Simplify the following using the formula:

Question:

Simplify the following using the formula: (a − b)(a + b) = a2 − b2:

(i) (82)2 − (18)2

(ii) (467)2 − (33)2

(iii) (79)2 − (69)2

(iv) 197 × 203

(v) 113 × 87

(vi) 95 × 105

(vii) 1.8 × 2.2

(viii) 9.8 × 10.2

Solution:

Here, we will use the identity $(a-b)(a+b)=a^{2}-b^{2}$

(i) Let us consider the following expression:

$(82)^{2}-(18)^{2}$

$=(82+18)(82-18)$

$=100 \times 64$

$=6400$

(ii) Let us consider the following expression:

$(467)^{2}-(33)^{2}$

$=(467+33)(467-33)$

$=500 \times 434$

$=217000$

(iii) Let us consider the following expression:

$(79)^{2}-(69)^{2}$

$=(79+69)(79-69)$

$=148 \times 10$

$=1480$

(iv) Let us consider the following product:

$197 \times 203$

$\because \frac{197+203}{2}=\frac{400}{2}=200$; therefore, we will write the above product as:

$197 \times 203$

$=(200-3)(200+3)$

$=(200)^{2}-(3)^{2}$

$=40000-9$

$=39991$

Thus, the answer is 39991 .

(v) Let us consider the following product:

$113 \times 87$

$\because \frac{113+87}{2}=\frac{200}{2}=100$; therefore, we will write the above product as:

$113 \times 87$

$=(100+13)(100-13)$

$=(100)^{2}-(13)^{2}$

$=10000-169$

$=9831$

Thus, the answer is 9831.

(vi) Let us consider the following product:

$95 \times 105$

$\because \frac{95+105}{2}=\frac{200}{2}=100$; therefore, we will write the above product as:

$95 \times 105$

$=(100+5)(100-5)$

$=(100)^{2}-(5)^{2}$

$=10000-25$

$=9975$

Thus, the answer is 9975.

(vii) Let us consider the following product:

$1.8 \times 2.2$

$\because \frac{1.8+2.2}{2}=\frac{4}{2}=2 ;$ therefore, we will write the above product as:

$1.8 \times 2.2$

$=(2-0.2)(2+0.2)$

$=(2)^{2}-(0.2)^{2}$

$=4-0.04$

$=3.96$

Thus, the answer is 3.96.

(viii) Let us consider the following product:

$9.8 \times 10.2$

$\because \frac{9.8+10.2}{2}=\frac{20}{2}=10 ;$ therefore, we will write the above product as:

$9.8 \times 10.2$

$=(10-0.2)(10+0.2)$

$=(10)^{2}-(0.2)^{2}$

$=100-0.04$

$=99.96$

Thus, the answer is 99.96.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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