How do you solve #(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# and solve for #z#:

Solution 1:

#4z + 2 = 0#

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

#4z + 0 = -2#

#4z = -2#

#(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}#