Question

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) ±1
D) 2

Answer 1

Hint: We have 2 quadratic equations with 1 common root. We will use Cramer’s rule to find the condition that satisfies the equations having one common root. Also, while solving we’ll keep in mind that the roots of a quadratic equation satisfy it.

Complete step by step solution: We have 2 equations,
$2{x^2} + 5\gamma x + 2 = 0$
$4{x^2} + 8\gamma x + 3 = 0$
If one of the roots is common, by Cramer’s rule,
$\dfrac{{{x^2}}}{{15\gamma - 16\gamma }} = \dfrac{x}{{8 - 6}} = \dfrac{1}{{16\gamma - 20\gamma }}$
On further simplification,
$\dfrac{{{x^2}}}{{ - \gamma }} = \dfrac{x}{2} = \dfrac{1}{{ - 4\gamma }}$
From the last 2 terms, we get
$x = \dfrac{2}{{ - 4\gamma }} = - \dfrac{1}{{2\gamma }}$ …….. (1)
From the 1st and last term, we get,
${x^2} = \dfrac{{ - \gamma }}{{ - 4\gamma }} = \dfrac{1}{4}$ …….. (2)
From equations (1) and (2), we get
\[\dfrac{1}{4} = {\left( { - \dfrac{1}{{2\gamma }}} \right)^2}\]
On simplification, we get
\[\dfrac{1}{4} = \dfrac{1}{{4{\gamma ^2}}}\]
Cross multiplying and canceling the common factors, we get,
\[{\gamma ^2} = 1\]
This implies,
\[\gamma = \pm 1\]
Thus, for the two quadratic equation to have a common root, \[ \gamma \]must be either 1 or -1.

Therefore, the correct answer is option C.

Note: Cramer’s rule is used for solving a system of quadratic equations. The solutions are expressed as determinants of matrices. While taking square roots, the possibility of having both positive and negative roots must be considered.
Cramer’s Rule
For 2 quadratic equations ${a_1}{x^2} + {b_1}x + {c_1} = 0$and ${a_2}{x^2} + {b_2}x + {c_2} = 0$ having one common root, by Cramer’s rule,
$\dfrac{{{x^2}}}{{\left| {\begin{array}{*{20}{c}}
  {{b_1}}&{{c_1}} \\
  {{b_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{{ - x}}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{c_1}} \\
  {{a_2}}&{{c_2}}
\end{array}} \right|}} = \dfrac{1}{{\left| {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right|}}$
Taking the determinants, we get,
$\dfrac{{{x^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{{ - x}}{{{a_1}{c_2} - {a_2}{c_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$… (3)
Take the 1st and last terms of equation (3),
$\dfrac{{{x^2}}}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
On rearranging, we get,
${x^2} = \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}$ … (4)
From the 2nd and 3rd term of equation (3),
$\dfrac{{ - x}}{{{a_1}{c_2} - {a_2}{c_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
On rearranging, we get,
$x = - \dfrac{{{a_1}{c_2} - {a_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}$
Taking the square, we get,
\[{x^2} = \dfrac{{{{\left( {{a_1}{c_2} - {a_2}{c_1}} \right)}^2}}}{{{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}}}\] … (5)
Comparing equation (4) and (5), we get,
\[\dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}} = \dfrac{{{{\left( {{a_1}{c_2} - {a_2}{c_1}} \right)}^2}}}{{{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}}}\]
On cross-multiplying and simplifying, we get,
\[{\left( {{a_1}{c_2} - {a_2}{c_1}} \right)^2} = \left( {{b_1}{c_2} - {b_2}{c_1}} \right)\left( {{a_1}{b_2} - {a_2}{b_1}} \right)\].
So according to Cramer’s rule, two quadratic equations whose coefficients satisfies this condition has one common root.