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# The probability that the roots of the equation ${{x}^{2}}+2nx+\left( 4n+\dfrac{5}{n} \right)=0$ are not real numbers where $n\in N$ such that $n\le 5$ is:A. $\dfrac{2}{5}$ B. $\dfrac{4}{5}$ C. $\dfrac{1}{5}$ D. $\dfrac{3}{5}$

Last updated date: 18th Sep 2024
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Hint: We here need to find the probability of the event that the equation ${{x}^{2}}+2nx+\left( 4n+\dfrac{5}{n} \right)=0$ has imaginary roots. For this, we will use the formula for probability given by: $\text{Probability=}\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$. After that we will find the total number of values n can take from the given conditions on n. Then we will find the favourable outcomes by keeping the value of discriminant of the given equation less than 0 as the roots have to be imaginary. After putting the value of discriminant less than 0, we will see how many natural numbers less than 5 are there in the interval in which n can lie so that the roots are imaginary. That will be our number of favourable outcomes. Then by putting the value of the number of favourable and total number of outcomes, we will get the required probability.

Now, we know that the probability of any event is given by the formula:
$\text{Probability=}\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
We will first find the total number of outcomes.
We have been given that n is a natural number and it is less than or equal to 5.
Thus, n can take values in the set:
$n\in \left\{ 1,2,3,4,5 \right\}$
Hence, the total number of outcomes for n will be 5.
Now, we will find the number of favourable outcomes.
We have been given that the equation ${{x}^{2}}+2nx+\left( 4n+\dfrac{5}{n} \right)=0$ should not have any real roots.
Thus, the roots of this equation should be imaginary.
Now, we know that for any quadratic equation to have imaginary roots, the value of their discriminant should be less than 0.
We also know that in a quadratic equation given by $a{{x}^{2}}+bx+c=0$, the discriminant is given by the formula:
$D={{b}^{2}}-4ac$
Here, in the equation ${{x}^{2}}+2nx+\left( 4n+\dfrac{5}{n} \right)=0$,
\begin{align} & a=1 \\ & b=2n \\ & c=\left( 4n+\dfrac{5}{n} \right) \\ \end{align}
Now, as mentioned above, $D < 0$ for the equation to have imaginary roots.
Thus, we get:
\begin{align} & D < 0 \\ & \Rightarrow {{\left( 2n \right)}^{2}}-4.1.\left( 4n+\dfrac{5}{n} \right) < 0 \\ \end{align}
Now, solving this inequality, we get:
\begin{align} & {{\left( 2n \right)}^{2}}-4.1.\left( 4n+\dfrac{5}{n} \right) < 0 \\ & \Rightarrow 4{{n}^{2}}-4\left( 4n+\dfrac{5}{n} \right) < 0 \\ & \Rightarrow 4{{n}^{2}}-16n-\dfrac{20}{n} < 0 \\ \end{align}
Now, if we multiply both sides of the inequality by n, the inequality will remain the same as n is a natural number hence it is a positive real number.
Thus, multiplying the inequality by n we get:
\begin{align} & 4{{n}^{2}}-16n-\dfrac{20}{n} < 0 \\ & \Rightarrow n\left( 4{{n}^{2}}-16n-\dfrac{20}{n} \right) < n\left( 0 \right) \\ & \Rightarrow 4{{n}^{3}}-16{{n}^{2}}-20 < 0 \\ \end{align}
Now, 4 can be taken as common on both sides (as 0 can be written as: $0=4\times 0$).
Thus, taking 4 common on both sides we get:
\begin{align} & 4{{n}^{3}}-16{{n}^{2}}-20 < 0 \\ & \Rightarrow 4\left( {{n}^{3}}-4{{n}^{2}}-5 \right) < 4\left( 0 \right) \\ & \Rightarrow {{n}^{3}}-4{{n}^{2}}-5 < 0 \\ \end{align}
Now, if we take the polynomial ${{n}^{3}}-4{{n}^{2}}-5$ on the LHS, we can see that n=-1 is its root. Hence, n+1=0 is a root of this polynomial.
Hence, if we factorise this polynomial, we will get:
\begin{align} & {{n}^{3}}-4{{n}^{2}}-5 < 0 \\ & \Rightarrow {{n}^{2}}\left( n+1 \right)-5n\left( n+1 \right)+5\left( n+1 \right) < 0 \\ & \Rightarrow \left( n+1 \right)\left( {{n}^{2}}-5n+5 \right) < 0 \\ \end{align}
Now, as we know that n is a natural number, hence, $n\ne -1$
Thus, we get:
\begin{align} & \left( n+1 \right)\left( {{n}^{2}}-5n+5 \right) < 0 \\ & \Rightarrow {{n}^{2}}-5n+5 < 0 \\ \end{align}
