# Problem with algebra homework

I don't know how to solve this question, can anyone help?

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

How do I solve for $x$? I'm confused. This is for Algebra 1, homework.

I don't understand how teacher said use substitution and all those stuff.

• Hints: What if $x=1$? What if you factor? You can use the quadratic formula if you cannot factor it. – Amzoti Jul 10 '13 at 0:43

## 4 Answers

It is key to learn how to recognize that

$$(a-b)^2=a^2-2ab+b^2\tag{1}$$

$$(a + b)^2 = a^2 + 2ab + b^2\tag{2}$$

Now, it looks like the left-hand side of your equation looks a bit like $(1)$, if we rewrite the equation $$x^2 - 2\cdot x\cdot 1 + (1)^2 = 0$$

Then our $a$ term here is $x$, and our $b$ term is $1$, which gives us $$x^2 - 2\cdot x\cdot 1 + (1)^2 = (x-1)^2 = (x - 1)(x - 1) = 0\tag{3}$$

Now, $$(x - 1)^2 = 0 \iff x = 1$$

We can "double check" our work by "plugging in" x = 1 into the original equation:

$$\text{At }\; x = 1 \implies x^2 - 2x + 1 = (1)^2 - 2\cdot 1 + 1 = 1 - 2 + 1 = 0$$

So our solution is, indeed, $x = 1$.

There is only one value of $x$ which makes the equation true, and $x = 1$ is called a "zero". It is also a repeated root of the polynomial $$f(x) = x^2 - 2x + 1 = (x - 1)(x-1)$$ and it's called a repeated root because of the repeated factor $(x - 1)(x - 1)$, each of which is zero exactly when $x = 1$.

• nice and clean + 1 – Amzoti Jul 11 '13 at 1:07

$(x-1)^2=(x-1)(x-1)=x^2-2x+1$

hence solving $\\$ $x^2-2x+1=0$ is equivalent to solving $(x-1)^2=0$. Taking the square root on both sides of the last equation you get : $x-1=0$.

so the solution is $x=1$

A general approach to your problem would go like this:

whenever you have a quadratic equation of the form: $ax^2+bx+c=0$

you can find $x$ using the formula: $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

In your special case you have : $a=1 \ \ \ b=-2 \ \ \ c=1$

You can either use the fact that $(x-y)^2=(x^2-2xy+y^2),$ or try the quadratic formula: $\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

Another method at your disposal is graphing. The graph shows that there is a single solution at $x=1$. You'll want to test this solution by plugging it into the original equation and verifying that the result is exactly zero. This is because a result obtained by graphing could be subject to a rounding error, unlike the algebraic methods already shown. 