Answer
Verified
439.8k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE