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secondary 4 | E Maths
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Date Posted: 4 years ago
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(2x - 1)⁴ + (y + 2)⁴ = 0
Since (2x - 1)⁴ ≥ 0 and (y + 2)⁴ ≥ 0 for all real values of x and y,
(2x - 1)⁴ + (y + 2)⁴ ≥ 0 for real values of x and y
It follows that (2x - 1)⁴ = 0 and (y + 2)⁴ = 0 in order to satisfy the equation.
2x - 1 = 0 and y + 2 = 0
2x = 1 and y = -2
x = ½
x - y = ½ - (-2)
= ½ + 2
= 2½ or 5/2 or 2.5
Since (2x - 1)⁴ ≥ 0 and (y + 2)⁴ ≥ 0 for all real values of x and y,
(2x - 1)⁴ + (y + 2)⁴ ≥ 0 for real values of x and y
It follows that (2x - 1)⁴ = 0 and (y + 2)⁴ = 0 in order to satisfy the equation.
2x - 1 = 0 and y + 2 = 0
2x = 1 and y = -2
x = ½
x - y = ½ - (-2)
= ½ + 2
= 2½ or 5/2 or 2.5
See 1 Answer
X=1/2 and Y = -2. Note that the both the powers are even. Any perfect square must be more than or equal to 0. So the only possible case for both powers to be 0 is when both are zeros.
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