How do you solve the equation: #5w^2-5w=0#?

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Answer 1 · 2018-03-14T17:35:50

Remove extra coefficients to reveal two solutions #w=1# and #w=0#

There is a common factor on the left-hand side that is equal to #5w#. If you pull that common factor you you see:

#5w(w-1)=0#

Since anything multiplied by zero is zero, we can look at the two scenarios where you get 0=0:

#(w-1)=0 rArr color(red)(w=1)#

#5w=0 rArr color(blue)(w=0)#

We can also treat this like a quadratic equation, where:

#a=5#
#b=-5#
#c=0#

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

#w=(-(-5)+-sqrt((-5)^2-4(5*0)))/(2(5))#

#w=(5+-sqrt(25))/10#

#w=(5+-5)/10 rArr w=1, w=0#