JEE Mathematics 2026 Question with Solution

Given: $\log_{(x+3)}(6x^2 + 28x + 30) = 5 - 2\log_{(6x+10)}(x^2 + 6x + 9)$

Find: The sum of all real solutions.

Factor the polynomial terms:

and

Substitute into the equation:

Using logarithmic properties:

Let

Then

using

So the equation becomes

Multiplying by $y$:

Hence,

Therefore,

which gives

So,

Check validity:

• For $x=1$, base is $4$ and argument is $64$, so it is valid.
• For $x=-1$, base is $2$ and argument is $8$, so it is valid.

Thus the sum of solutions is

Therefore, the correct option is B.

Given: $\log_{(x+3)}(6x^2 + 28x + 30) = 5 - 2\log_{(6x+10)}(x^2 + 6x + 9)$

Find: The sum of all real solutions.

After factoring,

so the equation reduces to expressions involving

and its reciprocal. This works because

Let

Then the equation directly becomes

which gives

Hence,

Now use the meaning of the logarithm:

So,

and the real solutions are $x=1$ and $x=-1$. Their sum is $0$. Therefore, the correct option is B.