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Question: If \(\alpha ,\beta \) be the roots of \({x^2} - px + q = 0\)and \(\alpha ',\beta '\) are the roots o...

If α,β\alpha ,\beta be the roots of x2−px+q=0{x^2} - px + q = 0and α′,β′\alpha ',\beta ' are the roots of x2−p′x+q′=0{x^2} - p'x + q' = 0, find the value of(α−α′)2+(β−α′)2+(α−β′)2+(β−β′)2{\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2}

Explanation

Solution

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

Given α,β\alpha ,\beta are the roots of the quadratic equation x2−px+q=0{x^2} - px + q = 0and α′,β′\alpha ',\beta 'are the roots of the quadratic equationx2−p′x+q′=0{x^2} - p'x + q' = 0.
So, for the quadratic equation x2−px+q=0{x^2} - px + q = 0
Sum of the roots is α+β=p\alpha + \beta = p and product of the roots is αβ=q\alpha \beta = q
Similarly, for the quadratic equation x2−p′x+q′=0{x^2} - p'x + q' = 0
Sum of the roots is α′+β′=p′\alpha ' + \beta ' = p' and product of the roots is α′β′=q′\alpha '\beta ' = q'
Now consider (α−α′)2+(β−α′)2+(α−β′)2+(β−β′)2{\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2}
=α2+α′2−2αα′+β2+α′2−2βα′+α2+β′2−2αβ′+β2+β′2−2ββ′= {\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(α2+α′2+β2+β′2)−2(α+β)(α′+β′)= 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[p2−2q+p′2−2q′−pp′]= 2\left[ {{p^2} - 2q + p{'^2} - 2q' - pp'} \right]
Hence the value of (α−α′)2+(β−α′)2+(α−β′)2+(β−β′)2{\left( {\alpha - \alpha '} \right)^2} + {\left( {\beta - \alpha '} \right)^2} + {\left( {\alpha - \beta '} \right)^2} + {\left( {\beta - \beta '} \right)^2} is=2[p2−2q+p′2−2q′−pp′] = 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 ax2+bx+c=0a{x^2} + bx + c = 0, the sum of the roots is −ba - \dfrac{b}{a} and the product of the roots is ca\dfrac{c}{a}.