Question

How do you solve ${{x}^{2}}-2x+2$ by completing the square?

Answer 1

Hint: We first equate the given equation with 0 to solve it. We then separate the variables and the constants. We add 1 to both sides of the equation ${{x}^{2}}-2x=-2$. We then form a square for the left side of the new equation. Then we take the square root on both sides of the equation. From that we add 1 to the both sides to find the value of $x$ for ${{x}^{2}}-2x+2$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}-2x+2$. We equate it with 0 to get ${{x}^{2}}-2x+2=0$.
We then separate the variables and the constants ${{x}^{2}}-2x=-2$.
We first add 1 to both sides of ${{x}^{2}}-2x=-2$. We use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$.
$\begin{align}
  & {{x}^{2}}-2x+1=-2+1 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}=-1 \\
\end{align}$
We interchanged the numbers for $a=x,b=1$.
Now we have a quadratic equation ${{\left( x-1 \right)}^{2}}=-1$.
We need to find the solution of the given equation ${{\left( x-1 \right)}^{2}}=-1$.
We take square roots on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\left( x-1 \right)}^{2}}}=\sqrt{-1}=\pm i \\
 & \Rightarrow \left( x-1 \right)=\pm i \\
\end{align}$
Here $i$ is the complex value.
Now we add 1 to the both sides of the equation $\left( x-1 \right)=\pm i$ to get value for variable $x$.
$\begin{align}
  & \left( x-1 \right)+1=\pm i+1 \\
 & \Rightarrow x=1\pm i \\
\end{align}$

The given quadratic equation has two solutions and they are $x=1\pm i$.

Note: We try to verify the value of the root of $x=1\pm i$ for the equation ${{x}^{2}}-2x+2$.
Putting the value $x=1+i$ in the left side of the equation we get
$\begin{align}
  & {{x}^{2}}-2x+2 \\
 & \Rightarrow {{\left( 1+i \right)}^{2}}-2\left( 1+i \right)+2 \\
 & \Rightarrow 1+{{i}^{2}}+2i-2-2i+2 \\
 & \Rightarrow 1-1 \\
 & \Rightarrow 0 \\
\end{align}$
Therefore, the value $x=1+i$ satisfies the equation ${{x}^{2}}-2x+2$ as $f\left( 1+i \right)=0$.
Same thing can be said for $x=1-i$.