If the sum of all real x such that (2x - 4)^3 (4x -2)^3=. (4x 2x - ...
Solution:
Given, (2x - 4)^3 (4x -2)^3 = (4x 2x - 6)^3
Step 1: Simplifying the given equation
Expanding (2x - 4)^3, (4x -2)^3 and (4x + 2x - 6)^3, we get
8(x - 2)^3 64(x - 1)^3 = 64(x - 1)(x - 3)^3
2(x - 2)^3 16(x - 1)^3 = 16(x - 1)(x - 3)^3
(x - 2)^3 8(x - 1)^3 = 8(x - 1)(x - 3)^3
(x - 2)^3 = (x - 1)(x - 3)^3 / 8
Step 2: Solving the equation
Taking cube root on both sides, we get
x - 2 = (x - 1)^(1/3) (x - 3)^(1/3) / 2^(1/3)
x - 2 = [(x - 1)(x - 3)]^(1/3) / 2^(1/3)
Let y = (x - 1)(x - 3), then the equation becomes
x - 2 = y^(1/3) / 2^(1/3)
2^(1/3) (x - 2) = y^(1/3)
Cubing both sides, we get
2(x - 2)^3 = (x - 1)(x - 3)
2(x^3 - 6x^2 + 12x - 8) = x^2 - 4x + 3
2x^3 - 13x^2 + 20x - 11 = 0
Solving this cubic equation, we get
x = 1, 11/4, 3
Hence, the sum of all real x is 1 + 11/4 + 3 = 19/4
Step 3: Finding pq
The product of p and q is 19*4 = 76
Since p and q are co-prime, the only possible value of p and q are 1 and 76 or 76 and 1
Therefore, the value of (p q) is 76. Answer: Option (D) 11.
If the sum of all real x such that (2x - 4)^3 (4x -2)^3=. (4x 2x - ...
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