Question

The number of integral values of 'm' less than 50, so that the roots of the quadratic equation $mx^2$ + (2m - 1)x + (m - 2) = 0 are rational, are

• A. 6
• B. 7
• C. 8
• D. 5

Answer 1

Answer: (A)

Option (A) 6

Answer 2

Solution:

Given the quadratic equation:

$$mx^2 + (2m-1)x + (m-2)=0,$$

with $m < 50$ and $m$ being an integer (with $m \neq 0$ so that it remains a quadratic). For the roots to be rational, the discriminant must be a perfect square. The discriminant ($D$) is

$$D = (2m-1)^2 - 4m(m-2) = 4m^2 - 4m + 1 - (4m^2 - 8m) = 4m + 1.$$

Set

$$4m + 1 = k^2,\quad k\in \mathbb{Z}.$$

So,

$$m = \frac{k^2 - 1}{4}.$$

For $m$ to be an integer, $k^2 -1$ must be divisible by 4. Notice that if $k$ is odd (say, $k = 2n+1$), then

$$k^2 = (2n+1)^2 = 4n^2 + 4n + 1 \quad \Rightarrow \quad k^2 - 1 = 4n(n+1),$$

which is divisible by 4. Thus, $k$ must be odd.

Now, we consider positive odd $k$ values such that $m < 50$:

$$m = \frac{k^2 - 1}{4} < 50 \quad \Rightarrow \quad k^2 < 201.$$

So, $k$ can be 1, 3, 5, 7, 9, 11, 13.

Computing $m$:

• For $k=1$: $m=\frac{1-1}{4}=0$ (discard $m=0$).
• For $k=3$: $m=\frac{9-1}{4}=2$.
• For $k=5$: $m=\frac{25-1}{4}=6$.
• For $k=7$: $m=\frac{49-1}{4}=12$.
• For $k=9$: $m=\frac{81-1}{4}=20$.
• For $k=11$: $m=\frac{121-1}{4}=30$.
• For $k=13$: $m=\frac{169-1}{4}=42$.

Thus, the valid values of $m$ are: 2, 6, 12, 20, 30, and 42. There are 6 such values.

Answer: Option (A) 6

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Explanation (minimal):

1. Find discriminant $D = 4m+1$.
2. Set $4m+1 = k^2$ for rational roots.
3. Require $m=\frac{k^2-1}{4}$ be an integer: $k$ must be odd.
4. Use $k=1,3,5,7,9,11,13$ (ensuring $m < 50$) and discard $m=0$.
5. Count valid $m$: 2, 6, 12, 20, 30, 42 (6 values).