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The value of $\gamma $ in order that the equations $2{x^2} + 5\gamma x + 2 = 0$ and $4{x^2} + 8\gamma x + 3 = 0$ have a common root is given by:
A) 1
B) -1
C) $ \pm 1$
D) 2

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Last updated date: 13th Jun 2024
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Answer
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Hint:
It is given two quadratic equations to us in the question which are having 1 common root. In order to get to the answer, we will be using the Crammer’s rule to find the required condition that satisfies the equations having one common root. Also, it is important to remember that the roots of any quadratic equation when substituted back in the equation, will satisfy it.

Complete step by step solution:
Let us consider the two given equations:
$2{x^2} + 5\gamma x + 2 = 0$
$4{x^2} + 8\gamma x + 3 = 0$

Let us suppose that one of the roots is common, then we will use the crammer’s rule to get,
$
   \Rightarrow \dfrac{{{x^2}}}{{15\gamma - 16\gamma }} = \dfrac{x}{{8 - 6}} = \dfrac{1}{{16\gamma - 20\gamma }} \\
   \Rightarrow \dfrac{{{x^2}}}{{ - \gamma }} = \dfrac{x}{2} = \dfrac{1}{{ - 4\gamma }} \\
 $

Consider the last 2 terms of the expression of the Crammer’s rule to get;
$ \Rightarrow x = \dfrac{2}{{ - 4\gamma }} = - \dfrac{1}{{2\gamma }}......(1)$
From the first and the last term of the expression of the Crammer’s rule, we get;
$ \Rightarrow {x^2} = \dfrac{{ - \gamma }}{{ - 4\gamma }} = \dfrac{1}{4}.....(2)$

Clearly from (1) and (2) we get;
$
   \Rightarrow \dfrac{1}{4} = {\left( {\dfrac{{ - 1}}{{2\gamma }}} \right)^2} \\
   \Rightarrow \dfrac{1}{4} = \dfrac{1}{{4{\gamma ^2}}} \\
   \Rightarrow {\gamma ^2} = 1 \\
   \Rightarrow \gamma = \pm 1 \\
 $

Hence, it can be concluded that the value of $\gamma $ in order that the equations $2{x^2} + 5\gamma x + 2 = 0$ and $4{x^2} + 8\gamma x + 3 = 0$ have a common root is $ \pm 1$, which is option (c).

Note:
Crammer’s rule is used for solving a system of quadratic equations, in which the solutions are expressed as the determinant of matrices. It can be noted that, when you take the square roots, the possibility of having both positive and negative roots must be considered.
Consider the quadratic equations,
$
  {a_1}{x^2} + {b_1}x + {c_1} = 0 \\
  {a_2}{x^2} + {b_2}x + {c_2} = 0 \\
 $
Having one common root, by Crammer’s rule;
$
  \dfrac{{{x^2}}}{{\left| {\begin{array}{*{20}{c}}
  {{b_1}}&{{c_1}} \\
  {{b_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{{ - x}}{{\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|}} \\
   \Rightarrow \dfrac{{{x^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{{ - x}}{{{a_1}{c_2} - {a_2}{c_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}......(3) \\
 $

Let us consider first and last terms of equation (3),
\[
  \dfrac{{{x^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}} \\
   \Rightarrow {x^2} = \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}......(4) \\
 \]
Let us consider second and third terms of equation (3),
$
  \dfrac{{ - x}}{{{a_1}{c_2} - {a_2}{c_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}} \\
   \Rightarrow x = - \dfrac{{{a_1}{c_2} - {a_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}} \\
   \Rightarrow {x^2} = {\left( {\dfrac{{{a_1}{c_2} - {a_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}} \right)^2}......(5) \\
 $

Combining the equations (4) and (5) we get;
\[
  \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}} = {\left( {\dfrac{{{a_1}{c_2} - {a_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}} \right)^2} \\
  {b_1}{c_2} - {b_2}{c_1} = \dfrac{{{{\left( {{a_1}{c_2} - {a_2}{c_1}} \right)}^2}}}{{{a_1}{b_2} - {a_2}{b_1}}} \\
   \Rightarrow \left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{b_1}{c_2} - {b_2}{c_1}} \right) = {\left( {{a_1}{c_2} - {a_2}{c_1}} \right)^2} \\
 \]
Thus, if this condition is satisfied by the coefficients of the two quadratic equations has one common root.