Question

The equations ${x^2} - 11x + a$ and ${x^2} - 14x + 2a$ will have a common factor, if $a = $
A. $24$
B. $0,24$
C. $3,24$
D. $0,3$

Answer 1

Hint: To find the value of an unknown in the given quadratics, we first use the cramer’s rule as it is already given in the question that the given equations have a common root or factor. After applying the method, we will get three equations (all equal to one another). One by one we have to equalize these equations i.e., two equations at a time. After that, solve the equations to get the value of the unknown.

Formula Used: Apply the following condition:
$\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
  { - {c_1}}&{{b_1}} \\
  { - {c_2}}&{{b_2}}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{ - {c_1}} \\
  {{a_2}}&{ - {c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$ , where $\alpha $ is the common root of the equations.

Complete step-by-step solution:
Given quadratic equation,
${x^2} - 11x + a = 0$
It gives the coefficient’s value on comparing with the equation ${a_1}{x^2} + {b_1}x + {c_1} = 0$ .
We have ${a_1} = 1$ , ${b_1} = - 11$ and ${c_1} = a$ .
${x^2} - 14x + 2a = 0$
In comparison with ${a_2}{x^2} + {b_2}x + {c_2} = 0$ .
We have ${a_2} = 1$ , ${b_2} = - 14$ and ${c_2} = 2a$ .
Substitute these values in the condition $\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
  { - {c_1}}&{{b_1}} \\
  { - {c_2}}&{{b_2}}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{ - {c_1}} \\
  {{a_2}}&{ - {c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$
We get, $\dfrac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{c}}
  { - a}&{ - 11} \\
  { - 2a}&{ - 14}
\end{array}} \right|}} = \dfrac{\alpha }{{\left| {\begin{array}{*{20}{c}}
  1&{ - a} \\
  1&{ - 2a}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  1&{ - 11} \\
  1&{ - 14}
\end{array}} \right|}}$
Solving the determinants in the denominator, we get
$\dfrac{{{\alpha ^2}}}{{ - 22a + 14a}} = \dfrac{\alpha }{{a - 2a}} = \dfrac{1}{{ - 14 + 11}}$
Solving further,
$\dfrac{{{\alpha ^2}}}{{ - 8a}} = \dfrac{\alpha }{{ - a}} = \dfrac{1}{{ - 3}}$
Considering the first and third part, we get
${\alpha ^2} = \dfrac{{8a}}{3}$
And taking second and third part, we get
$\alpha = \dfrac{a}{3}$
Putting the value of $\alpha $
${\left( {\dfrac{a}{3}} \right)^2} = \dfrac{{8a}}{3}$
Solving this equation, we get two values of $a$ which are $0,24$ .
Hence, the correct option is B.

Note: Students should not make the mistake of canceling a from both sides then we will get only one value of a. So instead of canceling, we need to take common.