Question

How do you simplify ${{\left( 2-\sqrt{2} \right)}^{2}}$?

Answer 1

Hint: In this problem we need to simplify the given expression. For this we are going to use some of the algebraic and exponential formulas. We have the expression ${{\left( 2-\sqrt{2} \right)}^{2}}$. So, we will first expand the term by using the exponential rule $a \times a \times a \times a \times a \times a....\text{ n times}={{a}^{n}}$. After expanding the given expression, we will use the algebraic formula $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$. Now we will simplify the equation by using the above formula. After simplification we will use the distribution law of multiplication on the terms in the obtained equation. After applying the distribution law of multiplication, we will simplify the obtained equation to get the required result.

Formulas Used:
1. $a.a.a.a.a.a....\text{ n times}={{a}^{n}}$.
2. $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$.
3. $x\left( y+z \right)=xy+xz$
4. $x\left( y-z \right)=xy-xz$
5. $\sqrt{2}\times \sqrt{2}=2$

Complete Step by Step Procedure:
Given expression is ${{\left( 2-\sqrt{2} \right)}^{2}}$.
We have the exponential rule $a.a.a.a.a.a....\text{ n times}={{a}^{n}}$. From this rule we can write ${{\left( 2-\sqrt{2} \right)}^{2}}$ as
${{\left( 2-\sqrt{2} \right)}^{2}}=\left( 2-\sqrt{2} \right)\left( 2-\sqrt{2} \right)$
We can observe that the above equation is in the form of $\left( a+b \right)\left( c+d \right)$. We know that $\left( a+b \right)\left( c+d \right)=a\left( c+d \right)+b\left( c+d \right)$. Applying this formula in the above equation, then we will get
${{\left( 2-\sqrt{2} \right)}^{2}}=2\left( 2-\sqrt{2} \right)-\sqrt{2}\left( 2-\sqrt{2} \right)$
From the distribution law of multiplication, we have that $x\left( y-z \right)=xy-xz$. Applying this formula in the above equation, then we will get
${{\left( 2-\sqrt{2} \right)}^{2}}=2\times 2-2\times \sqrt{2}-\sqrt{2}\times 2-\sqrt{2}\left( -\sqrt{2} \right)$
We know that when we multiplied a negative sign with a negative sign, we will get positive sign as the result. At the same time, we have the value $\sqrt{2}\times \sqrt{2}=2$. Applying these conditions and simplifying the above equation, then we will get
$\begin{align}
  & {{\left( 2-\sqrt{2} \right)}^{2}}=4-2\sqrt{2}-2\sqrt{2}+2 \\
 & \Rightarrow {{\left( 2-\sqrt{2} \right)}^{2}}=6-4\sqrt{2} \\
\end{align}$
Taking $2$ as common in the RHS, then we will have
$\Rightarrow {{\left( 2-\sqrt{2} \right)}^{2}}=2\left( 3-2\sqrt{2} \right)$.

Hence the simplified value of ${{\left( 2-\sqrt{2} \right)}^{2}}$ is $2\left( 3-2\sqrt{2} \right)$.

Note:
We can also follow another method to solve this problem. We have algebraic formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Simplifying the given expression by using the above formula, then we will get
$\begin{align}
  & {{\left( 2-\sqrt{2} \right)}^{2}}={{2}^{2}}+{{\left( \sqrt{2} \right)}^{2}}-2\times 2\times \sqrt{2} \\
 & \Rightarrow {{\left( 2-\sqrt{2} \right)}^{2}}=4+2-4\sqrt{2} \\
 & \Rightarrow {{\left( 2-\sqrt{2} \right)}^{2}}=6-4\sqrt{2} \\
 & \Rightarrow {{\left( 2-\sqrt{2} \right)}^{2}}=2\left( 3-2\sqrt{2} \right) \\
\end{align}$
Hence the answer from both the methods is the same.