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Hint: Use the properties of sum of the roots and product of the roots of the given quadratic polynomial equation.

Given $\alpha ,\beta $ are the roots of the quadratic equation ${x^2} - px + q = 0$and $\alpha ',\beta '$are the roots of the quadratic equation${x^2} - p'x + q' = 0$.

So, for the quadratic equation ${x^2} - px + q = 0$

Sum of the roots is $\alpha + \beta = p$ and product of the roots is $\alpha \beta = q$

Similarly, for the quadratic equation ${x^2} - p'x + q' = 0$

Sum of the roots is \[\alpha ' + \beta ' = p'\] and product of the roots is $\alpha '\beta ' = q'$

Now consider ${\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2}$

\[ = {\alpha ^2} + \alpha {'^2} - 2\alpha \alpha ' + {\beta ^2} + \alpha {'^2} - 2\beta \alpha ' + {\alpha ^2} + \beta {'^2} - 2\alpha \beta ' + {\beta ^2} + \beta {'^2} - 2\beta \beta '\]

Grouping the terms, we have

$ = 2\left( {{\alpha ^2} + \alpha {'^2} + {\beta ^2} + \beta {'^2}} \right) - 2\left( {\alpha + \beta } \right)\left( {\alpha ' + \beta '} \right)$

$ = 2\left[ {\left\{ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right\} + \left\{ {{{\left( {\alpha ' + \beta '} \right)}^2} - 2\alpha '\beta '} \right\}} \right] - 2\left( {\alpha + \beta } \right)\left( {\alpha ' + \beta '} \right)$

By the above relations we get

$ = 2\left[ {\left\{ {{{\left( p \right)}^2} - 2q} \right\} + \left\{ {{{\left( {p'} \right)}^2} - 2q'} \right\}} \right] - 2\left( {pp'} \right)$

$ = 2\left[ {{p^2} - 2q + p{'^2} - 2q' - pp'} \right]$

Hence the value of ${\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2}$ is$ = 2\left[ {{p^2} - 2q + p{'^2} - 2q' - pp'} \right]$.

Note: In this type of problem we tend to make mistakes in opening and closing the brackets of squaring and rooting. Always remember that for the quadratic polynomial \[a{x^2} + bx + c = 0\], the sum of the roots is \[ - \dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\].

Given $\alpha ,\beta $ are the roots of the quadratic equation ${x^2} - px + q = 0$and $\alpha ',\beta '$are the roots of the quadratic equation${x^2} - p'x + q' = 0$.

So, for the quadratic equation ${x^2} - px + q = 0$

Sum of the roots is $\alpha + \beta = p$ and product of the roots is $\alpha \beta = q$

Similarly, for the quadratic equation ${x^2} - p'x + q' = 0$

Sum of the roots is \[\alpha ' + \beta ' = p'\] and product of the roots is $\alpha '\beta ' = q'$

Now consider ${\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2}$

\[ = {\alpha ^2} + \alpha {'^2} - 2\alpha \alpha ' + {\beta ^2} + \alpha {'^2} - 2\beta \alpha ' + {\alpha ^2} + \beta {'^2} - 2\alpha \beta ' + {\beta ^2} + \beta {'^2} - 2\beta \beta '\]

Grouping the terms, we have

$ = 2\left( {{\alpha ^2} + \alpha {'^2} + {\beta ^2} + \beta {'^2}} \right) - 2\left( {\alpha + \beta } \right)\left( {\alpha ' + \beta '} \right)$

$ = 2\left[ {\left\{ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right\} + \left\{ {{{\left( {\alpha ' + \beta '} \right)}^2} - 2\alpha '\beta '} \right\}} \right] - 2\left( {\alpha + \beta } \right)\left( {\alpha ' + \beta '} \right)$

By the above relations we get

$ = 2\left[ {\left\{ {{{\left( p \right)}^2} - 2q} \right\} + \left\{ {{{\left( {p'} \right)}^2} - 2q'} \right\}} \right] - 2\left( {pp'} \right)$

$ = 2\left[ {{p^2} - 2q + p{'^2} - 2q' - pp'} \right]$

Hence the value of ${\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2}$ is$ = 2\left[ {{p^2} - 2q + p{'^2} - 2q' - pp'} \right]$.

Note: In this type of problem we tend to make mistakes in opening and closing the brackets of squaring and rooting. Always remember that for the quadratic polynomial \[a{x^2} + bx + c = 0\], the sum of the roots is \[ - \dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\].

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