Question

If a root of the equations ${x^2} + px + q = 0$ and ${x^2} + \alpha x + \beta = 0$ is common, then its value will be (where $p \ne \alpha $ and $q \ne \beta $ )
A. $\dfrac{{q - \beta }}{{\alpha - p}}$
B. $\dfrac{{p\beta - \alpha q}}{{q - \beta }}$
C. $\dfrac{{q - \beta }}{{\alpha - p}}$ or $\dfrac{{p\beta - \alpha q}}{{q - \beta }}$
D. None of these

Answer 1

Hint: Here, we have two quadratic equations which have a common root, which we have to find. We can easily compare these equations with the general quadratic equation. Hence, we get the values of the coefficients of both the equations and substitute those values to get the common root. Solve the quadratic equations using those conditions to obtain the answer.

Formula Used: We have one common root so, apply Crammer’s rule,
$\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
  { - {c_1}}&{{b_1}} \\
  { - {c_2}}&{{b_2}}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{ - {c_1}} \\
  {{a_2}}&{ - {c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$ .

Complete step-by-step solution:
We have equations ${x^2} + px + q = 0$ and , where $p \ne \alpha $ and $q \ne \beta $ .
We know the general quadratic equations such as
${a_1}{x^2} + {b_1}x + {c_1} = 0$
And ${a_2}{x^2} + {b_2}x + {c_2} = 0$ .
Comparing these equations with the given equations of the question, we get the coefficients such as,
${a_1} = 1$ , ${b_1} = p$ , ${c_1} = q$
Likewise, ${a_2} = 1$ , ${b_2} = \alpha $ , ${c_2} = \beta $
Suppose, the one common root that both the equations possess is $y$ . To find the value of this one common root, we will be using the cramer's rule
$\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
  { - {c_1}}&{{b_1}} \\
  { - {c_2}}&{{b_2}}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{ - {c_1}} \\
  {{a_2}}&{ - {c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$
Substituting the values of ${a_1}$ , ${b_1}$ , ${c_1}$ and ${a_2}$ , ${b_2}$ , ${c_2}$ in the cramer's rule, we get
$\dfrac{{{y^2}}}{{p\beta - q\alpha }} = \dfrac{y}{{q - \beta }} = \dfrac{1}{{\alpha - p}}$
On evaluating, the first and second part of this equation, we get
$\dfrac{{{y^2}}}{{p\beta - q\alpha }} = \dfrac{y}{{q - \beta }}$
$ \Rightarrow \dfrac{{{y^2}}}{y} = y = \dfrac{{p\beta - q\alpha }}{{q - \beta }}$

Hence, the correct option will be B.

Note: The common root can be located by creating the same ${x^2}$ coefficient for the quadratic equations that are given, and then removing them. Using the connection between the quadratic equation's roots and coefficients. Since imaginary and surd roots only occur in conjugate pairs, two separate quadratic equations with rational coefficients cannot share a single common root that is complex or irrational.