The sum of squares of all the real solutions of the equation is equal to ________. [2026]

Question

The sum of squares of all the real solutions of the equation $\log_{(x + 1)} (2 x^{2} + 5 x + 3) = 4 - \log_{(2 x + 3)} (x^{2} + 2 x + 1)$ is equal to ________. [2026]

Smart Achievers · Accepted Answer

(2)

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

$\Rightarrow \log_{(x + 1)} [(x + 1) (2 x + 3)] = 4 - \log_{(2 x + 3)} [(x + 1) (x + 1)]$

$\Rightarrow \log_{(x + 1)} (x + 1) | \log_{(x + 1)} (2 x + 3) = 4 - \log_{(2 x + 3)} {(x + 1)}^{2}$

$\Rightarrow \log_{(x + 1)} (2 x + 3) = 4 - \log_{(2 x + 3)} (x + 1)$

$\Rightarrow \log_{(x + 1)} (2 x + 3) + 2 \log_{(2 x + 3)} (x + 1) - 3 = 0$

Let $t = \log_{(x + 1)} (2 x + 3)$ , then $\log (2_{x} + 3) (x + 1) = \frac{1}{t}$

$\Rightarrow t + \frac{2}{t} - 3 = 0$

$\Rightarrow t^{2} - 3 t + 2 = 0 \Rightarrow (t - 1) (t - 2) = 0$

Case (i), t = 1;

$\log_{(x + 1)} (2 x + 3) = 1$

$\Rightarrow x + 1 = 2 x + 3 \Rightarrow x = - 2$

For $\log_{(x + 1)}, (x + 1) > 0$

Let x + 1 = –1, which is invalid.

Case (ii), t = 2;

$\log_{(x + 1)} (2 x + 3) = 2$

$\Rightarrow x^{2} + 2 x + 1 = 2 x + 3 \Rightarrow x^{2} = 2$

$\Rightarrow x = \pm \sqrt{2}$

$\Rightarrow x = \sqrt{2}$ (valid)

$\Rightarrow x = - \sqrt{2}$ (invalid)

$∴$ Sum of square of all real solution = ${(\sqrt{2})}^{2} = 2$

Solution

Q.
The sum of squares of all the real solutions of the equation $\log_{(x + 1)} (2 x^{2} + 5 x + 3) = 4 - \log_{(2 x + 3)} (x^{2} + 2 x + 1)$ is equal to ________. [2026]

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Solution

Q. The sum of squares of all the real solutions of the equation log(x+1)(2x2+5x+3)=4–log(2x+3)(x2+2x+1) is equal to ________. [2026]

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Q.
The sum of squares of all the real solutions of the equation $\log_{(x + 1)} (2 x^{2} + 5 x + 3) = 4 - \log_{(2 x + 3)} (x^{2} + 2 x + 1)$ is equal to ________. [2026]