Algebra – Quadratic Equations – JEE Main 03 April 2025 Shift 2

Solution:

$$x(x+2)(12-k)=2$$

$$(12-k)x^2 + 2(12-k)x – 2 = 0$$

Let $12 – k = \lambda$

$$\lambda x^2 + 2\lambda x – 2 = 0$$

For equal roots, $D = 0$

$$b^2 – 4ac = 0$$

$$(2\lambda)^2 – 4(\lambda)(-2) = 0$$

$$4\lambda^2 + 8\lambda = 0$$

$$4\lambda(\lambda + 2) = 0$$

$$\lambda = -2 \quad (\text{Since } \lambda \neq 0, \text{ otherwise equation reduces to } -2 = 0)$$

$$12 – k = -2$$

$$k = 14$$

$$P\left(k, \frac{k}{2}\right) \equiv P\left(14, \frac{14}{2}\right) \equiv P(14, 7)$$

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

$$d = \frac{|3(14) + 4(7) + 5|}{\sqrt{3^2 + 4^2}}$$

$$d = \frac{|42 + 28 + 5|}{\sqrt{25}}$$

$$d = \frac{75}{5}$$

$$d = 15$$

Ans. (1)