How do you solve #(x-1)^2=4#?

2 Answers
Mar 12, 2018

Answer:

Using the quadratic formula, you can find that x=-1 and x=3

Explanation:

On my first reaction, i thought that you could just take the square root of both sides and have your answer... which is 3. However, I remembered at the last minute that the square of a negative number is positive, so that means there are two possible solutions.

Without using the quadratic formula, I guessed that x-1 has to equal 2 or -2, meaning that x could be 3 (3-1=2) or -1 (-1-1=-2).

Since this is a method for solving a quadratic, we'll also use the quadratic formula. First we have to expand the exponential via FOIL:

#(x-1)^2 = x^2-2x+1#

Now we re-write the equation:

#x^2-2x+1=4#

Now we move 4 to the Left-hand Side (LHS) to set up our quadratic:

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

our coefficients for the quadratic equation are:
#a=1#
#b=-2#
#c=-3#

plug them into the quadratic equation:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(2+-sqrt(4-(-12)))/(2)=(2+-sqrt(16))/(2)#

#x=(2+-4)/(2)#

and you solve once for "-b+..." and again for "-b-..."

#x=(2+4)/(2)=6/2=color(red)3#

#x=(2-4)/(2)=-2/2=color(red)(-1)#

Mar 12, 2018

Answer:

#x=-1# or #x=-3#

Explanation:

#(x-1)^2=4#
#color(white)("XXX")rarr (x-1)=+-2#

#{: ("If ", (x-1)=-2,color(white)("xxxxxxxx")"If ",(x-1)=+2), (,rarr x=-1,,rarrx=3) :}#