How do you solve #7( x ^ { 2} - 2x + 1) = 0#?

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Answer 1 · 2018-01-30T16:04:52

See a solution process below:

First, divide each side of the equation by #color(red)(7)# to eliminate the need for parenthesis while keeping the equation balanced:

#(7(x^2 - 2x + 1))/color(red)(7) = 0/color(red)(7)#

#(color(red)(cancel(color(black)(7)))(x^2 - 2x + 1))/cancel(color(red)(7)) = 0#

#x^2 - 2x + 1 = 0#

Next, factor the left side of the equation as:

#(x - 1)(x - 1) = 0#

Because both factors on the left side are the same there will be only one solution. We can solve the factor for #0#:

#x - 1 = 0#

#x - 1 + color(red)(1) = 0 + color(red)(1)#

#x - 0 = 1#

#x = 1#