Maths Q21 – JEE-MAIN 2026

Home Maths Relations and FunctionsQuestion #15

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 _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Domain Restrictions

Always start by identifying the domain for logarithmic expressions to ensure valid solutions. Remember that the base must be positive and not equal to 1, and the argument must be positive.

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Step 1: Determine the Domain of the Equation✦ Active

For the logarithmic terms $\log_{(x + 1)} (2 x^{2} + 5 x + 3)$ and $\log_{(2 x + 3)} (x^{2} + 2 x + 1)$ to be defined, we must satisfy the following conditions:

x + 1 > 0 ⟹ x > - 1

x + 1 \neq 1 ⟹ x \neq 0

2 x^{2} + 5 x + 3 > 0 ⟹ (2 x + 3) (x + 1) > 0

2 x + 3 > 0 ⟹ x > - 3 / 2

2 x + 3 \neq 1 ⟹ x \neq - 1

x^{2} + 2 x + 1 > 0 ⟹ (x + 1)^{2} > 0 ⟹ x \neq - 1

Combining all conditions, the valid domain for $x$ is $x \in (- 1, \infty)$ and $x \neq 0$ .

Step 2: Simplify the Logarithmic Equation○ Expand

Rewrite the arguments using factorization: $2 x^{2} + 5 x + 3 = (2 x + 3) (x + 1)$ and $x^{2} + 2 x + 1 = (x + 1)^{2}$ . The given equation is:

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

Using logarithm properties $\log_{b} (M N) = \log_{b} M + \log_{b} N$ and $\log_{b} (M^{k}) = k \log_{b} M$ :

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

Since $\log_{(x + 1)} (x + 1) = 1$ , the equation simplifies to:

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

💡 Teacher's Secret Hint

Ensure all logarithm properties are applied correctly, especially for the base and argument transformations.

Step 3: Solve the Transformed Equation and Verify Solutions○ Expand

Let $y = \log_{(x + 1)} (2 x + 3)$ . Then, using the change of base formula $\log_{b} a = \frac{1}{\log_{a} b}$ , we have $\log_{(2 x + 3)} (x + 1) = \frac{1}{y}$ . Substitute these into the simplified equation:

y + 1 = 4 - \frac{2}{y}

y^{2} + y = 4 y - 2

y^{2} - 3 y + 2 = 0

(y - 1) (y - 2) = 0

This gives two possible values for $y$ : $y = 1$ or $y = 2$ .

Case 1: $y = 1$

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

This solution $x = - 2$ is not in the domain $x \in (- 1, \infty)$ , so it is rejected.

Case 2: $y = 2$

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

2 x + 3 = x^{2} + 2 x + 1

x^{2} = 2 ⟹ x = \pm \sqrt{2}

We check these solutions against the domain $x \in (- 1, \infty)$ and $x \neq 0$ :

- For $x = \sqrt{2} \approx 1.414$ : This satisfies the domain conditions. So $x = \sqrt{2}$ is a valid solution.

- For $x = - \sqrt{2} \approx - 1.414$ : This does not satisfy $x > - 1$ . So $x = - \sqrt{2}$ is rejected.

The only real solution to the equation is $x = \sqrt{2}$ . The sum of squares of all real solutions is $(\sqrt{2})^{2} = 2$ .

💡 Teacher's Secret Hint

Always verify the obtained solutions against the domain restrictions established in Step 1. This is a common pitfall in logarithmic equations.

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Maths Question 21 – JEE-MAIN 2026

Clue 1 of 3: Domain Restrictions

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