The sum of all possible values of x satisfying the equation 2^(4x^2−22) × 2^(x^2+x+16) + 2^(2x+30) = 0, is:

Question

UPSC Quizroom · Accepted Answer

Step 1: Combine powers of 2

2^(4x² − 22 + x² + x + 16) + 2^(2x + 30) = 0

2^(5x² + x − 6) + 2^(2x + 30) = 0

Step 2: Substitution

Let y = 2^(2x² − x − 16)

Then the equation reduces to: y + 1/y = 1 → quadratic: 4y² − 4y + 1 = 0

Factor: (2y − 1)² = 0 → y = 1/2

Step 3: Back-substitute y

2^(2x² − x − 16) = 1/2 = 2^(−1)

2x² − x − 16 = −1 → 2x² − x − 15 = 0

Step 4: Solve quadratic

2x² − x − 15 = 0 → (2x + 5)(x − 3) = 0

x = 3 or x = −5/2

Step 5: Sum of all possible values

3 + (−5/2) = 1/2

Answer: 1/2

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