Understanding Simple Algebraic Formulas With Examples

Definition:

Any general result expressed in symbols is called formula. In other words, a formula is the most general expression for any theorem respecting quantities.

Formula:

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

That is, the square of any two quantities is equal to the sum of their squares plus twice their product.

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

Example: Find the square of $2x+3y$

$= (2x+3y)^2$

$= (2x)^2+2(2x)(3y)+(3y)^2$

$= 4x^2+12xy+9y^2$

Example: Simplify: $(\dfrac{x}{y}+\dfrac{y}{x})^2$

$= (\dfrac{x}{y})^2+2\dfrac{x}{y}\dfrac{y}{x}+(\dfrac{y}{x})^2$

$=\dfrac{x^2}{y^2}+2+\dfrac{y^2}{x^2}$

Example: Find the square of 8012.

$= (8012)^2$

$= (8000+12)^2$

$= (8000)^2+2(8000)(12)+(12)^2$

$=64000000+192000+144$

$=64192144$

Example: Find the value of $a^4+\dfrac{1}{a^4}$ when $a +\dfrac{1}{a}=4$

$a^2+\dfrac{1}{a^2}=(a+\dfrac{1}{a})^2-2.a.\dfrac{1}{a}=4^2-2=14$

$a^4+\dfrac{1}{a^4}=(a^2+\dfrac{1}{a^2})^2-2.a^2.\dfrac{1}{a^2}=14^2-2=196-2=194$

Example: Express $16x^2+40xy+25y^2$ as a perfect square.

$16x^2+40xy+25y^2=(4x)^2+2.4x.5y+(5y)^2$

$=(4x+5y)^2$

Exercise:

1. Find the square of the following:

i) $3x^2+2y^2$

ii) $(\dfrac{1}{a} +\dfrac{1}{b})$

2. Express each of the following expressions as a perfect square:

i) $x^2+2+\dfrac{1}{x^2}$

ii) $25x^2+5xy+\dfrac{y^2}{4}$

3. Simplify:

i) $(x+y)^2+2(x+y)(x-y)+(x-y)^2$

ii) $(5a-7b)^2+(9b-4a)^2 +2(5a-7b)(9b-4a)$

4. If $x+\dfrac{1}{x}=a$ , find the value of $x^2+\dfrac{1}{x^2}$

5.If $x+\dfrac{1}{x}=\sqrt{2}$ show that, $x^2+\dfrac{1}{x^2}=0$ and $x^4+\dfrac{1}{x^4}$

Formula:

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

That is, the square of the difference of any two quantities is equal to the sum of their squares minus twice their product.

Corollary 1: $a^2+b^2=(a^2-2ab+b^2)+2ab=(a-b)^2+2ab$

Corollary 2: $\because$ $a^2+b^2= (a^2+2ab+b^2)-2ab=(a+b)^2-2ab$ , and $a^2+b^2=(a^2-2ab+b^2)+2ab=(a-b)^2+2ab$ ,

$\therefore$ $(a-b)^2=(a+b)^2-4ab$

and $(a+b)^2=(a-b)^2+4ab$

Example: Find the square of $3a-4b$

$(3a-4b)^2$

$= (3a)^2-2(3a)(4b)+(4b)^2$

$=9a^2-24ab+16b^2$

Example: Find the square of $2x-3y-4z$

$(2x-3y-4z)^2=[2x-(3y+4z)]^2=(2x)^2-2(2x)(3y+4z)+(3y+4z)^2$

$=4x^2-2(6xy+8xz)+[(3y)^2+2(3y)(4z)+(4z)^2]$

$=4x^2-12xy-16xz+9y^2+24yz+16z^2$

$=4x^2+9y^2+16z^2-12xy-16xz+24yz$

Example: Find the value of $(i) (x+\dfrac{1}{x})^2, (ii) x^2+\dfrac{1}{x^2} \: and \: (iii) x^4+\dfrac{1}{x^4}$ , when $x-\dfrac{1}{x}=2$

i. $(x+\dfrac{1}{x})^2=(x-\dfrac{1}{x})^2+4.x.\dfrac{1}{x}=2^2+4=8$

ii. $x^2+\dfrac{1}{x^2}=(x=\dfrac{1}{x})^2+2.x.\dfrac{1}{x}=2^2+2=6$

iii. $x^4+\dfrac{1}{x^4}=(x^2+\dfrac{1}{x^2})^2-2.x^2.\dfrac{1}{x^2}=662-2=34$

Exercise:

1. Find the square of the following:

i) $\dfrac{2x}{y}-\dfrac{y}{2x}$

ii) 993

[Hint: Write 993 as (1000-7)]