Now, we will use the complete square method to solve this inequality.
Solving this equation by completing the square method we get:
\begin{align} & {{n}^{2}}-5n+5 < 0 \\ & \Rightarrow {{n}^{2}}-2.\dfrac{5}{2}n+5 < 0 \\ & \Rightarrow {{n}^{2}}-2.\dfrac{5}{2}n+{{\left( \dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{5}{2} \right)}^{2}}+5 < 0 \\ & \Rightarrow \left( {{n}^{2}}-2.\dfrac{2}{5}.n+\dfrac{25}{4} \right)-\dfrac{25}{4}+5 < 0 \\ \end{align}
\begin{align} & \Rightarrow {{\left( n-\dfrac{5}{2} \right)}^{2}}+\dfrac{-25+20}{4} < 0 \\ & \Rightarrow {{\left( n-\dfrac{5}{2} \right)}^{2}}-\dfrac{5}{4} < 0 \\ \end{align}
Now, we can write $\dfrac{5}{4}$ as:
$\dfrac{5}{4}={{\left( \dfrac{\sqrt{5}}{2} \right)}^{2}}$
Hence, putting this value in the inequality we get:
\begin{align} & {{\left( n-\dfrac{5}{2} \right)}^{2}}-\dfrac{5}{4} < 0 \\ & \Rightarrow {{\left( n-\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{5}}{2} \right)}^{2}} < 0 \\ \end{align}
Now, we can see that the LHS of the inequality is in the form of ${{a}^{2}}-{{b}^{2}}$ which we know factorizes as:
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Hence, factorizing the LHS of the inequality we get:
\begin{align} & {{\left( n-\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{5}}{2} \right)}^{2}} < 0 \\ & \Rightarrow \left( \left( n-\dfrac{5}{2} \right)-\dfrac{\sqrt{5}}{2} \right)\left( \left( n-\dfrac{5}{2} \right)+\dfrac{\sqrt{5}}{2} \right) < 0 \\ & \Rightarrow \left( n-\dfrac{5}{2}-\dfrac{\sqrt{5}}{2} \right)\left( n-\dfrac{5}{2}+\dfrac{\sqrt{5}}{2} \right) < 0 \\ & \Rightarrow \left( n-\left( \dfrac{5}{2}+\dfrac{\sqrt{5}}{2} \right) \right)\left( n-\left( \dfrac{5}{2}-\dfrac{\sqrt{5}}{2} \right) \right) < 0 \\ \end{align}
Now, the LHS is in the form of $\left( x-a \right)\left( x-b \right)$ and we know that if $\left( x-a \right)\left( x-b \right) < 0$, where b < a then x is given as:
$x\in \left( b,a \right)$
Hence, we get the value of our inequality becomes:
\begin{align} & \left( n-\left( \dfrac{5}{2}+\dfrac{\sqrt{5}}{2} \right) \right)\left( n-\left( \dfrac{5}{2}-\dfrac{\sqrt{5}}{2} \right) \right) < 0 \\ & \Rightarrow n\in \left( \dfrac{5}{2}-\dfrac{\sqrt{5}}{2},\dfrac{5}{2}+\dfrac{\sqrt{5}}{2} \right) \\ \end{align}
Now, we know that the value of $\sqrt{5}$ is somewhere near 2.23.
Thus, taking $\sqrt{5}=2.23$and putting it into the interval on n we get:
\begin{align} & n\in \left( \dfrac{5}{2}-\dfrac{\sqrt{5}}{2},\dfrac{5}{2}+\dfrac{\sqrt{5}}{2} \right) \\ & \Rightarrow n\in \left( \dfrac{5-\sqrt{5}}{2},\dfrac{5+\sqrt{5}}{2} \right) \\ & \Rightarrow n\in \left( \dfrac{5-2.23}{2},\dfrac{5+2.23}{2} \right) \\ & \Rightarrow n\in \left( \dfrac{2.77}{2},\dfrac{7.23}{2} \right) \\ & \Rightarrow n\in \left( 1.385,3.615 \right) \\ \end{align}
Now, we know that n is a natural number and in the interval $\left( 1.385,3.615 \right)$, there are only 2 natural numbers: 2 and 3
Thus, for the roots of the equation ${{x}^{2}}+2nx+\left( 4n+\dfrac{5}{n} \right)=0$ to be imaginary, n must lie in the set given as:
$n\in \left\{ 2,3 \right\}$
Thus, the number of favourable outcomes is 2.
Now, that we know both the number of total outcomes (5) and the number of favourable outcomes (2), we can calculate the probability by putting these values in the formula for probability.
Putting these values in the formula for probability we get:
\begin{align} & \text{Probability=}\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\ & \text{Probability=}\dfrac{2}{\text{5}} \\ \end{align}
Hence, the required probability is $\dfrac{2}{5}$

So, the correct answer is “Option A”.

Note: Here, we have used the discriminant to find the required values for n. We should know what a discriminant is used for.
A discriminant of a quadratic equation helps us determine if the equation has real or imaginary roots. The discriminant of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given by $D={{b}^{2}}-4ac$.
Now, if D < 0, the roots are imaginary
If D > 0, the roots are real and distinct
If D=0, the roots are real and equal.