Which of the following values of \( x \) will satisfy the equation \( |x-1|^{2}+3|x-1|-4=0 \) ?
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To solve \( |x-1|^{2}+3|x-1|-4=0 \), let \( y = |x-1| \). Then we can rewrite the equation as: \[ y^2 + 3y - 4 = 0 \] Now, we can factor this quadratic equation: \[ (y + 4)(y - 1) = 0 \] This gives us the solutions \( y + 4 = 0 \) or \( y - 1 = 0 \), which leads to: \[ y = -4 \quad (\text{not valid since } y = |x-1| \geq 0) \] \[ y = 1 \] Now we substitute back \( y = |x-1| \): \[ |x-1| = 1 \] This results in two cases: 1. \( x - 1 = 1 \) which gives \( x = 2 \) 2. \( x - 1 = -1 \) which gives \( x = 0 \) Thus, the values of \( x \) that satisfy the equation are \( x = 2 \) and \( x = 0 \).