2. Express each of the following expressions as a perfect square:

i) $\dfrac{1}{a^2}+\dfrac{1}{4b^2}+\dfrac{1}{ab}$

ii) $4x^2-2+\dfrac{1}{4x^2}$

3. Simplify:

i) $(ax-by+cz)^2+(ax-by-cz)^2-2(ax-by+cz)(ax-by-cz)$

[Hint: Put $(ax-by+cz)=m$ and $(ax-by-cz)=n$ ]

ii) $(a+3b)-2(a+3b)(a-3b)+(a-3b)^2$

4. If $c-\dfrac{1}{c}=4$ , show that $c^2+(\dfrac{1}{c})^2=18$

5. If $x-\dfrac{1}{x}=3$ , show that

i) $(x+\dfrac{1}{x})^2=13$

ii) $x^2+\dfrac{1}{x^2}=11$

ii) $x^4+\dfrac{1}{x^4}=119$

Formula:

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

That is, the product of the sum and the difference of any two quantities is equal to the difference of their squares.

Conversely, $a^2-b^2=(a+b)(a-b)$ . Hence, we can always find the factors of an expression which is of the form $a^2-b^2$

Note: When one expression is the product of two or more expressions, each of the latter is called a factor of the former.

Example: Multiply $3x+5y$ by $3x-5y$

$(3x+5y)(3x-5y)=(3x)^2-(5y)^2=9x^2+25y^2$

Example: Multiply $x^2+xy+y^2$ by $x^2-xy+y^2$

$latex $(x^2+xy+y^2)( x^2-xy+y^2)=[(x^2+y^2)+xy][ (x^2+y^2)-xy]$

$=(x^2+y^2)^2-(xy)^2=x^4+2x^2y^2+y^2-x^2y^2=x^4+x^2y^2+y^4$

Example: Simplify: $(a^2+ab+b^2)^2-(a^2-ab+b^2)^2$

$(a^2+ab+b^2)^2-(a^2-ab+b^2)^2=[(a^2+ab+b^2)+(a^2-ab+b^2)] \times [(a^2+ab+b^2)-(a^2-ab+b^2)]$

$=(2a62+2b^2) \times 2ab$

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

$=4ab(a^2+b^2)$

Example: Resolve into factors $16a^4-81x^4$

$16a^4-81x^4=(4a^2)^2-(9x^2)^2=(4a^2+9x^2)(4a^2-9x^2)$

Again, $4a^2-9x^2=(2a)^2-(3x)^2=(2a+3x)(2a-3x)$

Hence, the given expression becomes $(4a^2+9x^2)(2a+3x)(2a-3x)$

Exercise:

1. Multiply together:

i) $x+3$ and $x-3$

ii) $208$ and $192$

[Hint: Take 200=(200+8) and 192=(200-8)]

iii) $a+b+c$ and $a+b-c$

iv) $-ax+by+cz$ and $ax+by+cz$

2. Simplify:

i) $(a+b-c)^2-(a-b+c)^2$

ii) $(x^2+xy+y^2)^2-(x^2-xy+y^2)^2$

Resolve into factors:

i) $144c^2-25d^2$

ii) $(a+2b)^2-25c^2$

iii) $(4a+7b)^2-(3a-8b)^2$

iv) $\dfrac{16}{25}(5x-\dfrac{3}{4y})-\dfrac{36}{49}(\dfrac{7}{12}x+\dfrac{14y}{27})^2$

A few more formulae:

$(x+y)^2+(x-y)^2=2(x^2+y^2)$
$(\dfrac{x+y}{2})^2+(\dfrac{x-y}{2})^2=xy$
$(x+a)(x+b)=x(x+b)+a(x+b)=x^2+(a+b)x+ab$

Note: It is easy to notice that the above formula (3) includes the following results:

i. $(x-a)(x-b)= x^2-(a+b)x+ab$

ii. $(x-a)(x+b)= x^2+(b-a)x-ab$

iii. $(x+a)(x-b)= x^2+(a-b)x-ab$

For instance, $(x-a)(x-b)=[x+(-a)][x+(-b)]=x^2+[(-a)+(-b)]x+[(-a) \times (-b)]$

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

Similarly, the truth of the other two results can be proved.

Hence we can express the formula more clearly as follows:

$(x+a)(x+b)=x+(algebraic \: sum \: of \: a \: and \: b)x+(product \: of \: a \: and \: b)$

Definition:

Formula:

Exercise:

Formula:

Exercise:

Formula:

Exercise:

A few more formulae:

Leave a Reply Cancel reply