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Hint: Use identities of the Quadratic polynomial and simply solve them.

Given: ${x^4} - p{x^3} + q{x^2} - rx + s = 0{\text{ }} \ldots \left( 1 \right)$

Let the four roots of the given quadratic polynomial be $\alpha ,\beta ,\gamma ,\delta $

Case 1: Sum of two roots is equal to the sum of other two roots,

$\therefore \alpha + \delta = \beta + \gamma {\text{ }} \ldots \left( 2 \right)$

General quadratic equation: $a{x^4} + b{x^3} + c{x^2} + dx + e = 0{\text{ }} \ldots \left( 3 \right)$

Comparing equation (1) and (3), we get:

$a = 1,{\text{ }}b = - p,{\text{ }}c = q,{\text{ }}d = - r,{\text{ }}e = s{\text{ }} \ldots \left( 4 \right)$

As we know, the sum of roots of a quadratic equation is $ - \dfrac{b}{a}$ .

$

\Rightarrow \alpha + \beta + \gamma + \delta = - \dfrac{b}{a} \\

\Rightarrow \alpha + \beta + \gamma + \delta = - \left( {\dfrac{{ - p}}{1}} \right){\text{ }}\left( {{\text{using }}\left( 4 \right)} \right) \\

\Rightarrow \alpha + \beta + \gamma + \delta = p \\

\Rightarrow \alpha + \beta = \gamma + \delta = \dfrac{p}{2}{\text{ }}\left( {{\text{using }}\left( 2 \right)} \right) \ldots \ldots \left( 5 \right) \\

$

Also, the sum of products of roots of a quadratic equation is $\dfrac{c}{a}$ .

$

\Rightarrow \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \dfrac{c}{a} \\

\Rightarrow \left( {\alpha + \delta } \right)\left( {\beta + \gamma } \right) + \alpha \delta + \beta \gamma = q{\text{ }}\left( {{\text{using }}\left( 4 \right)} \right) \\

$

Now, substituting values of $\left( {\alpha + \delta } \right)$and $\left( {\beta + \gamma } \right)$from equation (5), we get:

$

\Rightarrow \left( {\dfrac{p}{2}} \right)\left( {\dfrac{p}{2}} \right) + \alpha \delta + \beta \gamma = q{\text{ }}\left( {{\text{using }}\left( 5 \right)} \right) \\

\Rightarrow \alpha \delta + \beta \gamma = q - \dfrac{{{p^2}}}{4}{\text{ }} \ldots \left( 6 \right) \\

$

Now, the sum of products of roots of a quadratic equation is $r$.

$

\Rightarrow \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = r \\

\Rightarrow \beta \gamma \left( {\alpha + \delta } \right) + \alpha \delta \left( {\beta + \gamma } \right) = r \\

$

Now, substituting values of $\left( {\alpha + \delta } \right)$and $\left( {\beta + \gamma } \right)$from equation (5), we get:

$ \Rightarrow \beta \gamma \left( {\dfrac{p}{2}} \right) + \alpha \delta \left( {\dfrac{p}{2}} \right) = r{\text{ }}\left( {{\text{using }}\left( 5 \right)} \right)$

Taking $\dfrac{p}{2}$ common, we get:

$ \Rightarrow \dfrac{p}{2}\left( {\beta \gamma + \alpha \delta } \right) = r{\text{ }} \ldots \left( 7 \right)$

Putting value of $\left( {\beta \gamma + \alpha \delta } \right)$from equation (6), we get:

$

\Rightarrow \dfrac{p}{2}\left( {q - \dfrac{{{p^2}}}{4}} \right) = r{\text{ }}\left( {{\text{using }}\left( 6 \right)} \right) \\

\Rightarrow {p^3} - 4pq + 8r = 0 \\

$

Case 2: Product of two roots is equal to the product of other two roots, therefore $\alpha \delta = \beta \gamma {\text{ }} \ldots \left( 8 \right)$

We know: $\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = r$

$

\Rightarrow \beta \gamma \left( {\alpha + \delta } \right) + \alpha \delta \left( {\beta + \gamma } \right) = r \\

\Rightarrow \alpha \delta \left( {\alpha + \beta + \gamma + \delta } \right) = r \\

\Rightarrow \alpha \delta = \beta \gamma = \dfrac{r}{p}{\text{ }}\left( {\because {\text{sum of roots is p}}} \right) \\

$

Now, $\alpha \beta \gamma \delta = \left( {\alpha \delta } \right)\left( {\beta \gamma } \right)$

Putting values of $\alpha \delta $and $\beta \gamma $, we get:

$\alpha \beta \gamma \delta = \left( {\dfrac{r}{p}} \right)\left( {\dfrac{r}{p}} \right) = s$

$

\therefore \dfrac{{{r^2}}}{{{p^2}}} = s \\

{r^2} = {p^2}s \\

$

Hence Proved.

Note: Whenever there are polynomial equations and you have to prove something related to their roots, always try to use the polynomial identities like sum of roots, product of roots etc. and try to obtain relations between them.

