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

Apr 10, 2016

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

$x = 4 , - 2$

#### 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

$\rightarrow x - 1 = \sqrt{9}$

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

Solve for both

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

($3$)

$\rightarrow x - 1 = 3$

Add $1$ both sides

$\rightarrow x \cancel{- 1 \textcolor{p u r p \le}{+ 1}} = 3 \textcolor{p u r p \le}{+ 1}$

$\rightarrow x = 3 + 1$

color(green)(rArrx=4

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

($- 3$)

$\rightarrow x - 1 = - 3$

Add $1$ both sides

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

color(green)(rArrx=-2

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

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