How do you solve $(4 z + 2) (1 - z) = 0$ $(4z + 2) ( 1- z ) = 0$ ?

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Answer 1 · 2018-03-20T16:39:51

See a solution process below:

To solve this and find the roots of the equation equate each term on the left to $0$ $0$ and solve for $z$ $z$ :

Solution 1:

$4 z + 2 = 0$ $4z + 2 = 0$

$4 z + 2 - 2 = 0 - 2$ $4z + 2 - color(red)(2) = 0 - color(red)(2)$

$4 z + 0 = - 2$ $4z + 0 = -2$

$4 z = - 2$ $4z = -2$

$\frac{4 z}{4} = - \frac{2}{4}$ $(4z)/color(red)(4) = -2/color(red)(4)$

$(color(red)(cancel(color(black)(4)))z)/cancel(color(red)(4)) = -1/2$

$z = -1/2$

Solution 2:

$1 - z = 0$

$1 - z + color(red)(z) = 0 + color(red)(z)$

$1 - 0 = z$

$1 = z$

$z = 1$

The Solution Set Is: $z = {-1/2, 1}$

How do you solve (4z + 2) ( 1- z ) = 0(4z+2)(1−z)=0(4z + 2) ( 1- z ) = 0?