How do you solve $x^{2} - 2 x + 1 = 0$ $x^2-2x+1=0$ by completing the square?

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Answer 1 · 2017-09-28T22:13:12

${(x - 1)}^{2} = 0$ $(x-1)^2=0$

$x^{2} - 2 x + 1 = 0$ $x^2-2x+1=0$

Let's start by subtracting $1$ $1$ from both sides so that the $x$ $x$ terms are isolated:

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

Now, let's compute ${(\frac{b}{2})}^{2}$ $(b/2)^2$ and add it to both sides of the equation:

${(\frac{b}{2})}^{2} = {(- \frac{2}{2})}^{2} = 1$ $(b/2)^2=(-2/2)^2=1$

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

Let's factorize the left hand side and simplify:

$(x - 1) (x - 1) = 0$ $(x-1)(x-1)=0$

${(x - 1)}^{2} = 0$ $(x-1)^2=0$

How do you solve x2−2x+1=0x^2-2x+1=0 by completing the square?