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

If α,β\alpha ,\beta are the roots of x2+px+q=0{x^2} + px + q = 0 and γ,δ\gamma ,\delta are the roots of x2+rx+s=0{x^2} + rx + s = 0; evaluate (α−λ)(α−δ)(β−γ)(β−δ)(\alpha - \lambda )(\alpha - \delta )(\beta - \gamma )(\beta - \delta ) in terms of p,q,rp,q,r and ss. Deduce the condition that the equation has a common root.

Explanation

Solution

Hint: We know that for any quadratic equation x2+bx+c=0{x^2} + bx + c = 0, we have sum of roots x1+x2=−b{x_1} + {x_2} = - b and product of roots x1x2=c{x_1}{x_2} = c. Also, α\alpha is the root of x2+px+q=0{x^2} + px + q = 0, so it will satisfy α2+pα+q=0{\alpha ^2} + p\alpha + q = 0. This will be followed by other roots of the quadratic equation as well.

Complete step-by-step answer:

According to question
α,β\alpha ,\beta are the roots of x2+px+q=0{x^2} + px + q = 0, ∴\therefore α+β=−p\alpha + \beta = - p and α.β=q\alpha .\beta = q
γ,δ\gamma ,\delta are the roots of x2+rx+s=0{x^2} + rx + s = 0, ∴γ+δ=−r\therefore \gamma + \delta = - r and γ.δ=s\gamma .\delta = s.

We have been given the expression that,
⇒(α−λ)(α−δ)(β−γ)(β−δ)\Rightarrow (\alpha - \lambda )(\alpha - \delta )(\beta - \gamma )(\beta - \delta ) …(1)
⇒[α2−α(γ+δ)+γδ][β2−β(γ+δ)+γδ]\Rightarrow \left[ {{\alpha ^2} - \alpha (\gamma + \delta ) + \gamma \delta } \right]\left[ {{\beta ^2} - \beta (\gamma + \delta ) + \gamma \delta } \right]
Now replacing γ+δ\gamma + \delta by rr and γδ\gamma \delta by ss, then we obtain
⇒[α2+rα+s][β2+rβ+s]\Rightarrow \left[ {{\alpha ^2} + r\alpha + s} \right]\left[ {{\beta ^2} + r\beta + s} \right] …(2)

We know α\alpha is root of x2+px+q=0{x^2} + px + q = 0, ∴α2+pα+q=0\therefore {\alpha ^2} + p\alpha + q = 0, or we can write α2=−pα−q{\alpha ^2} = - p\alpha - q and similarly, we can write β2=−pβ−q{\beta ^2} = - p\beta - q.
Now replacing α2{\alpha ^2} by −pα−q - p\alpha - q and β2{\beta ^2} by −pβ−q - p\beta - q in equation (2), we will get
⇒(−pα−q+rα+s)(−pβ−q+rβ+s)\Rightarrow ( - p\alpha - q + r\alpha + s)( - p\beta - q + r\beta + s)
⇒[(r−p)α−(q−s)][(r−p)β−(q−s)]\Rightarrow [(r - p)\alpha - (q - s)][(r - p)\beta - (q - s)]
⇒(r−p)2αβ−(r−p)(q−s)(α+β)+(q−s)2\Rightarrow {(r - p)^2}\alpha \beta - (r - p)(q - s)(\alpha + \beta ) + {(q - s)^2}
⇒(r−p)2q−(r−p)(q−s)(−p)+(q−s)2\Rightarrow {(r - p)^2}q - (r - p)(q - s)( - p) + {(q - s)^2}
⇒(r−p)[(qr−pq)+(pq−ps)]+(q−s)2\Rightarrow (r - p)[(qr - pq) + (pq - ps)] + {(q - s)^2}
⇒(q−s)2−(p−r)(qr−ps)\Rightarrow {(q - s)^2} - (p - r)(qr - ps) …(2)

Now, we have been asked to deduce the condition when equations have common root i.e. α=γ  \alpha = \gamma \; and α=δ\alpha = \delta or we can write it as α−γ=0  \alpha - \gamma = 0\; and   α−δ=0\;\alpha - \delta = 0, Then equation (1) becomes zero when we put α−γ=0  \alpha - \gamma = 0\;.
Therefore, ⇒(q−s)2−(p−r)(qr−ps)=0 \Rightarrow {(q - s)^2} - (p - r)(qr - ps) = 0
⇒(q−s)2=(p−r)(qr−ps)\Rightarrow {(q - s)^2} = (p - r)(qr - ps). This is the required condition when roots will be equal.

Note- For calculating the roots of any quadratic equation, we can use Sridharacharya formula as shown below:
x1,x2=−b±b2−4ac2a{x_{1,}}{x_2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}where x1{x_1}and x2{x_2}are roots of the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0.
This method is a little lengthy, and in our case roots are already given, we only need to calculate the sum of roots and product of roots.