JEE Mathematics 2025 Question with Solution

Given: $x^2 + |2x - 3| - 4 = 0$

Find: The sum of the squares of all the roots.

Split the equation into cases using the absolute value.

For Case 1, when $x \geq \frac{3}{2}$, we have:

So,

From the solution, the valid root taken is

For Case 2, when $x < \frac{3}{2}$, we have:

So,

From the solution, the valid root taken is

Now compute the sum of squares:

Expanding as shown in the solution,

Therefore, the sum of the squares of all the roots is $6(2 - \sqrt{2})$. The correct option is B.

The solution's also displays "The Correct Option is D", but the worked solution concludes $6(2 - \sqrt{2})$, which matches option B. Following the solution working, B is the defensible answer.

Given: $x^2 + |2x - 3| - 4 = 0$

Find: The sum of the squares of all valid roots.

The expression inside the modulus changes sign at

So the equation must be solved in two intervals.

If $x \geq \frac{3}{2}$, then

Hence,

The root retained in the solution is

which satisfies $x \geq \frac{3}{2}$.

If $x < \frac{3}{2}$, then

Hence,

The root retained in the solution is

which satisfies $x < \frac{3}{2}$.

Now add the squares of these roots:

So the worked result corresponds to option B.