In this chapter we will learn to expand binomials with power 2 such as \mathtt{( a+b)^{2} \ or\ ( a-b)^{2}} with solved examples.

At the end of the chapter some problems are also given for your practice.

Before learning the formulas, let us revise the basics first.

## What are binomials ?

T**he algebraic expression containing two terms separated by addition or subtraction signs **are called **binomials**.

Given below are examples of binomial with power 2.

\mathtt{\Longrightarrow \ ( x+6)^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 2x\ +\ 5y){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ \left(\frac{1}{2} -x\right)^{2}}

I hope you understood the concept of binomials. Given below the formula and proof of binomial expansion with power 2.

## Expanding \mathtt{( a+b)^{2}}

Here \mathtt{( a+b)^{2}} can be written as;

\mathtt{\Longrightarrow \ ( a+b)( a+b)}

Multiplying the individual terms.

\mathtt{\Longrightarrow \ a^{2} +ab+ab+b^{2}}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +2ab+b^{2}}

Hence, **the formula can be written as;**

\mathtt{( a+b)^{2} =\ a^{2} +2ab+b^{2}}

Please remember the above formula as it would help you solve the questions faster.

Let us now move towards another binomial expression.

## Expanding \mathtt{( a-b)^{2}}

Here \mathtt{( a-b)^{2}} can be expressed as;

\mathtt{\Longrightarrow \ ( a-b)( a-b)}\\\ \\ \mathtt{\Longrightarrow \ a^{2} -ab-ab+b^{2}}\\\ \\ \mathtt{\Longrightarrow \ a^{2} -2ab+b^{2}}

Hence,** the binomial can be expressed as formula**;

\mathtt{( a-b)^{2} =\ a^{2} -2ab+b^{2}}

This is another important formula that needs to be remembered.

## Adding / Subtracting two binomial expression square

(i) \mathtt{( a+b)^{2} +( a-b)^{2}}

Solving the expression by applying above mentioned formula.

\mathtt{\Longrightarrow \ \ a^{2} +2ab+b^{2} +\ a^{2} -2ab+b^{2}}\\\ \\ \mathtt{\Longrightarrow \ 2a^{2} +2b^{2}}\\\ \\ \mathtt{\Longrightarrow \ 2\ \left( a^{2} +b^{2}\right)}

Hence, we get the following formula;

\mathtt{( a+b)^{2} +( a-b)^{2} \Longrightarrow \ 2\ \left( a^{2} +b^{2}\right)}

Now let’s move on to subtracting the above expressions.

(ii) \mathtt{( a+b)^{2} -( a-b)^{2}}

Expanding each of the binomials using above formula;

\mathtt{\Longrightarrow \ \ a^{2} +2ab+b^{2} -\left( \ a^{2} -2ab+b^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ a^{2} +2ab+b^{2} -a^{2} +2ab-b^{2})}\\\ \\ \mathtt{\Longrightarrow \ 4ab}

Hence, the formula can be represented as;

\mathtt{( a+b)^{2} -( a-b)^{2} =\ 4ab}

### Other derived formulas from square of binomials

Using the above four formulas, one can derive other different formulas by moving variables from one side to another.

Given below are some list of such formulas.

Note that these formulas are **not important** and you don’t have to remember each of them. I am making the list here, just to given you an idea.

\mathtt{( i) \ \ \ \ a^{2} +b^{2} =\ ( a+b)^{2} -2ab}\\\ \\ \mathtt{( ii) \ \ \ a^{2} +b^{2} =( a+b)^{2} +\ 2ab}\\\ \\ \mathtt{( iii) \ \ \ ( a+b)^{2} =( a-b)^{2} +4ab}\\\ \\ \mathtt{( iv) \ \ \ ( a-b)^{2} =( a+b)^{2} -4ab}

I hope you understood all the formulas. Given below are some solved problems which will help you to learn how to apply these formulas.

## Square of binomial – solved problems

**Example 01**Expand the expression \mathtt{( 2x-7)^{2}}

**Solution**

Referring the formula;

\mathtt{( a-b)^{2} =\ a^{2} -2ab+b^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ ( 2x)^{2} -2\ ( 2x)( 7) \ +\ 7^{2}}\\\ \\ \mathtt{\Longrightarrow \ 4x^{2} -28x\ +\ 49}

Hence, the above expression is the solution.

**Example 02**

Expand \mathtt{\left( x+\frac{1}{3}\right)^{2}}

**Solution**

Referring the formula;

\mathtt{( a+b)^{2} =\ a^{2} +2ab+b^{2}}

Using the formula, we get;

\mathtt{\Longrightarrow \ ( x)^{2} +2\ ( x)\left(\frac{1}{3}\right) \ +\ \left(\frac{1}{3}\right)^{2}}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +\frac{2x}{3} +\frac{1}{9}}

**Example 03**

Expand \mathtt{\left(\frac{1}{5} x-\frac{3}{7} y\right)^{2}}

**Solution**

Referring the formula.

\mathtt{( a-b)^{2} =\ a^{2} -2ab+b^{2}}

Putting the values, we get;

\mathtt{\Longrightarrow \ \left(\frac{1}{5} x\right)^{2} -2\ \left(\frac{x}{5}\right)\left(\frac{3y}{7}\right) \ +\ \left(\frac{3y}{7}\right)^{2}}\\\ \\ \mathtt{\Longrightarrow \ \frac{x^{2}}{25} -\frac{6xy}{35} +\frac{9y^{2}}{49}}

Hence, the above expression is the solution.

**Example 04**

If \mathtt{x+\frac{1}{x} =\ 5} . Find the value of \mathtt{x^{2} +\frac{1}{x^{2}}}

**Solution**

Squaring the given equation on both sides.

\mathtt{\left( x+\frac{1}{x}\right)^{2} =\ 5}

Referring the formula;

\mathtt{( a+b)^{2} =\ a^{2} +2ab+b^{2}}

Using the formula, we get;

\mathtt{x^{2} +2( x)\left(\frac{1}{x}\right) +\frac{1}{x^{2}} =\ 25}\\\ \\ \mathtt{x^{2} +2+\frac{1}{x^{2}} =\ 25}\\\ \\ \mathtt{x^{2} +\frac{1}{x^{2}} =\ 25-2}\\\ \\ \mathtt{x^{2} +\frac{1}{x^{2}} =\ 23}

Hence, **23 is the required answer.**

**Example 05**

Find the value of xy, if following expression is given;

x + y = 10;

x – y = 6

Solution

Referring the formula;

\mathtt{( a+b)^{2} -( a-b)^{2} =\ 4ab}

Using the formula;

\mathtt{( x+y)^{2} -\ ( x-y)^{2} =\ 4xy}\\\ \\ \mathtt{( 10)^{2} -\ ( 6)^{2} =\ 4xy}\\\ \\ \mathtt{100\ -\ 36\ =\ 4xy}\\\ \\ \mathtt{64\ =\ 4xy}\\\ \\ \mathtt{xy\ =\ \frac{64}{4}}\\\ \\ \mathtt{xy\ =\ 16}

Hence, **the value of xy is 16.**

**Next chapter :** **Trinomial formula with power 2**