Question

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

Answer 1

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

Complete step-by-step answer:

According to question
$\alpha ,\beta $ are the roots of ${x^2} + px + q = 0$, $\therefore $$\alpha + \beta = - p$ and $\alpha .\beta = q$
$\gamma ,\delta $ are the roots of ${x^2} + rx + s = 0$, $\therefore \gamma + \delta = - r$ and $\gamma .\delta = s$.

We have been given the expression that,
$ \Rightarrow (\alpha - \lambda )(\alpha - \delta )(\beta - \gamma )(\beta - \delta )$ …(1)
\[ \Rightarrow \left[ {{\alpha ^2} - \alpha (\gamma + \delta ) + \gamma \delta } \right]\left[ {{\beta ^2} - \beta (\gamma + \delta ) + \gamma \delta } \right]\]
Now replacing $\gamma + \delta $ by $r$ and $\gamma \delta $ by $s$, then we obtain
\[ \Rightarrow \left[ {{\alpha ^2} + r\alpha + s} \right]\left[ {{\beta ^2} + r\beta + s} \right]\] …(2)

We know $\alpha $ is root of ${x^2} + px + q = 0$, $\therefore {\alpha ^2} + p\alpha + q = 0$, or we can write ${\alpha ^2} = - p\alpha - q$ and similarly, we can write ${\beta ^2} = - p\beta - q$.
Now replacing ${\alpha ^2}$ by $ - p\alpha - q$ and ${\beta ^2}$ by $ - p\beta - q$ in equation (2), we will get
\[ \Rightarrow ( - p\alpha - q + r\alpha + s)( - p\beta - q + r\beta + s)\]
\[ \Rightarrow [(r - p)\alpha - (q - s)][(r - p)\beta - (q - s)]\]
$ \Rightarrow {(r - p)^2}\alpha \beta - (r - p)(q - s)(\alpha + \beta ) + {(q - s)^2}$
$ \Rightarrow {(r - p)^2}q - (r - p)(q - s)( - p) + {(q - s)^2}$
$ \Rightarrow (r - p)[(qr - pq) + (pq - ps)] + {(q - s)^2}$
$ \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. \[\alpha = \gamma \;\] and \[\alpha = \delta \]or we can write it as \[\alpha - \gamma = 0\;\] and \[\;\alpha - \delta = 0\], Then equation (1) becomes zero when we put \[\alpha - \gamma = 0\;\].
Therefore, $ \Rightarrow {(q - s)^2} - (p - r)(qr - ps) = 0$
 $ \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:
${x_{1,}}{x_2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$where ${x_1}$and ${x_2}$are roots of the quadratic equation $a{x^2} + bx + c = 0$.
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.