JEE Main 2026 — 8 April, Evening Shift — Question 38

$\log _{(x+1)}(2 x+3)(x+1)+\log _{(2 x+3)}(x+1)^{2}=4$ $1+\log _{(x+1)}(2 x+3)+2 \log _{(2 x+3)}(x+1)=4$ $\log _{\mathrm{x}+1}(2 \mathrm{x}+3)=\mathrm{t} \mathrm{t}+\frac{2}{\mathrm{t}}=3$ $\mathrm{t}^{2}-3 \mathrm{t}+2=0 \Rightarrow \mathrm{t}=1,2$ for $\mathrm{t}=1 \Rightarrow \log _{(\mathrm{x}+1)}(2 \mathrm{x}+3)=1 \Rightarrow 2 \mathrm{x}+3=\mathrm{x}+1$ $\Rightarrow \mathrm{x}=-2$ (rejected) for $\mathrm{t}=2 \Rightarrow 2 \mathrm{x}+3=\mathrm{x}^{2}+2 \mathrm{x}+1 \Rightarrow \mathrm{x}^{2}=2$ $\Rightarrow \mathrm{x}= \pm \sqrt{2}$ rejecting $\mathrm{x}=-\sqrt{2}$, we get $\mathrm{x}=\sqrt{2}$ ∴ sum of squares of all the roots $=2$