
Let S be the set of all non- zero real numbers $\alpha $ such that the quadratic equation $\alpha {x^2} - x + \alpha = 0$ has two distinct real roots ${x_1}$ and ${x_2}$ satisfying the inequality $\left| {{x_1} - {x_2}} \right| < 1.$ Which of the following intervals is(are) subset(s) of S?
This question has multiple correct options
A. $\left( {\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{{\sqrt 5 }}} \right)$
B. $\left( {\dfrac{{ - 1}}{{\sqrt 5 }},0} \right)$
C. $\left( {0,\dfrac{1}{{\sqrt 5 }}} \right)$
D. $\left( {\dfrac{1}{{\sqrt 5 }},\dfrac{1}{2}} \right)$
Answer
614.7k+ views
Hint: In order to solve this question we will make use two properties of quadratic equation first one is that for two real distinct real roots ${b^2} - 4ac$ of quadratic equation $\left( {a{x^2} + bx + c = 0} \right)$ must be greater than 0, and second one is that sum of given equation roots can be written as $\dfrac{{ - b}}{{2a}}$ and product of roots $\dfrac{c}{a}.$
Complete Step-by-Step solution:
$\alpha {x^2} - x + \alpha = 0$ has two distinct real roots ${x_1}$ and ${x_2};\left| {{x_1} - {x_2}} \right| < 1.$
As we know that for real and distinct roots $D > 0$
Where $D = {b^2} - 4ac$
$
D > 0 \Rightarrow 1 - 4{\alpha ^2} > 0 \\
\Rightarrow {\alpha ^2} < \dfrac{1}{4} \\
\Rightarrow \alpha \in \left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)...........\left( 1 \right) \\
$
Also,
$
{\left| {{x_1} - {x_2}} \right|^2} = {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} < 1 \\
\Rightarrow 1 > {\left( {\dfrac{1}{\alpha }} \right)^2} - 4 \\
\Rightarrow \dfrac{1}{{{\alpha ^2}}} < 5 \\
\Rightarrow {\alpha ^2} > \dfrac{1}{5} \\
\Rightarrow \alpha \in \left( { - \infty ,\dfrac{{ - 1}}{{\sqrt 5 }}} \right) \cup \left( {\dfrac{1}{{\sqrt 5 }},\infty } \right)..........\left( 2 \right) \\
$
Therefore, intersection of equation (1) and (2) gives
$\alpha \in \left( {\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{{\sqrt 5 }}} \right) \cup \left( {\dfrac{1}{{\sqrt 5 }},\dfrac{1}{2}} \right)$
Hence the correct option is “A” and “D”.
Note: In order to solve these types of questions, remember the basic properties of quadratic equations such as the complex roots of a quadratic equation always exist in pairs. Also remember that when the value of D is equal to zero equal roots exist and when D is less than zero complex roots exist and when D is greater than zero real and distinct roots exist.
Complete Step-by-Step solution:
$\alpha {x^2} - x + \alpha = 0$ has two distinct real roots ${x_1}$ and ${x_2};\left| {{x_1} - {x_2}} \right| < 1.$
As we know that for real and distinct roots $D > 0$
Where $D = {b^2} - 4ac$
$
D > 0 \Rightarrow 1 - 4{\alpha ^2} > 0 \\
\Rightarrow {\alpha ^2} < \dfrac{1}{4} \\
\Rightarrow \alpha \in \left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)...........\left( 1 \right) \\
$
Also,
$
{\left| {{x_1} - {x_2}} \right|^2} = {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} < 1 \\
\Rightarrow 1 > {\left( {\dfrac{1}{\alpha }} \right)^2} - 4 \\
\Rightarrow \dfrac{1}{{{\alpha ^2}}} < 5 \\
\Rightarrow {\alpha ^2} > \dfrac{1}{5} \\
\Rightarrow \alpha \in \left( { - \infty ,\dfrac{{ - 1}}{{\sqrt 5 }}} \right) \cup \left( {\dfrac{1}{{\sqrt 5 }},\infty } \right)..........\left( 2 \right) \\
$
Therefore, intersection of equation (1) and (2) gives
$\alpha \in \left( {\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{{\sqrt 5 }}} \right) \cup \left( {\dfrac{1}{{\sqrt 5 }},\dfrac{1}{2}} \right)$
Hence the correct option is “A” and “D”.
Note: In order to solve these types of questions, remember the basic properties of quadratic equations such as the complex roots of a quadratic equation always exist in pairs. Also remember that when the value of D is equal to zero equal roots exist and when D is less than zero complex roots exist and when D is greater than zero real and distinct roots exist.
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