The sum of squares of all the real…

Numerical answer+4 / -1Answer matches NTA’s final key

The sum of squares of all the real solutions of the equation

$\log_{(x+1)}\left(2x^2 + 5x + 3\right) = 4 - \log_{(2x+3)}\left(x^2 + 2x + 1\right)$

is equal to ______.

Your answer

Solution

Domain. Bases $x + 1$ and $2x + 3$ must be positive and not $1$ : $x > -1$ and $x \ne 0$ .

Factor. $2x^2 + 5x + 3 = (x + 1)(2x + 3)$ and $x^2 + 2x + 1 = (x + 1)^2$ . Let $t = \log_{(x+1)}(2x + 3)$ . Then

$1 + t = 4 - \frac{2}{t} \;\Rightarrow\; t^2 - 3t + 2 = 0 \;\Rightarrow\; t = 1 \text{ or } t = 2.$

$t = 1$ : $2x + 3 = x + 1$ , so $x = -2$ , outside the domain.
$t = 2$ : $2x + 3 = (x + 1)^2$ , so $x^2 = 2$ and $x = \sqrt2$ (the root $-\sqrt2$ is outside the domain).

The only real solution is $\sqrt2$ , and the sum of squares is $2$ .

Answer: 2.

Answer checked against the official answer key. The solution is our own working.