For the function $f (x) = x^{2} - 2 x + 1$ $f(x) = x^2 - 2x + 1$ , how do you solve f(x) = 0?

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For the function $f (x) = x^{2} - 2 x + 1$ $f(x) = x^2 - 2x + 1$ , how do you solve f(x) = 0?

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Answer 1 · 2016-06-30T20:20:35

$x = 1$ $x=1$

Explanation:

If $f (x) = x^{2} - 2 x + 1$ $f(x)=x^2-2x+1$
when $f (x) = 0$ $f(x)=0$
then $x^{2} - 2 x + 1 = 0$ $x^2-2x+1=0$
and, factoring the left side,:
${(x - 1)}^{2} = 0$ $(x-1)^2=0$
$\Rightarrow x - 1 = 0$ $rArr x-1=0$
$\Rightarrow x = 1$ $rArr x=1$

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For the function $f (x) = x^{2} - 2 x + 1$ $f(x) = x^2 - 2x + 1$ , how do you solve f(x) = 0?

1 Answer

Explanation:

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For the function f(x)=x2−2x+1f(x) = x^2 - 2x + 1, how do you solve f(x) = 0?

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For the function $f (x) = x^{2} - 2 x + 1$ $f(x) = x^2 - 2x + 1$ , how do you solve f(x) = 0?