To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=1\times 1=1$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.
$$a=-1$$ $$b=-1$$
Rewrite $x^{2}-2x+1$ as $\left(x^{2}-x\right)+\left(-x+1\right)$.
$$\left(x^{2}-x\right)+\left(-x+1\right)$$
Factor out $x$ in the first and $-1$ in the second group.
$$x\left(x-1\right)-\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
$$\left(x-1\right)\left(x-1\right)$$
Rewrite as a binomial square.
$$\left(x-1\right)^{2}$$
To find equation solution, solve $x-1=0$.
$$x=1$$
Steps Using the Quadratic Formula
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$7x^{2}-14x+7=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $7$ for $a$, $-14$ for $b$, and $7$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$x^{2}-2x+1=-1+1$$
Add $-1$ to $1$.
$$x^{2}-2x+1=0$$
Factor $x^{2}-2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-1\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$
Simplify.
$$x-1=0$$ $$x-1=0$$
Add $1$ to both sides of the equation.
$$x=1$$ $$x=1$$
The equation is now solved. Solutions are the same.
