Anonymous

|2x-1|^2 = |1-2x| |2x-1|=0--->x=1/2 |2x-1|=1--->x=1 , x=0 --->x=1 , x=0 , x=1/2

Anonymous

(2x - 1)² = |1 - 2x| ← the contents of an absolute value can be positive or negative (2x - 1)² = ± (1 - 2x) 4x² - 4x + 1 = ± (1 - 2x) First case: 4x² - 4x + 1 = + (1 - 2x) 4x² - 4x + 1 = 1 - 2x 4x² - 2x = 0 2x.(2x - 1) = 0 x.(2x - 1) = 0 x = 0 x = 1/2 Second case: 4x² - 4x + 1 = - (1 - 2x) 4x² - 4x + 1 = - 1 + 2x 4x² - 6x = - 2 x² - (6/4).x = - 2/4 x² - (3/2).x + (3/4)² = - (2/4) + (3/4)² x² - (3/2).x + (3/4)² = 1/16 [x - (3/4)]² = (± 1/4)² x - (3/4) = ± 1/4 x = (3/4) ± (1/4) x = (3 ± 1)/4 x = (3 + 1)/4 → x = 1 x = (3 - 1)/4 → x = 1/2 → Solution = { 0 ; 1/2 ; 1 }

Anonymous

4x^2 - 4x + 1 = 1 - 2x 4x^2 - 2x = 0 [ 2x ] [ 2x - 1 ] = 0 2x = 0 , 2x = 1 x = 0 , x = 1/2

Anonymous

Note • (2x-1)²=|2x-1|² • |1-2x|=|2x-1| hence |2x-1|²-|2x-1|=0 |2x-1|(|2x-1|-1|=0 2 cases: ► 2x-1=0 ⇒ x=1/2 ► |2x-1|-1=0 ⇒ |2x-1|=1⇒ 2x-1=±1 2 subcases ..► 2x-1=-1 ⇒ x=0 ..► 2x-1=1 ⇒ x=1 3 solutions x₁=0 ; x₂=1/2; x₃=1