for an open interval (0,1) of a set a that is a subset of x belonging to real numbers, is a a closed subset?
1 answer:
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Answer:
4(x - )² = 0
Step-by-step explanation:
Given
4x² - 4x + 1 = 0
To complete the square the coefficient of the x² term must be 1
Factor out 4 from 4x² - 4x
= 4(x² - x) + 1 = 0
add/subtract ( half the coefficient of the x- term )² to x² - x
4(x² + 2(- )x +
-
) + 1 = 0
4(x - )² - 1 + 1 = 0
4(x - )² = 0