Perfect Squares

The identities $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2\text{:}$ why they are true, how to expand with them, and how they make squares such as $99^2$ quick to work out.

Before you start

A square means a number or expression multiplied by itself: $(a + b)^2 = (a + b)(a + b)\text{.}$

In lesson 3.3 you expanded two brackets by multiplying each term in the first by each term in the second. A perfect square is two brackets that are the same, and it always expands in the same pattern.

$(a + b)^2$ Is Not $a^2 + b^2$

A common guess is that $(a + b)^2 = a^2 + b^2$ . Try numbers: with $a = 2$ and $b = 3$ , $(2 + 3)^2 = 5^2 = 25$ , but $2^2 + 3^2 = 4 + 9 = 13$ . The guess misses $12$ , and the rest of this lesson shows where the $12$ comes from: $2 \times 2 \times 3 = 12$ .

The Identity for $(a + b)^2$

By expanding. $(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2\text{.}$ Since $ba = ab$ , the two middle terms are like terms: $a^2 + 2ab + b^2$ .

By the frame. The two equal products $ab$ sit on one diagonal:

×	$a$	$+b$
$a$	$a^2$	$+ab$
$+b$	$+ab$	$+b^2$

$a^2 + ab + ab + b^2 = a^2 + 2ab + b^2$

By area. Draw a square of side $(a + b)$ and cut each side into $a$ and $b$ . The square is cut into a square $a$ by $a$ , a square $b$ by $b$ , and two rectangles $a$ by $b$ . That is why the middle term is $2ab$ , not $ab$ .

The first perfect square identity

$(a + b)^2 = a^2 + 2ab + b^2$

In words: the square of the first term, plus twice the product of the two terms, plus the square of the second term.

It is an identity: true for every value of $a$ and $b$ , and for any terms in place of $a$ and $b$ , such as $3x$ or $\frac{1}{2}$ .

The Identity for $(a - b)^2$

By expanding. $(a - b)^2 = (a - b)(a - b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2\text{.}$ The last term is $+b^2$ , because $(-b) \times (-b) = +b^2\text{.}$

The same comes from the first identity with $-b$ in place of $b$ : $(a + (-b))^2 = a^2 + 2a(-b) + (-b)^2 = a^2 - 2ab + b^2\text{.}$

By area (for $a$ greater than $b$ ). Start with a square of side $a$ , area $a^2$ . Cut a strip $b$ wide off the right and another off the bottom. What is left in the top left is a square of side $a - b$ .

The strip down the right runs the whole side: it is $b$ by $a$ , area $ab$ . The strip along the bottom also runs the whole side: it is $a$ by $b$ , area $ab$ . Both strips contain the corner square $b^2$ . Taking both away from $a^2$ takes the corner square $b^2$ away twice, once with each strip, so it must be added back once:

$(a - b)^2 = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2$

The second perfect square identity

$(a - b)^2 = a^2 - 2ab + b^2$

Only the middle term changes sign. The last term is always $+b^2$ .

Expanding with the Identities

Name the first term and the second term of the bracket, then write the three terms of the identity. Square the whole of each term, number and letter: $(4a)^2 = 16a^2$ .

Example 1: One letter

Expand (a) $(x + 4)^2$ ; (b) $(x - 3)^2$ .

Solution

(a) First term $x$ , second term $4$ : $x^2 + 2(x)(4) + 4^2 = x^2 + 8x + 16\text{.}$

(b) First term $x$ , second term $3$ , with a minus: $x^2 - 2(x)(3) + 3^2 = x^2 - 6x + 9\text{.}$

(a) $x^2 + 8x + 16$ ; (b) $x^2 - 6x + 9$

Example 2: Coefficients and two letters

Expand (a) $(4a + 3b)^2$ ; (b) $(5a - 2b)^2$ .

Solution

(a) The first term is $4a$ and the second is $3b$ .

$(4a)^2 + 2(4a)(3b) + (3b)^2 = 16a^2 + 24ab + 9b^2$

(b) The first term is $5a$ and the second is $2b$ .

