How do you solve #3( x - 2 ) ^ { 2 } - 2 0 = 5 5#?

1 Answer
Oct 12, 2016

#x_1 = 7#
#x_2 = -3#

Explanation:

This equation can be solved in a number of ways, but I think this way is the simplest and leads to the fewest errors.

#1.#Evaluate the brackets:

#3(x-2)^2-20=55#

In order to evaluate #(x-2)^2#, we just need to remember that any number squared can just be written as being multiplied by itself.

#3(x-2)(x-2)-20=55#

We can evaluate #(x-2)(x-2)# using the FOIL method

#3(x^2-4x+4)-20=55#

#2.#Expand the brackets:

#3(x^2-4x+4)-20=55#
#3x^2-12x+12-20=55#

#3.# Set the equation equal to #0# and collect like terms:

#3x^2-12x+12-20=55#
#3x^2-12x+12-20-55=0#
#3x^2-12x-63=0#

#4.#Solve for the unknown:

#3x^2-12x-63=0#

The easiest way to solve this is to have a coefficient of #1# for #x^2#, as this will make factorising the equation easier:

#x^2-4x-21=0#

At this stage, we have a simplified equation equal to #0#. In order to find the roots of #x#, we need to find two factors which multiply to give #c# (in this case #-21#) and add to give #b# (in this case #-4#):

#x^2-7x+3x-21=0#

Then we factorise each 'pair' of terms in the equation:

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

Since we are multiplying #(x-7)# by #x# and by #+3#, we can put these two multipliers in one bracket to make solving the equation easier:

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

#5.#Set each bracket equal to #0# to get values for #x#:

Since we know that if #a*b=0#, then #a# or #b# must be equal to #0#, we can apply that rule here. But since we don't know which factor equals #0#, we set both equal to #0# and solve for both.

#x-7=0#
#x=7#

#x+3=0#
#x=-3#

#x_1 = 7#
#x_2 = -3#