JEE Mathematics 2025 Question with Solution

Given: The equation is $x^2 + 4x - n = 0$ with $n \in [20,100]$ and $n$ a natural number.

Find: The number of distinct values of $n$ for which the equation has integral roots.

Rewrite the equation by completing the square:

$$x^2 + 4x + 4 = n + 4$$ $$(x+2)^2 = n+4$$

Hence,

$$x = -2 \pm \sqrt{n+4}$$

For the roots to be integers, $\sqrt{n+4}$ must be an integer. Let

$$n+4 = k^2$$

where $k \in \mathbb{Z}$. Then

$$n = k^2 - 4$$

Now use the range $20 \le n \le 100$:

$$20 \le k^2 - 4 \le 100$$ $$24 \le k^2 \le 104$$

The perfect squares in this interval are

$$25, 36, 49, 64, 81, 100$$

so

$$k = 5, 6, 7, 8, 9, 10$$

Therefore the corresponding values of $n$ are

$$21, 32, 45, 60, 77, 96$$

There are $6$ distinct values of $n$.

Therefore, the correct option is C.

Given: $x^2 + 4x - n = 0$ with $20 \le n \le 100$.

Find: How many natural numbers $n$ make the roots integral.

For a quadratic equation $ax^2 + bx + c = 0$ to have integral roots, its discriminant must be a perfect square. Here,

$$D = b^2 - 4ac = 16 + 4n = 4(n+4)$$

The roots are

$$x = \frac{-4 \pm \sqrt{D}}{2} = -2 \pm \sqrt{n+4}$$

So the roots are integers exactly when $n+4$ is a perfect square. Let

$$n+4 = k^2$$

Then

$$n = k^2 - 4$$

Using the condition $20 \le n \le 100$,

$$20 \le k^2 - 4 \le 100$$ $$24 \le k^2 \le 104$$

Thus $k$ can be

$$5, 6, 7, 8, 9, 10$$

This gives the six values

$$21, 32, 45, 60, 77, 96$$

Hence the number of distinct values of $n$ is $6$.

Therefore, the correct option is C.