How do you solve #z( z - 2) = 2#?

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Answer 1 · 2017-04-25T00:15:42

#z=1+√3 or z=1−√3#

Let's solve your equation step-by-step.
#z(z−2)=2#

Step 1: Simplify both sides of the equation.
#z2−2z=2#

Step 2: Subtract 2 from both sides.
#z2−2z−2=2−2#
#z2−2z−2=0#

Step 3: Use quadratic formula with a=1, b=-2, c=-2.

#z=−b±√(b2−4ac)/"2a"#

#z=−(−2)±√((−2)2−4(1)(−2))/"2(1)"#

#z=(2±√12)/"2"#

#z=1+√3 or z=1−√3#

hope this helps!

Answer 2 · 2017-04-25T00:21:17

#z=1+sqrt3#

#z=1-sqrt3#

Order of operations.
Multiply first.

#z( z - 2) = 2#

#z^2 - 2z= 2#

#z^2 - 2z-2= 0#

Solve the quadratic equation with your preferred method.
I will use the quadratic formula.

#z=(-b+-sqrt(b^2-4ac))/(2a)#

#z=(-(-2)+-sqrt((-2)^2-4(1)(-2)))/(2(1))#

#z=(2+-sqrt(4+8))/(2)#

#z=(2+-sqrt(12))/(2)#

#z=(2+-sqrt(4*3))/(2)#

#z=(2+-2sqrt(3))/(2)#

#z=(cancel2+-cancel2sqrt(3))/(cancel2)#

#z=1+-sqrt(3)#

So,

#z=1+sqrt3#

#z=1-sqrt3#