Question

How do you solve $2{{x}^{2}}-8x+16=0$ by completing the square?

Answer 1

Hint: In the completing the square method, we make a perfect square on the LHS of a given quadratic equation. For this, we first need to simplify the given equation by dividing it by $2$ to get ${{x}^{2}}-4x+4=0$. Then we have to divide and multiply the coefficient of the second term $-4x$ of the ${{x}^{2}}-4x+4=0$ by $2$. And then, we have to add and subtract the square of the coefficient of $2x$ in the equation obtained. Then applying the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we will be able to generate a perfect square on the LHS. Finally using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we will be able to factorise the LHS and hence obtain the solutions.

Complete step by step solution:
The equation given in the above question is
$\Rightarrow 2{{x}^{2}}-8x+16=0$
Since the factor $2$ is common to all the terms, we divide the above equation by $2$ to get
$\begin{align}
  & \Rightarrow \dfrac{2{{x}^{2}}-8x+16}{2}=\dfrac{0}{2} \\
 & \Rightarrow {{x}^{2}}-4x+4=0 \\
\end{align}$
Now, we divide and multiply the coefficient of $x$ by $2$ to get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+\dfrac{2\left( -4 \right)}{2}x+4=0 \\
 & \Rightarrow {{x}^{2}}-\left( 2 \right)2x+4=0 \\
\end{align}\]
The coefficient of $-2x$ in the above equation is equal to $2$. Adding and subtracting its square, we get
$\begin{align}
  & \Rightarrow {{x}^{2}}-\left( 2 \right)2x+{{\left( 2 \right)}^{2}}+4-{{\left( 2 \right)}^{2}}=0 \\
 & \Rightarrow {{x}^{2}}-2\left( 2 \right)x+{{\left( 2 \right)}^{2}}+4-4=0 \\
 & \Rightarrow {{x}^{2}}-2\left( 2 \right)x+{{\left( 2 \right)}^{2}}=0 \\
\end{align}$
Now, using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we write the above equation as
\[\begin{align}
  & \Rightarrow {{\left( x-2 \right)}^{2}}=0 \\
 & \Rightarrow \left( x-2 \right)\left( x-2 \right)=0 \\
 & \Rightarrow x=2,x=2 \\
\end{align}\]

Hence, we have solved the given equation using the completing the square method and obtained the solutions as $x=2$ and $x=2$.

Note: We must note that after the first step in the above solution, that is, after we had divided the given equation by $2$ and obtained the equation ${{x}^{2}}-4x+4=0$, we can observe that the LHS is already a perfect square. It can be written as ${{\left( x-2 \right)}^{2}}=0$. But since we had to solve the given equation using the completing the square method, we had to perform all of the steps of this method. We cannot skip any of the steps, even if we can simplify the equation in any of the intermediate steps.