quadratic equations (3t – 1) (3t + 1) = 0
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( 3T - 1 ) ( 3T + 1 ) = 0
(3T)² - (1)² = 0
9T² - 1 = 0
9T² = 1
T² = 1/9
T = √1/9 = 1/3 or -1/3
(3T)² - (1)² = 0
9T² - 1 = 0
9T² = 1
T² = 1/9
T = √1/9 = 1/3 or -1/3
Answered by
4
(3t - 1 ) ( 3t + 1 ) = 0
(3t)^2 - (1)^2 = 0
9t² - 1 = 0
t² = 1/9
t = 1/3 or -1/3
(3t)^2 - (1)^2 = 0
9t² - 1 = 0
t² = 1/9
t = 1/3 or -1/3
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