Question

Consider the quadratic equation \[n{{x}^{2}}+7\sqrt{n}x+n=0\] where \[n\] is a positive integer.
Which of the following are necessarily correct?
(1) For any \[n\] the roots are distinct
(2) There are infinitely many values of \[n\] for which both the roots are real
(3) The product of roots is necessarily an integer
(a) (3) only
(b) (1) and (3) only
(c) (2) and (3) only
(d) (1), (2) and (3)

Answer 1

Hint: We solve this problem by using the discriminant and the product of the roots formula of a quadratic equation.
If the quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] then the discriminant and product of roots is given as
\[\Rightarrow D={{b}^{2}}-4ac\]
\[\Rightarrow \text{Product of roots}=\dfrac{c}{a}\]
We have the condition for real and distinct roots that is \[D>0\]

Complete step by step answer:
We are given that the quadratic equation of \[x\] as
\[\Rightarrow n{{x}^{2}}+7\sqrt{n}x+n=0\]
Now, let us check each and every statement one by one.
(1) For any \[n\] the roots are distinct

Now, let us find the discriminant for the given quadratic equation
We know that the quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] then the discriminant and product of roots is given as
\[\Rightarrow D={{b}^{2}}-4ac\]
By using the above condition we get the value of discriminant of given quadratic equation as
\[\begin{align}
  & \Rightarrow D={{\left( 7\sqrt{n} \right)}^{2}}-4\left( n \right)\left( n \right) \\
 & \Rightarrow D=49n-4{{n}^{2}} \\
\end{align}\]
We know that the condition for real and distinct roots that is \[D>0\]
Let us assume that the discriminant is less than 0 then we get
\[\begin{align}
  & \Rightarrow 49n-4{{n}^{2}}\le 0 \\
 & \Rightarrow n\left( 49-4n \right)\le 0 \\
\end{align}\]
Here we can see that we get an integer \[n=0\] where the discriminant is 0
We know that if the discriminant is 0 then, the roots are equal.
But, we are given that for any \[n\] the roots are distinct
So, we can conclude that statement (1) is wrong
(2) There are infinitely many values of \[n\] for which both the roots are real
Now, let us take the discriminant greater than 0 then we get
\[\begin{align}
  & \Rightarrow 49n-4{{n}^{2}}>0 \\
 & \Rightarrow 4{{n}^{2}}-49n<0 \\
 & \Rightarrow 4n\left( n-\dfrac{49}{n} \right)<0 \\
 & \Rightarrow \left( n-0 \right)\left( n-\dfrac{49}{4} \right)<0 \\
\end{align}\]
We know that the condition that is \[\left( x-a \right)\left( x-b \right)<0\] then the values of \[x\] are given as
\[x=\left( a,b \right)\]
By using this condition in above equation we get
\[\Rightarrow n=\left( 0,\dfrac{49}{4} \right)\]
We are given that \[n\] is a positive integer.
Here we can see that there are only 1 to 12 numbers that satisfy the condition that discriminant is greater than 0
Therefore we can see that there are only 12 values of \[n\] for which the roots are real
But we are given that there are infinite numbers so, we can conclude that statement (2) is wrong
(3) The product of roots is necessarily an integer
We know that if the quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] then the discriminant and product of roots is given as
\[\Rightarrow \text{Product of roots}=\dfrac{c}{a}\]
By using the above formula we get the product of roots of the given equation as
\[\Rightarrow P=\dfrac{n}{n}=1\]
Here we can see that the product of roots is 1 which is an integer.
Therefore, we can conclude that statement (3) is only correct
So, option (a) is the correct answer.


Note:
 Students may make mistakes in considering the condition for distinct roots.
In the first statement, we have the discriminant as
\[\Rightarrow D=49n-4{{n}^{2}}\]
Here, we can see that for \[n=0\] the discriminant \[D=0\] which gives that the roots are equal
But in the first statement, we are given that for all values of \[n\] the roots are distinct.
This statement is wrong because the distinct roots mean that the roots are unequal
But students may miss this point that the meaning is distinct and consider that the given statement is correct.