$(5a)^2 - 2(5a)(2b) + (2b)^2 = 25a^2 - 20ab + 4b^2$

(a) $16a^2 + 24ab + 9b^2$ ; (b) $25a^2 - 20ab + 4b^2$

Example 3: Fractions

Expand (a) $(3y + \frac{1}{3})^2$ ; (b) $(x - \frac{1}{2})^2$ .

Solution

(a) $(3y)^2 + 2(3y)(\frac{1}{3}) + (\frac{1}{3})^2 = 9y^2 + 2y + \frac{1}{9}\text{,}$ since $2 \times 3y \times \frac{1}{3} = 2y$ .

(b) $x^2 - 2(x)(\frac{1}{2}) + (\frac{1}{2})^2 = x^2 - x + \frac{1}{4}$

(a) $9y^2 + 2y + \frac{1}{9}$ ; (b) $x^2 - x + \frac{1}{4}$

Example 4: A negative first term

Expand (a) $(-x - 3)^2$ ; (b) $(3 - 2x)^2$ . Write each answer in descending powers of $x$ .

Solution

(a) The first term is $-x$ and the second is $-3$ : $(-x)^2 + 2(-x)(-3) + (-3)^2 = x^2 + 6x + 9\text{.}$

This is $(x + 3)^2$ . It must be: $-x - 3 = -(x + 3)$ , and squaring removes the minus sign.

(b) $3^2 - 2(3)(2x) + (2x)^2 = 9 - 12x + 4x^2 = 4x^2 - 12x + 9\text{.}$

This is also $(2x - 3)^2$ : $3 - 2x$ and $2x - 3$ differ only in sign, and their squares are equal.

(a) $x^2 + 6x + 9$ ; (b) $4x^2 - 12x + 9$

Example 5: Spot the error

Nadia wrote (a) $(x - 4)^2 = x^2 - 16$ and (b) $(2x + 3)^2 = 2x^2 + 12x + 9\text{.}$ Check each with $x = 1$ , then correct it.

Solution

(a) $(1 - 4)^2 = 9$ , but $1 - 16 = -15$ . The middle term $-2(x)(4) = -8x$ is missing: $(x - 4)^2 = x^2 - 8x + 16\text{,}$ and $1 - 8 + 16 = 9$ . ✓

(b) $(2 + 3)^2 = 25$ , but $2 + 12 + 9 = 23$ . The first term is $(2x)^2 = 4x^2$ , not $2x^2$ : $(2x + 3)^2 = 4x^2 + 12x + 9\text{,}$ and $4 + 12 + 9 = 25$ . ✓

(a) $x^2 - 8x + 16$ ; (b) $4x^2 + 12x + 9$

Expanding and Simplifying

Example 6: Sums and differences of squares

Simplify (a) $(x + 3)^2 - (x - 3)^2$ ; (b) $(x + 2)^2 + (x - 2)^2$ ; (c) $2(x - 1)^2 - x(x - 4)\text{.}$

Solution

(a) $(x^2 + 6x + 9) - (x^2 - 6x + 9) = x^2 + 6x + 9 - x^2 + 6x - 9 = 12x$

(b) $(x^2 + 4x + 4) + (x^2 - 4x + 4) = 2x^2 + 8$

In each, the square was expanded inside a bracket first, so that the sign in front reached every term.

(a) $12x$ ; (b) $2x^2 + 8$ ; (c) $x^2 + 2$

Squaring Numbers Quickly

Split a number into a round number and a small part, then use an identity. The working is short enough to do in your head.

Example 7: Squares without a calculator

Without a calculator, find (a) $35^2$ ; (b) $99^2$ ; (c) $1001^2$ ; (d) $4.9^2$ .

Solution

(a) $(30 + 5)^2 = 30^2 + 2(30)(5) + 5^2 = 900 + 300 + 25 = 1225$

(b) $(100 - 1)^2 = 100^2 - 2(100)(1) + 1^2 = 10\,000 - 200 + 1 = 9801$

(d) $(5 - 0.1)^2 = 25 - 2(5)(0.1) + 0.01 = 25 - 1 + 0.01 = 24.01$

(a) $1225$ ; (b) $9801$ ; (c) $1\,002\,001$ ; (d) $24.01$

Numbers ending in 5

$45^2 = (40 + 5)^2 = 1600 + 400 + 25 = 2025\text{.}$ The first two terms, $40^2 = 1600$ and $2 \times 40 \times 5 = 400$ , are both whole hundreds, so a whole number ending in $5$ always has a square ending in $25$ .

