How do you solve #b^2=-3b#?

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Answer 1 · 2017-03-09T23:35:21

See the entire solution process below:

First, add #color(red)(3b)# to each side of the equation to put the equation in standard form:

#b^2 + color(red)(3b) = -3b + color(red)(3b)#

#b^2 + 3b = 0#

Next, factor a #b# out of each term on the left:

#(b * b) + (3 * b) = 0#

#b(b + 3) = 0#

Now, solve each term on the left for #0# to determine all of the solutions:

Solution 1:

#b = 0#

Solution 2:

#b + 3 = 0#

#b + 3 - color(red)(3) = 0 - color(red)(3)#

#b + 0 = -3#

#b = -3#

The Solutions Are:

#b = [-3, 0}#