \right)}{1} \\ & \Rightarrow -10{{d}^{2}}=-\left( 3m+2 \right) \\ & \Rightarrow {{d}^{2}}=\dfrac{\left( 3m+2 \right)}{10} \\ \end{align}
Now by using the product of roots formula we get
\begin{align} & \Rightarrow -3d\times -d\times d\times 3d=\dfrac{{{m}^{2}}}{1} \\ & \Rightarrow 9{{\left( {{d}^{2}} \right)}^{2}}={{m}^{2}} \\ \end{align}
Now, by substituting the value of $'{{d}^{2}}'$ in above equation we get
\begin{align} & \Rightarrow 9{{\left( \dfrac{3m+2}{10} \right)}^{2}}={{m}^{2}} \\ & \Rightarrow 9\left( 9{{m}^{2}}+12m+4 \right)=100{{m}^{2}} \\ & \Rightarrow 19{{m}^{2}}-108m-36=0 \\ \end{align}
Now by using the factorisation method that is rewriting the middle in such a way that we get factors then we get
\begin{align} & \Rightarrow 19{{m}^{2}}-114m+6m-36=0 \\ & \Rightarrow 19m\left( m-6 \right)+6\left( m-6 \right)=0 \\ & \Rightarrow \left( m-6 \right)\left( 19m+6 \right)=0 \\ \end{align}
We know that if $a\times b=0$ then either of $a,b$ will be zero.
By using this result to above equation we get the first term as
$\Rightarrow m=6$
Similarly, by taking the second term we get
$\Rightarrow m=\dfrac{-6}{19}$

We are given that $m>0$ so, the value of $'m'$ is 6.

Note: Students may make mistakes in taking the terms of an A.P.
The general representation of A.P is
$a,\left( a+d \right),\left( a+2d \right),\left( a+3d \right),.......$
Students may take these four terms which gives the complex solution.
But if we consider the terms as
$\left( a-3d \right),\left( a-d \right),\left( a+d \right),\left( a+3d \right)$
Here, we get the solution easily because while adding the terms we get only one variable remains which will be easy to solve.
This point needs to be taken care of.