Question

The equation ${{x}^{n}}+p{{x}^{2}}+qx+r=0$, where $n\ge 5$ & $r\ne 0$ has roots ${{\alpha }_{1}}$, ${{\alpha }_{2}}$, ${{\alpha }_{3}}$,……, ${{\alpha }_{n}}$. Denoting $\sum\limits_{i=1}^{n}{\alpha _{i}^{k}}$ by ${{S}_{k}}$, then
(a) ${{S}_{2}}=0$, ${{S}_{n}}=nr$.
(b) ${{S}_{2}}=0$, ${{S}_{n}}=-nr$.
(c) ${{S}_{2}}=1$, ${{S}_{n}}=-nr$.
(d) ${{S}_{2}}=1$, ${{S}_{n}}=nr$.

Answer 1

We start solving the problem by the finding the sum of the roots of the given equation using the fact the sum of the roots $\dfrac{-1\times \text{coefficient of }{{x}^{n-1}}}{\text{coefficient of }{{x}^{n}}}$, which will be the value of ${{S}_{1}}$. We use this value and the fact that ${{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}}=\dfrac{\text{coefficient of }{{x}^{n-2}}}{\text{coefficient of }{{x}^{n}}}$ to find the value of ${{S}_{2}}$. We then substitute each root into the given equation one by one and add all of them at once to get the relation between ${{S}_{1}}$, ${{S}_{2}}$, ${{S}_{n}}$ and r. We then make necessary substitutions and calculations to get the required value of ${{S}_{n}}$.

Complete step-by-step solution
According to the problem, we are given that ${{\alpha }_{1}}$, ${{\alpha }_{2}}$, ${{\alpha }_{3}}$,……, ${{\alpha }_{n}}$ are the roots of ${{x}^{n}}+p{{x}^{2}}+qx+r=0$, where $n\ge 5$ & $r\ne 0$. We need to find the values of ${{S}_{2}}$, ${{S}_{n}}$ if ${{S}_{k}}=\sum\limits_{i=1}^{n}{\alpha _{i}^{k}}$.
We know that the sum of the roots ${{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}+......+{{\alpha }_{n}}=\dfrac{-1\times \text{coefficient of }{{x}^{n-1}}}{\text{coefficient of }{{x}^{n}}}$.
$\Rightarrow {{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}+......+{{\alpha }_{n}}=\dfrac{-1\times 0}{1}$.
$\Rightarrow {{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}+......+{{\alpha }_{n}}=0$ ---(1).
Now, let us find the value of ${{S}_{1}}$.
We have ${{S}_{1}}=\sum\limits_{i=1}^{n}{{{\alpha }_{i}}}={{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}+......+{{\alpha }_{n}}$.
From equation (1), we get
$\Rightarrow {{S}_{1}}=0$ ---(2).
Now, let us find the value of ${{S}_{2}}$.
${{S}_{2}}=\sum\limits_{i=1}^{n}{\alpha _{i}^{2}}=\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+......+\alpha _{n}^{2}$ ---(3).
We know that $\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+......+\alpha _{n}^{2}={{\left( {{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}+......+{{\alpha }_{n}} \right)}^{2}}-2\left( {{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}} \right)$.
From equation (1), we get
$\Rightarrow \alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+......+\alpha _{n}^{2}={{\left( 0 \right)}^{2}}-2\left( {{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}} \right)$.
$\Rightarrow \alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+......+\alpha _{n}^{2}=-2\left( {{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}} \right)$ ---(4).
We know that the ${{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}}=\dfrac{\text{coefficient of }{{x}^{n-2}}}{\text{coefficient of }{{x}^{n}}}$.
$\Rightarrow {{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}}=\dfrac{0}{1}$.
$\Rightarrow {{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}+.............+{{\alpha }_{n-2}}{{\alpha }_{n}}+{{\alpha }_{n-1}}{{\alpha }_{n}}=0$. We substitute this in equation (5).
$\Rightarrow \alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+......+\alpha _{n}^{2}=-2\left( 0 \right)$.
$\Rightarrow \alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+......+\alpha _{n}^{2}=0$. We substitute this in equation (3).
So, we get ${{S}_{2}}=0$ ---(5).
Now, we know that the roots ${{\alpha }_{1}}$, ${{\alpha }_{2}}$, ${{\alpha }_{3}}$,……, ${{\alpha }_{n}}$ satisfy the given equation ${{x}^{n}}+p{{x}^{2}}+qx+r=0$.
Let us substitute each of the roots in the equation ${{x}^{n}}+p{{x}^{2}}+qx+r=0$.
On substituting ${{\alpha }_{1}}$, we get ${{\left( {{\alpha }_{1}} \right)}^{n}}+p{{\left( {{\alpha }_{1}} \right)}^{2}}+q\left( {{\alpha }_{1}} \right)+r=0$.
$\Rightarrow \alpha _{1}^{n}+p\alpha _{1}^{2}+q{{\alpha }_{1}}+r=0$ ---(6).
On substituting ${{\alpha }_{2}}$, we get ${{\left( {{\alpha }_{2}} \right)}^{n}}+p{{\left( {{\alpha }_{2}} \right)}^{2}}+q\left( {{\alpha }_{2}} \right)+r=0$.
$\Rightarrow \alpha _{2}^{n}+p\alpha _{2}^{2}+q{{\alpha }_{2}}+r=0$ ---(7).
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On substituting ${{\alpha }_{n}}$, we get ${{\left( {{\alpha }_{n}} \right)}^{n}}+p{{\left( {{\alpha }_{n}} \right)}^{2}}+q\left( {{\alpha }_{n}} \right)+r=0$.
$\Rightarrow \alpha _{n}^{n}+p\alpha _{n}^{2}+q{{\alpha }_{n}}+r=0$ ---(8).
Let us add the equations (6), (7), (8), and the other similar equations we get on substituting the other roots.
So, we get $\sum\limits_{i=1}^{n}{\alpha _{i}^{n}}+p\sum\limits_{i=1}^{n}{\alpha _{i}^{2}}+q\sum\limits_{i=1}^{n}{{{\alpha }_{i}}}+\sum\limits_{i=1}^{n}{r}=0$.
$\Rightarrow {{S}_{n}}+p{{S}_{2}}+q{{S}_{1}}+nr=0$.
From equations (2) and (5), we get
$\Rightarrow {{S}_{n}}+p\left( 0 \right)+q\left( 0 \right)+nr=0$.
$\Rightarrow {{S}_{n}}+0+0+nr=0$.
$\Rightarrow {{S}_{n}}+nr=0$.
$\Rightarrow {{S}_{n}}=-nr$.
So, we have found the values of ${{S}_{2}}$ and ${{S}_{n}}$ as 0 and $-nr$.
$\therefore$ The correct option for the given problem is (b).

Note: Whenever we get the quadratic equations, we need not always find the roots and do the squaring or sum. We can sometimes use the sum and product of the roots while solving this type of problem. We should make sure that the sum of the product of two roots, the sum of the product of three roots, etc is performed without any mistake. We can see that the sum of the squares of the roots of given equations is zero which is possible only if all the roots are zero but according to the problem it is given $r\ne 0$, which is the product of all the roots. So, this clearly tells us that the given equation doesn’t have any real roots.