Given: ${x^4} - p{x^3} + q{x^2} - rx + s = 0{\text{ }} \ldots \left( 1 \right)$

Let the four roots of the given quadratic polynomial be $\alpha ,\beta ,\gamma ,\delta $

Case 1: Sum of two roots is equal to the sum of other two roots,

$\therefore \alpha + \delta = \beta + \gamma {\text{ }} \ldots \left( 2 \right)$

General quadratic equation: $a{x^4} + b{x^3} + c{x^2} + dx + e = 0{\text{ }} \ldots \left( 3 \right)$

Comparing equation (1) and (3), we get:

$a = 1,{\text{ }}b = - p,{\text{ }}c = q,{\text{ }}d = - r,{\text{ }}e = s{\text{ }} \ldots \left( 4 \right)$

As we know, the sum of roots of a quadratic equation is $ - \dfrac{b}{a}$ .

$

\Rightarrow \alpha + \beta + \gamma + \delta = - \dfrac{b}{a} \\

\Rightarrow \alpha + \beta + \gamma + \delta = - \left( {\dfrac{{ - p}}{1}} \right){\text{ }}\left( {{\text{using }}\left( 4 \right)} \right) \\

\Rightarrow \alpha + \beta + \gamma + \delta = p \\

\Rightarrow \alpha + \beta = \gamma + \delta = \dfrac{p}{2}{\text{ }}\left( {{\text{using }}\left( 2 \right)} \right) \ldots \ldots \left( 5 \right) \\

$

Also, the sum of products of roots of a quadratic equation is $\dfrac{c}{a}$ .

$

\Rightarrow \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \dfrac{c}{a} \\

\Rightarrow \left( {\alpha + \delta } \right)\left( {\beta + \gamma } \right) + \alpha \delta + \beta \gamma = q{\text{ }}\left( {{\text{using }}\left( 4 \right)} \right) \\

$

Now, substituting values of $\left( {\alpha + \delta } \right)$and $\left( {\beta + \gamma } \right)$from equation (5), we get:

$

\Rightarrow \left( {\dfrac{p}{2}} \right)\left( {\dfrac{p}{2}} \right) + \alpha \delta + \beta \gamma = q{\text{ }}\left( {{\text{using }}\left( 5 \right)} \right) \\

\Rightarrow \alpha \delta + \beta \gamma = q - \dfrac{{{p^2}}}{4}{\text{ }} \ldots \left( 6 \right) \\

$

Now, the sum of products of roots of a quadratic equation is $r$.

$

\Rightarrow \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = r \\

\Rightarrow \beta \gamma \left( {\alpha + \delta } \right) + \alpha \delta \left( {\beta + \gamma } \right) = r \\

$

Now, substituting values of $\left( {\alpha + \delta } \right)$and $\left( {\beta + \gamma } \right)$from equation (5), we get:

$ \Rightarrow \beta \gamma \left( {\dfrac{p}{2}} \right) + \alpha \delta \left( {\dfrac{p}{2}} \right) = r{\text{ }}\left( {{\text{using }}\left( 5 \right)} \right)$

Taking $\dfrac{p}{2}$ common, we get:

$ \Rightarrow \dfrac{p}{2}\left( {\beta \gamma + \alpha \delta } \right) = r{\text{ }} \ldots \left( 7 \right)$

Putting value of $\left( {\beta \gamma + \alpha \delta } \right)$from equation (6), we get:

$

\Rightarrow \dfrac{p}{2}\left( {q - \dfrac{{{p^2}}}{4}} \right) = r{\text{ }}\left( {{\text{using }}\left( 6 \right)} \right) \\

\Rightarrow {p^3} - 4pq + 8r = 0 \\

$

Case 2: Product of two roots is equal to the product of other two roots, therefore $\alpha \delta = \beta \gamma {\text{ }} \ldots \left( 8 \right)$

We know: $\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = r$

$

\Rightarrow \beta \gamma \left( {\alpha + \delta } \right) + \alpha \delta \left( {\beta + \gamma } \right) = r \\

\Rightarrow \alpha \delta \left( {\alpha + \beta + \gamma + \delta } \right) = r \\

\Rightarrow \alpha \delta = \beta \gamma = \dfrac{r}{p}{\text{ }}\left( {\because {\text{sum of roots is p}}} \right) \\

$

Now, $\alpha \beta \gamma \delta = \left( {\alpha \delta } \right)\left( {\beta \gamma } \right)$

Putting values of $\alpha \delta $and $\beta \gamma $, we get:

$\alpha \beta \gamma \delta = \left( {\dfrac{r}{p}} \right)\left( {\dfrac{r}{p}} \right) = s$

$

\therefore \dfrac{{{r^2}}}{{{p^2}}} = s \\

{r^2} = {p^2}s \\

$

Hence Proved.

Note: Whenever there are polynomial equations and you have to prove something related to their roots, always try to use the polynomial identities like sum of roots, product of roots etc. and try to obtain relations between them.

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