JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

From the first equation x^2 - 3x + r = 0, the sum and product of the roots are: + = 3 = r From the second equation x^2 + 3x + r = 0, the sum and product of the roots are: 2 + 2 = -3 ( 2 )(2 ) = r = r We have a system of linear equations in terms of and : + = 3 + 4 = -6 Subtracting the first equation from the second gives: 3 = -9 = -3 Substituting = -3 into + = 3 gives: = 6 Now, we can find r: r = = 6 (-3) = -18 The roots of the third equation x^2 + 6x - m = 0 are given as 2 + + 2r and - 2 - r 2 . Let us calculate their values: R_1 = 2(6) + (-3) + 2(-18) = 12 - 3 - 36 = -27 R_2 = 6 - 2(-3) - -18 2 = 6 + 6 + 9 = 21 The product of the roots of the equation x^2 + 6x - m = 0 is -m. Therefore: -m = R_1 R_2 = (-27)(21) -m = -567 m = 567 Answer: 567