First slide

Theory of equations

Question

The integral values of m for which the roots of the equation ${mx}^{2} + (2 m - 1) x + (m - 2) = 0$ are rational are given by the expression [where n is integer]

Moderate

A

$n^{2}$
B

$n (n + 2)$
C

$n (n + 1)$
D

none of these

Solution

Discriminant $D = (2 m - 1)^{2} - 4 (m - 2) m = 4 m + 1$
must be perfect square. Hence,
$4 m + 1 = k^{2}, for some k \in I$
$\Rightarrow m = \frac{(k - 1) (k + 1)}{4}$
Clearly, k must be odd. Let $k = 2 n + 1$
$∴ m = \frac{2 n (2 n + 2)}{4} = n (n + 1), n \in I$

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