How do you solve #-7x ^ { 2} + 21x = 0#?

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Answer 1 · 2017-07-15T17:59:48

See a solution process below:

First, factor a #-7x# from each term on the left side of the equation:

#(-7x * x) + (-7x * -3) = 0#

#-7x(x + (-3)) = 0#

#-7x(x - 3) = 0#

Now, solve each term on the left for #0# to find the solutions:

Solution 1

#-7x = 0#

#(-7x)/color(red)(-7) = 0/color(red)(-7)#

#(color(red)(cancel(color(black)(-7)))x)/cancel(color(red)(-7)) = 0#

#x = 0#

Solution 2

#x - 3 = 0#

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

#x - 0 = 3#

#x = 3#

The solutions are:

#x = 0# and #x = 3#