JEE Advanced Maths: Theory of Equations questions with solutions

Q1. Solve the simultaneous equations: (x + y)/(x - y) + (x - y)/(x + y) = 13/6 and x*y = 5.

1. (x, y) = (5, 1) or (1, 5) or (-5, -1) or (-1, -5)
2. (x, y) = (5, 1) or (1, 5) only
3. (x, y) = (2.5, 2) or (2, 2.5)
4. No real solution

Answer: (x, y) = (5, 1) or (1, 5) or (-5, -1) or (-1, -5)

Setting t = (x+y)/(x-y) turns equation one into a solvable quadratic. Each value of t gives a linear ratio between x and y, which with xy = 5 yields the pairs.

Q2. Solve for real x: |x² + 3x + 2| + x + 1 = 0.

1. x = -1 and x = -3
2. x = -1 only
3. x = -2 and x = -3
4. No real solution

Answer: x = -1 and x = -3

Removing the absolute value requires casework on the sign of (x+1)(x+2). Each case gives a quadratic whose roots must be checked against the case's interval.

Q3. Solve the equation x⁴ - 2x³ + 3x² - 2x + 1 = 0.

1. x = 1
2. x = -1
3. x = (1 +- sqrt5)/2
4. x = (3 +- sqrt5)/2

Answer: x = (1 +- sqrt5)/2

The coefficients are palindromic, so divide by x² and substitute y = x + 1/x. This reduces to a quadratic in y. Solving gives y, then solving x + 1/x = y for each y produces the roots. The real roots are x = (1 +- sqrt5)/2 (the others are complex).