How do you solve #m^ { 2} = 21m - 108#?

1 Answer
May 27, 2017

See a solution process below:

Explanation:

First, subtract #color(red)(21m)# and add #color(blue)(108)# to both sides of the equation to form a quadratic on the left side of the equation while keeping the equation balanced:

#m^2 - color(red)(21m) + color(blue)(108) = 21m - 108 - color(red)(21m) + color(blue)(108)#

#m^2 - 21m + 108 = 21m - color(red)(21m) - 108 + color(blue)(108)#

#m^2 - 21m + 108 = 0 - 0#

#m^2 - 21m + 108 = 0#

Next, factor the quadratic on the left side of the equation as:

#(m - 12)(m - 9) = 0#

Now, solve each term on the left side of the equation for #0#:

Solution 1)

#m - 12 = 0#

#m - 12 + color(red)(12) = 0 + color(red)(12)#

#m - 0 = 12#

#m = 12#

Solution 2)

#m - 9 = 0#

#m - 9 + color(red)(9) = 0 + color(red)(9)#

#m - 0 = 9#

#m = 9#

The solution is: #m = 12# and #m = 9#