Using $a + b$ and $ab$

Rearranging the identities gives $a^2 + b^2 = (a + b)^2 - 2ab$ and $a^2 + b^2 = (a - b)^2 + 2ab\text{.}$ So $a^2 + b^2$ can be found from $a + b$ (or $a - b$ ) and $ab$ , without knowing $a$ and $b$ .

Example 8: Finding $a^2 + b^2$

(a) Given that $a + b = 7$ and $ab = 10$ , find $a^2 + b^2$ . (b) Given that $a - b = 3$ and $ab = 4$ , find $a^2 + b^2$ .

Solution

(a) $(a + b)^2 = a^2 + 2ab + b^2\text{,}$ so $a^2 + b^2 = (a + b)^2 - 2ab = 49 - 20 = 29\text{.}$

Check with numbers that fit, $a = 5$ and $b = 2$ : $25 + 4 = 29$ . ✓

(b) $(a - b)^2 = a^2 - 2ab + b^2\text{,}$ so $a^2 + b^2 = (a - b)^2 + 2ab = 9 + 8 = 17\text{.}$

Check with $a = 4$ and $b = 1$ : $16 + 1 = 17$ . ✓

(a) $29$ ; (b) $17$

Unknowns and Areas

Example 9: Find the unknowns

Given that $(x + k)^2 = x^2 + 10x + c$ for all values of $x$ , find $k$ and $c$ .

Solution

$(x + k)^2 = x^2 + 2kx + k^2\text{.}$

Compare the $x$ terms: $2k = 10$ , so $k = 5$ . Compare the number terms: $c = k^2 = 25$ .

$k = 5$ , $c = 25$

Example 10: A growing square

A square has sides of $x$ m. Each side is made $2$ m longer. (a) Find the new area. (b) By how much has the area grown?

Solution

(a) $(x + 2)^2 = x^2 + 4x + 4$

(b) $(x^2 + 4x + 4) - x^2 = 4x + 4$

Check with $x = 3$ : the area grows from $9$ m² to $25$ m², by $16$ m²; and $4(3) + 4 = 16$ . ✓

(a) $(x^2 + 4x + 4)$ m²; (b) $(4x + 4)$ m²

Common Mistakes

Leaving out the middle term: $(x - 4)^2 \ne x^2 - 16$ . It is $x^2 - 8x + 16$ .
Squaring only the letter: $(2x + 3)^2 \ne 2x^2 + 12x + 9\text{.}$ The first term is $(2x)^2 = 4x^2$ .
Writing the middle term as $ab$ instead of $2ab$ : $(x + 5)^2 = x^2 + 10x + 25\text{,}$ not $x^2 + 5x + 25$ .
A minus last term in $(a - b)^2$ : the last term is $(-b)^2 = +b^2$ .
Forgetting a bracket when subtracting a square: $-(x - 3)^2 = -(x^2 - 6x + 9) = -x^2 + 6x - 9\text{.}$

Practice

Try each question before you open its solution.

Question 1

Expand (a) $(x + 6)^2$ ; (b) $(7 + 3a)^2$ ; (c) $(y + 1)^2$ .

ShowHide solution

(a) $x^2 + 2(x)(6) + 36 = x^2 + 12x + 36$

(b) $7^2 + 2(7)(3a) + (3a)^2 = 49 + 42a + 9a^2$

(a) $x^2 + 12x + 36$ ; (b) $9a^2 + 42a + 49$ ; (c) $y^2 + 2y + 1$

Question 2

Expand (a) $(x - 9)^2$ ; (b) $(3a - 4b)^2$ ; (c) $(2p - 5)^2$ .

