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

Mar 12, 2018

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:

${\left(x - 1\right)}^{2} = {x}^{2} - 2 x + 1$

Now we re-write the equation:

${x}^{2} - 2 x + 1 = 4$

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

${x}^{2} - 2 x - 3 = 0$

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

plug them into the quadratic equation:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{2 \pm \sqrt{4 - \left(- 12\right)}}{2} = \frac{2 \pm \sqrt{16}}{2}$

$x = \frac{2 \pm 4}{2}$

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

$x = \frac{2 + 4}{2} = \frac{6}{2} = \textcolor{red}{3}$

$x = \frac{2 - 4}{2} = - \frac{2}{2} = \textcolor{red}{- 1}$

Mar 12, 2018

$x = - 1$ or $x = - 3$
${\left(x - 1\right)}^{2} = 4$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow \left(x - 1\right) = \pm 2$
{: ("If ", (x-1)=-2,color(white)("xxxxxxxx")"If ",(x-1)=+2), (,rarr x=-1,,rarrx=3) :}