JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

The sum of all the real solutions of the equation $lo g_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 lo g_{(6 x + 10)} (x^{2} + 6 x + 9)$ is equal to :

Solution

Note $6 x^{2} + 28 x + 30 = (x + 3) (6 x + 10)$ and $x^{2} + 6 x + 9 = (x + 3)^{2}$ . Substitute:

$lo g_{x + 3} [(x + 3) (6 x + 10)] = 5 - 2 lo g_{6 x + 10} (x + 3)^{2}$

$1 + lo g_{x + 3} (6 x + 10) = 5 - 4 lo g_{6 x + 10} (x + 3)$

Let $A = lo g_{x + 3} (6 x + 10)$ , so $lo g_{6 x + 10} (x + 3) = 1/ A$ .

$1 + A = 5 - \frac{4}{A} \Rightarrow A + \frac{4}{A} = 4 \Rightarrow A^{2} - 4 A + 4 = 0 \Rightarrow A = 2$ .

Therefore $(x + 3)^{2} = 6 x + 10 \Rightarrow x^{2} + 6 x + 9 = 6 x + 10 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$ .

Both satisfy domain requirements (bases and arguments positive, bases $\neq = 1$ ).

Sum of solutions $= 1 + (- 1) = 0$ .

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