Algebraic formulae and their applications with examples

Formulae and their applications



Problem: 1 Find the product of the following

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

Solution:

Let $abc$ $=x$  and  $d^2$ $=y$

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

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

$=(x^4-y^4)$

Replacing the values of $x$ and $d$

$(a^4b^4c^4$ $-d^8)$


$ii)$  $(a+b-c+d)$ $(a+b+c-d)$

Solution:

$=\{(a+b)-(c-d)\}$ $\{((a+b)+(c-d)\}$

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

$=a^2$ $+2ab$ $+b^2$ $-(c^2-2cd+d^2)$

$=a^2$ $+2ab$ $+b^2$ $-c^2$ $+2cd$ $-d^2$

$=a^2$ $+b^2$ $-c^2$ $-d^2$ $+2ab$ $+2cd$    Ans.


Problem 2. Find the values by using the appropriate formula

$i)$ $(1104$ $\times 1104)$

Solution:

$(1104$ $\times 1104)$ $=(1104)^2$

$=(1100+4)^2$
$=(1100)^2$ $+2(1100)(4)$ $+(4)^2$
$=12,10,000$ $+8800$ $+16$
$=12,18,816$    Ans.

$ii)$ $989$ $\times 989$
Solution:
$989$ $\times 989$ $=(989)^2$
$=(1,000$ $-1)$
$=(1,000)^2$ $-2(1,000)(11)$ $+(11)^2$
$=1,000,000$ $-22,000$ $+121$
$989$ $\times 989$ $=978,121$    Ans.

$6)$ $(a+b)^2$ $=(a-b)^2$ $+4ab$
Verification:
L.H.S  $=(a+b)^2$ $=a^2$ $+2ab$ $+b^2$
$=a^2$ $+2ab$ $+b^2$ $-2ab$ $+2ab$
$=a^2$ $-2ab$ $+b^2$ $+2ab$ $+2ab$
$=(a^2$ $-2ab$ $+b^2)$ $+4ab$
$=(a-b)^2$ $+4ab$ $=$ R.H.S    Proved.

$7)$ $(a-b)^2$ $=(a+b)^2$ $-4ab$
Verification:
R.H.S $=(a+b)^2$ $-4ab$
$=a^2$ $+2ab$ $+b^2$ $-4ab$
$=a^2$ $+b^2$ $-2ab$
$(a+b)^2$ $-4ab$ $=(a-b)^2$ $=$ R.H.S    Proved.

$8)$ $(a+b)^2$ $-(a-b)^2$ $=4ab$
Verification:
R.H.S $=4ab$
add and subtract $a^2$ and $b^2$    
$=4ab$ $+a^2$ $-a^2$ $+b^2$ $-b^2$
$=a^2$ $+2ab$ $+b^2$ $-a^2$ $+2ab$ $-b^2$
$=(a^2+2ab+b^2)$ $-(a^2-2ab+b^2)$
$4ab$ $=(a+b)^2$ $-(a-b)^2$ $=$ L.H.S    Proved

$9.$    $(a+b)^2+(a-b)^2=2(a^2+b^2)$
Verification:
R.H.S $=2(a^2+b^2)$
$=(a^2+b^2)+(a^2+b^2)$
add and subtract $2ab$
$=(a^2$ $+2ab$ $+b^2)$ $+(a^2$ $-2ab$ $+b^2)$
$=(a+b)^2$ $+(a-b)^2=$ L.H.S    Proved

Problem: $1$
Find the value of $a^2+b^2$ when $a-b=5$, and $a+b=-9$
Solution:
∴ $\left( a+b \right)^2$ $-\left( a-b \right)^2$ $=2\left( a^2+b^2 \right)$
$\left( -9 \right)^2$ $-\left( 5 \right)^2$ $=2\left( a^2+b^2 \right)$
$81$ $+25$ $=2(a^2$ $+b^2)$
${a^2+b^2}$ $=\frac {106}{2}$ $\Rightarrow $ $\left( a+b \right)^2$ $=53$


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