question_answer

Using\[\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab\], find

(a) \[103\times 104\]

(b) \[5.1\text{ }\times \text{ }5.2~\]

(c) \[103\text{ }\times \text{ }98\]

(d) \[9.7\text{ }\times \text{ }9.8~\]

Answer:

(a)\[103\times 104=\left( 100+3 \right)\text{ }\left( 100+4 \right)\]

\[={{\left( 100 \right)}^{2\text{ }}}+\left( 3+4 \right)\times 100+3\text{ }\times 4\]

= 10000 + 700 + 12

= 10712

(b)    \[5.1\times 5.2=\left( 5+0.1 \right)\text{ }\left( 5+0.2 \right)\]

\[={{\left( 5 \right)}^{2}}+\left( 0.1+0.2 \right)\times 5+\left( 0.1\times 0.2 \right)\]

\[=25+0.3\times 5+0.02\]

\[=25+1.5+0.02\]

= 26.52

(c)    \[103\times 98=\left( 100+3 \right)\text{ }\left( 100-2 \right)\]

\[={{\left( 100 \right)}^{2}}+\left\{ 3+\left( -2 \right) \right\}\times 100+\left( 3\times \left( -2 \right) \right)\]

\[=\text{ }10000+\left( 3-2 \right)\times 100-6\]

\[=\text{ }10000+1\times 100-6\]

\[=10000+100-6\]

= 10094

(d)   \[9.7\times 9.8=\left( 10-0.3 \right)\text{ }\left( 10-0.2 \right)\]

\[={{\left( 10 \right)}^{2}}+\left( -\text{ }0.3\text{ }+-\text{ }0.2 \right)\times 10+\left\{ \left( -\text{ }0.3 \right)\text{ }\left( -\text{ }0.2 \right) \right\}\]

\[=100+\left( -\text{ }0.3-0.2 \right)-10\times \left\{ +\text{ }0.06 \right\}\]

\[=100-0.5\times 10-0.06\]

\[=100-5+0.06\]

=95.06