Question

The integral values of m for which the roots of the equation $m{x^2} + \left( {2m - 1} \right)x + \left( {m - 2} \right) = 0$ are rational are given by expression [where m is integer]
$\left( a \right){n^2}$
$\left( b \right)n\left( {n + 2} \right)$
$\left( c \right)n\left( {n + 1} \right)$
$\left( d \right)$ None of these.

Answer 1

Hint: In this particular question use the concept that for any quadratic equation if roots of the equation is rational then the value of the discriminant (D) must be a complete square, so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given equation:
$m{x^2} + \left( {2m - 1} \right)x + \left( {m - 2} \right) = 0$
Now it is given that the roots of the above equation are rational.
So for this the discriminant (D) should be a perfect square.
So discriminant (D) is given as,
$ \Rightarrow D = {b^2} - 4ac$, where $b = (2m – 1)$, $a = m$, and $c = (m – 2)$
Now substitute the values we have,
$ \Rightarrow D = {\left( {2m - 1} \right)^2} - 4\left( m \right)\left( {m - 2} \right)$
Now simplify the above equation we have,
$ \Rightarrow D = 4{m^2} + 1 - 4m - 4{m^2} + 8m$
$ \Rightarrow D = 1 + 4m$
Now it is given that m is integer so 4 multiply by integer always given an even number.
Therefore, $(4m + 1)$ is an odd number.
So discriminant D is an odd number.
But discriminant should be a perfect square for roots to be rational.
So let, D = ${k^2}$, where K is an odd number and general form of an odd number is, $(2n + 1)$, where $n \in I$.
$ \Rightarrow 4m + 1 = {k^2}$
Now simplify for (m) we have,
$ \Rightarrow 4m = {k^2} - 1$
$ \Rightarrow 4m = \left( {k + 1} \right)\left( {k - 1} \right)$
$ \Rightarrow m = \dfrac{{\left( {k + 1} \right)\left( {k - 1} \right)}}{4}$
Now substitute the value of k we have,
$ \Rightarrow m = \dfrac{{\left( {2n + 1 + 1} \right)\left( {2n + 1 - 1} \right)}}{4}$
Now simplify the above equation we have,
$ \Rightarrow m = \dfrac{{\left( {2n + 2} \right)\left( {2n} \right)}}{4} = \dfrac{{4n\left( {n + 1} \right)}}{4} = n\left( {n + 1} \right)$
So this is the required value of m such that the roots of the given equation is rational.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that any integer when multiplied by an even number it becomes an even number and ever number plus one become an odd number, and the general form of an odd number is $(2n + 1)$, where n belongs to integer value.