Chapter 6.4, Problem 3OE

(a)

To determine

To Simplify:

  ${\left(x-y\right)}^2$

Answer to Problem 3OE

  $x^2-2xy+y^2$

Explanation of Solution

Given:

The expression is :

  ${\left(x-y\right)}^2$

Calculation:

  ${\left(x-y\right)}^2=\left(x-y\right)\left(x-y\right)$

Use FOIL method:

F represents the multiplication of first terms of each binomial = $x\cdot x=x^2$

O represents the multiplication of the outermost terms of the product = $x\cdot (-y)=-xy$

I represents the multiplication of the innermost terms of the product = $\text{(-}y)\cdot x=-xy$

L represents the multiplication of the last terms of each binomial = $\text{(-}y)\cdot (-y)=y^2$

Add all the products:

  $\begin{array}{l}{x}^2-xy-xy+y^2\\ =x^2-2xy+y^2\\ \text{So, }\\ {\left(x-y\right)}^2=x^2-2xy+y^2\end{array}$

(b)

To determine

To Simplify:

  ${\left(\sqrt{5}-\sqrt{2}\right)}^2$

Answer to Problem 3OE

  $7-2\sqrt{10}$

Explanation of Solution

Given:

The expression is :

  ${\left(\sqrt{5}-\sqrt{2}\right)}^2$

Calculation:

Given:

  ${\left(\sqrt{5}-\sqrt{2}\right)}^2\mathrm{..............}(1)$

From part (a),

  ${\left(x-y\right)}^2=x^2-2xy+y^2\mathrm{............}(2)$

Comparing equation (1) with (2),

  $x=\sqrt{5}\text{ , }y=\sqrt{2}$

Substitute the values in equation (2),

  $\begin{array}{l}{\left(x-y\right)}^2=x^2-2xy+y^2\\ \Rightarrow {\left(\sqrt{5}-\sqrt{2}\right)}^2={\left(\sqrt{5}\right)}^2-2\cdot \sqrt{5}\cdot \sqrt{2}+{\left(\sqrt{2}\right)}^2=5-2\sqrt{10}+2=7-2\sqrt{10}\end{array}$

So,

  ${\left(\sqrt{5}-\sqrt{2}\right)}^2=7-2\sqrt{10}$

(c)

To determine

To Simplify:

  ${\left(2-\sqrt{3}\right)}^2$

Answer to Problem 3OE

  $7-4\sqrt{3}$

Explanation of Solution

Given:

The expression is :

  ${\left(2-\sqrt{3}\right)}^2$

Calculation:

Given:

  ${\left(2-\sqrt{3}\right)}^2\mathrm{..............}(1)$

From part (a),

  ${\left(x-y\right)}^2=x^2-2xy+y^2\mathrm{............}(2)$

Comparing equation (1) with (2),

  $x=2\text{ , }y=\sqrt{3}$

Substitute the values in equation (2),

  $\begin{array}{l}{\left(x-y\right)}^2=x^2-2xy+y^2\\ \Rightarrow {\left(2-\sqrt{3}\right)}^2={\left(2\right)}^2-2\cdot 2\cdot \sqrt{3}+{\left(\sqrt{3}\right)}^2=4-4\sqrt{3}+3=7-4\sqrt{3}\end{array}$

So,

  ${\left(2-\sqrt{3}\right)}^2=7-4\sqrt{3}$