Find the value of (102) $^{2}$ + (98) $^{2}$ .

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Solution

(102) $^{2}$ can be written as (100+2) $^{2}$ ,
(98) $^{2}$ can be written as (100-2) $^{2}$ .
Using the identity $(a + b)^{2} + (a - b)^{2} = 2 (a^{2} + b^{2})$
$(102)^{2} + (98)^{2} = 2 (100^{2} + 2^{2}) = 20008$ .

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Question 7

The value of $\frac{10^{22} + 10^{20}}{10^{20}}$ is
(a) 10
(b) $10^{42}$
(c) 101
(d) $10^{22}$

The value of $(102)^{2} + (98)^{2}$ is:

\frac{10^{22} + 10^{20}}{10^{20}}

^{2}

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