ShowHide solution

(a) $x^2 - 2(x)(9) + 81 = x^2 - 18x + 81$

(b) $(3a)^2 - 2(3a)(4b) + (4b)^2 = 9a^2 - 24ab + 16b^2$

(a) $x^2 - 18x + 81$ ; (b) $9a^2 - 24ab + 16b^2$ ; (c) $4p^2 - 20p + 25$

Question 3

Expand (a) $(x + \frac{1}{2})^2$ ; (b) $(2y - \frac{1}{3})^2$ .

ShowHide solution

(a) $x^2 + 2(x)(\frac{1}{2}) + \frac{1}{4} = x^2 + x + \frac{1}{4}$

(b) $(2y)^2 - 2(2y)(\frac{1}{3}) + \frac{1}{9} = 4y^2 - \frac{4}{3}y + \frac{1}{9}$

(a) $x^2 + x + \frac{1}{4}$ ; (b) $4y^2 - \frac{4}{3}y + \frac{1}{9}$

Question 4

Expand (a) $(-2x + 1)^2$ ; (b) $(-a - 5b)^2$ .

ShowHide solution

(a) $(-2x)^2 + 2(-2x)(1) + 1^2 = 4x^2 - 4x + 1\text{.}$ (It is the same as $(2x - 1)^2$ .)

(b) $-a - 5b = -(a + 5b)\text{,}$ so the square is $(a + 5b)^2 = a^2 + 10ab + 25b^2\text{.}$

(a) $4x^2 - 4x + 1$ ; (b) $a^2 + 10ab + 25b^2$

Question 5

Simplify (a) $(x + 5)^2 - (x - 5)^2$ ; (b) $(2x + 1)^2 - 4x(x + 1)\text{;}$ (c) $3(x + 2)^2 - (x + 6)^2\text{.}$

ShowHide solution

(a) $(x^2 + 10x + 25) - (x^2 - 10x + 25) = 20x$

(b) $(4x^2 + 4x + 1) - (4x^2 + 4x) = 1$

(a) $20x$ ; (b) $1$ ; (c) $2x^2 - 24$

Question 6

Without a calculator, find (a) $51^2$ ; (b) $98^2$ ; (c) $65^2$ ; (d) $10.1^2$ .

ShowHide solution

(a) $(50 + 1)^2 = 2500 + 100 + 1 = 2601$

(b) $(100 - 2)^2 = 10\,000 - 400 + 4 = 9604$

(d) $(10 + 0.1)^2 = 100 + 2 + 0.01 = 102.01$

(a) $2601$ ; (b) $9604$ ; (c) $4225$ ; (d) $102.01$

Question 7

(a) Given that $a + b = 9$ and $ab = 14$ , find $a^2 + b^2$ . (b) Given that $a - b = 5$ and $ab = 6$ , find $a^2 + b^2$ .

ShowHide solution

(a) $a^2 + b^2 = (a + b)^2 - 2ab = 81 - 28 = 53\text{.}$ Check with $a = 7$ , $b = 2$ : $49 + 4 = 53$ . ✓

(b) $a^2 + b^2 = (a - b)^2 + 2ab = 25 + 12 = 37\text{.}$ Check with $a = 6$ , $b = 1$ : $36 + 1 = 37$ . ✓

(a) $53$ ; (b) $37$

Question 8

(a) Given that $(x - k)^2 = x^2 - 14x + c$ for all $x$ , find $k$ and $c$ . (b) Given that $(px + 3)^2 = 16x^2 + qx + 9$ for all $x$ , where $p > 0$ , find $p$ and $q$ .

ShowHide solution

(a) $(x - k)^2 = x^2 - 2kx + k^2\text{.}$ So $2k = 14$ , $k = 7$ , and $c = 49$ .

(b) $(px + 3)^2 = p^2x^2 + 6px + 9\text{.}$ So $p^2 = 16$ ; with $p > 0$ , $p = 4$ . Then $q = 6p = 24$ .

(a) $k = 7$ , $c = 49$ ; (b) $p = 4$ , $q = 24$

Question 9

A square tile has sides of $(2x + 1)$ cm. (a) Find its area in expanded form. (b) Find the area when $x = 3$ .

