Question

The number of integers ‘n’ such that the equation $n{{x}^{2}}+\left( n+1 \right)x+\left( n+2 \right)=0$ has rational roots only is
(A) 1 (B) 2 (C) 3 (D) 4

Answer 1

Hint: Use the fact that, to a quadratic equation if the discriminant is a perfect square of a rational number then the root will also be rational.

Complete step-by-step answer:
For a given quadratic equation
$a{{x}^{2}}+bx+c=0$
The discriminant d is given by
\[d={{b}^{2}}-4ac\]
Now by comparing the equation
$n{{x}^{2}}+\left( n+1 \right)x+n+2=0$
by the above equation we get
$\begin{align}
  & a=n \\
 & b=n+1 \\
 & c=n+2 \\
\end{align}$
Now we are going to complete the discriminant by substituting the values of a, b and c in the discriminant equation.
$d={{b}^{2}}-4ac$
$\Rightarrow d={{(n+1)}^{2}}-4\times n\times (n+2)$
$\Rightarrow d={{n}^{2}}+1+2n-4{{n}^{2}}-8n$
$\Rightarrow d=1-6n-3{{n}^{2}}$
The only possibility for d to be a perfect square of a rational no and be greater than 0 is
$n=0$
$\therefore $ There is only 1 integer n for which the quadratic equation gives rational roots.

Correct answer is A.

Note: It is helpful to know all the conditions related to discriminate and roots . Ex- If discriminant is less than zero then the roots of quadratic equation is irrational