Quadratic Equations – Logarithmic Equations – JEE Main 23 Jan 2026 Shift 2

Solution:

Simplify the arguments of the logarithms:
$6x^2 + 28x + 30 = 2(3x^2 + 14x + 15) = 2(3x+5)(x+3)$
$x^2 + 6x + 9 = (x+3)^2$

The equation becomes:
$$ \log_{(x+3)}[2(3x+5)(x+3)] = 5 – 2\log_{2(3x+5)}[(x+3)^2] $$

Using log properties $\log_a(mn) = \log_a m + \log_a n$ and $\log_b(a^k) = k \log_b a$:
$$ \log_{(x+3)}(x+3) + \log_{(x+3)}(2(3x+5)) = 5 – 4\log_{2(3x+5)}(x+3) $$
$$ 1 + \log_{(x+3)}(6x+10) = 5 – \frac{4}{\log_{(x+3)}(6x+10)} $$

Let $t = \log_{(x+3)}(6x+10)$.
$$ 1 + t = 5 – \frac{4}{t} $$
$$ t – 4 + \frac{4}{t} = 0 \Rightarrow t^2 – 4t + 4 = 0 $$
$$ (t-2)^2 = 0 \Rightarrow t = 2 $$

Substituting $t$ back:
$$ \log_{(x+3)}(6x+10) = 2 $$
$$ 6x + 10 = (x+3)^2 $$
$$ 6x + 10 = x^2 + 6x + 9 $$
$$ x^2 = 1 \Rightarrow x = 1, -1 $$

Check domain validity ($x > -5/3$ and base $\neq 1$):
For $x=1$: Base $4 \neq 1$, Arg $>0$. Valid.
For $x=-1$: Base $2 \neq 1$, Arg $>0$. Valid.

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

Ans. (3)