ShowHide solution

(a) $(2x + 1)^2 = 4x^2 + 4x + 1$

(b) $4(9) + 12 + 1 = 49$ . Check: the side is $7$ cm and $7^2 = 49$ . ✓

(a) $(4x^2 + 4x + 1)$ cm²; (b) $49$ cm²

Question 10

A square garden has sides of $x$ m. Each side is made $3$ m longer, and the area grows by $69$ m². Write an expression for the increase in area, and find $x$ .

ShowHide solution

Increase $= (x + 3)^2 - x^2 = x^2 + 6x + 9 - x^2 = 6x + 9\text{.}$

$6x + 9 = 69$ , so $6x = 60$ and $x = 10$ .

Check: $13^2 - 10^2 = 169 - 100 = 69\text{.}$ ✓

The increase is $(6x + 9)$ m², and $x = 10$ .

Question 11

Each line has one mistake. Check each with the letter equal to $2$ , then correct it. (a) $(x + 8)^2 = x^2 + 8x + 64\text{;}$ (b) $(x - 6)^2 = x^2 - 12x - 36\text{;}$ (c) $(3y - 1)^2 = 9y^2 - 3y + 1$

ShowHide solution

(a) $10^2 = 100$ , but $4 + 16 + 64 = 84$ . The middle term is $2 \times x \times 8 = 16x$ : $x^2 + 16x + 64$ , and $4 + 32 + 64 = 100$ . ✓

(b) $(-4)^2 = 16$ , but $4 - 24 - 36 = -56$ . The last term is $(-6)^2 = +36$ : $x^2 - 12x + 36$ , and $4 - 24 + 36 = 16$ . ✓

(c) $5^2 = 25$ , but $36 - 6 + 1 = 31$ . The middle term is $-2(3y)(1) = -6y$ : $9y^2 - 6y + 1$ , and $36 - 12 + 1 = 25$ . ✓

(a) $x^2 + 16x + 64$ ; (b) $x^2 - 12x + 36$ ; (c) $9y^2 - 6y + 1$

Question 12

Show that $(n + 1)^2 - n^2 = 2n + 1\text{.}$ Use it, with $30^2 = 900$ , to find $31^2$ .

ShowHide solution

$(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$

With $n = 30$ : $31^2 - 30^2 = 61$ , so $31^2 = 900 + 61 = 961\text{.}$

$31^2 = 961$

Key Points

$(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2\text{.}$
$(a + b)^2$ is not $a^2 + b^2$ : the square of side $a + b$ holds two rectangles $ab$ as well.
Square the whole of each term: $(4a)^2 = 16a^2$ ; the last term of $(a - b)^2$ is $+b^2$ .
A bracket and its negative have the same square: $(3 - 2x)^2 = (2x - 3)^2\text{.}$
Squares of numbers: split into a round number and a small part, such as $99 = 100 - 1$ .
$a^2 + b^2 = (a + b)^2 - 2ab = (a - b)^2 + 2ab\text{.}$

Perfect Squares

Before you start

(a+b)2(a + b)^2(a+b)2 Is Not a2+b2a^2 + b^2a2+b2

The Identity for (a+b)2(a + b)^2(a+b)2

The first perfect square identity

The Identity for (a−b)2(a - b)^2(a−b)2

The second perfect square identity

Expanding with the Identities

Example 1: One letter

Example 2: Coefficients and two letters

Example 3: Fractions

Example 4: A negative first term

Example 5: Spot the error

Expanding and Simplifying

Example 6: Sums and differences of squares

Squaring Numbers Quickly

Example 7: Squares without a calculator

Numbers ending in 5

Using a+ba + ba+b and ababab

Example 8: Finding a2+b2a^2 + b^2a2+b2

Unknowns and Areas

Example 9: Find the unknowns

Example 10: A growing square

Common Mistakes

Practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

$(a + b)^2$ Is Not $a^2 + b^2$

The Identity for $(a + b)^2$

The Identity for $(a - b)^2$

Using $a + b$ and $ab$

Example 8: Finding $a^2 + b^2$