1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hints: What if $x=1$? What if you factor? You can use the quadratic formula if you cannot factor it. $\endgroup$ – Amzoti Jul 10 '13 at 0:43
6
$\begingroup$

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 $x = 1$.

$\endgroup$
  • $\begingroup$ nice and clean + 1 $\endgroup$ – Amzoti Jul 11 '13 at 1:07
2
$\begingroup$

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

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 : $a=1 \ \ \ b=-2 \ \ \ c=1$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$x^2-2x+1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.