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

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

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Answer 1 · 2015-06-25T15:44:07

$x = 1$ $x=1$

Explanation:

Given
$XXXX$ $color(white)("XXXX")$ $x^{2} - 2 x + 1 = 0$ $x^2-2x+1 = 0$
Factor the left side to get
$XXXX$ $color(white)("XXXX")$ $(x - 1) (x - 1) = 0$ $(x-1)(x-1) = 0$
From which it follows that
$XXXX$ $color(white)("XXXX")$ $(x - 1) = 0$ $(x-1) = 0$
and
$XXXX$ $color(white)("XXXX")$ $x = 1$ $x=1$

Answer 2 · 2015-06-25T17:48:58

This is a quadratic equation, that you can factorise.

Explanation:

Find a pair of numbers that add up to $- 2 x$ $-2x$ and multiplies to $+ 1$ $+1$

Maybe you'll see that that would be $- 1 and - 1$ $-1and-1$

Now you can factorise:
$= (x - 1) \cdot (x - 1) = 0 \to x - 1 = 0 \to x = 1$ $=(x-1)*(x-1)=0->x-1=0->x=1$

Check your answer:
$1^{2} - 2 \cdot 1 + 1 = 0$ $1^2-2*1+1=0$
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How do you solve $x^{2} - 2 x + 1 = 0$ $x^2 - 2x +1=0$ ?

2 Answers

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