
If every pair of the equations \[{x^2} + px + qr = 0,\;{x^2} + qx + rp = 0,\;{x^2} + rx + pq = 0\] have a common root, then the sum of three common roots is
A. $\dfrac{{ - \left( {p + q + r} \right)}}{2}$
B. $\dfrac{{ - p + q + r}}{2}$
C. $ - \left( {p + q + r} \right)$
D. $ - p + q + r$
Answer
162.6k+ views
Hint: In order to solve this type of question, first we will consider the given equations. Then, we will assume the roots of each equation and find the sum of those roots. Next, we will add all the roots together to get the desired correct answer.
Formula used:
$\alpha + \beta = \dfrac{{ - Coefficient\;of\;a}}{{Coefficient\;of\;{a^2}}} = \dfrac{{ - p}}{1} = - p$
$\beta + \gamma = \dfrac{{ - Coefficient\;of\;a}}{{Coefficient\;of\;{a^2}}} = \dfrac{{ - q}}{1} = - q$
$\gamma + \alpha = \dfrac{{ - Coefficient\;of\;a}}{{Coefficient\;of\;{a^2}}} = \dfrac{{ - r}}{1} = - r$
Complete step by step solution:
We are given the equations,
\[{x^2} + px + qr = 0\]
\[{x^2} + qx + rp = 0\]
\[{x^2} + rx + pq = 0\]
Let the roots be $\alpha ,\beta ;\;\beta ,\gamma $ and $\gamma ,\alpha $ respectively.
Sum of roots of each equation is,
$\alpha + \beta = \dfrac{{ - Coefficient\;of\;x}}{{Coefficient\;of\;{x^2}}} = \dfrac{{ - p}}{1} = - p$
$\beta + \gamma = \dfrac{{ - Coefficient\;of\;x}}{{Coefficient\;of\;{x^2}}} = \dfrac{{ - q}}{1} = - q$
$\gamma + \alpha = \dfrac{{ - Coefficient\;of\;x}}{{Coefficient\;of\;{x^2}}} = \dfrac{{ - r}}{1} = - r$
Adding all the three equations,
$2\left( {\alpha + \beta + \gamma } \right) = - p - q - r$
$\left( {\alpha + \beta + \gamma } \right) = \dfrac{{ - \left( {p + q + r} \right)}}{2}$
Thus, the correct option is A.
Note: The key concept of solving this type of question is to be careful while finding the sum of roots of the equation using the coefficients. The sum of roots of a quadratic equation is given by $\dfrac{{ - b}}{a}.$ The product of the roots of a quadratic equation is given by $\dfrac{c}{a}.$
Formula used:
$\alpha + \beta = \dfrac{{ - Coefficient\;of\;a}}{{Coefficient\;of\;{a^2}}} = \dfrac{{ - p}}{1} = - p$
$\beta + \gamma = \dfrac{{ - Coefficient\;of\;a}}{{Coefficient\;of\;{a^2}}} = \dfrac{{ - q}}{1} = - q$
$\gamma + \alpha = \dfrac{{ - Coefficient\;of\;a}}{{Coefficient\;of\;{a^2}}} = \dfrac{{ - r}}{1} = - r$
Complete step by step solution:
We are given the equations,
\[{x^2} + px + qr = 0\]
\[{x^2} + qx + rp = 0\]
\[{x^2} + rx + pq = 0\]
Let the roots be $\alpha ,\beta ;\;\beta ,\gamma $ and $\gamma ,\alpha $ respectively.
Sum of roots of each equation is,
$\alpha + \beta = \dfrac{{ - Coefficient\;of\;x}}{{Coefficient\;of\;{x^2}}} = \dfrac{{ - p}}{1} = - p$
$\beta + \gamma = \dfrac{{ - Coefficient\;of\;x}}{{Coefficient\;of\;{x^2}}} = \dfrac{{ - q}}{1} = - q$
$\gamma + \alpha = \dfrac{{ - Coefficient\;of\;x}}{{Coefficient\;of\;{x^2}}} = \dfrac{{ - r}}{1} = - r$
Adding all the three equations,
$2\left( {\alpha + \beta + \gamma } \right) = - p - q - r$
$\left( {\alpha + \beta + \gamma } \right) = \dfrac{{ - \left( {p + q + r} \right)}}{2}$
Thus, the correct option is A.
Note: The key concept of solving this type of question is to be careful while finding the sum of roots of the equation using the coefficients. The sum of roots of a quadratic equation is given by $\dfrac{{ - b}}{a}.$ The product of the roots of a quadratic equation is given by $\dfrac{c}{a}.$
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