How do you solve #(x - 1)^2 = 9#?

2 Answers
Apr 10, 2016

Answer:

x1= 4 and x2=-2

Explanation:

x^2 - 2x*1 +1^2 = 9
x^2-2x+1=9

since this is a quadratic equation it has to be equal to 0, so:

x^2-2x+1-9=0
x^2-2x-8=0

we can solve this in two ways: the first way is to find the factors of x^2-2x-8 and the second one is to solve it by using the quadratic formula.

Let me factorize it.
(x-4)(x+2)=0

In order for this to be equal zero:
x-4=0 => x=4
and
x+2=0 => x=-2

The quadratic equation has always two roots or two solutions. Consequently, x1= 4 and x2=-2.

Apr 11, 2016

Answer:

#x=4,-2#

Explanation:

#color(blue)((x-1)^2=9#

Use the golden rule of Algebra: What you do on one side must be done on the other side also

Take the square root of both sides (#sqrt#)

#rarrsqrt((x-1)^2=sqrt9#

Use: #color(brown)(sqrt(x^2)=x#

#rarrx-1=sqrt9#

But, #color(brown)(sqrt9=+-sqrt9=(3,-3)#

Solve for both

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

(#3#)

#rarrx-1=3#

Add #1# both sides

#rarrxcancel(-1color(purple)(+1))=3color(purple)(+1)#

#rarrx=3+1#

#color(green)(rArrx=4#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

(#-3#)

#rarrx-1=-3#

Add #1# both sides

#rarrxcancel(-1color(purple)(+1))=-3color(purple)(+1#

#color(green)(rArrx=-2#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

#color(blue)( ul bar |x=(4,-2)|#