question_answer

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

(a) \[103\times 104~\]

(b) \[5.1\times 5.2~\]

(c) \[103\times 98~\]

(d) \[9.7\times 9.8\]

Answer:

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

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

\[=\text{ }10000+700+12\]

= 10712

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

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

\[=\text{ }25+0.3\times 5+0.02\]

\[=25+1.5+0.02\]

= 26.52

(c) \[103\times 98=\left( 100+3 \right)\left( 1002 \right)\]

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

\[=\text{ }10000+(32)\times 1006\]

\[=10000+1\times 1006\]

\[=\text{ }10000+1006\]

= 10094

(d) \[9.7\times 9.8=\left( 100.3 \right)\left( 100.2 \right)\]

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

\[=100+\left( \text{ }0.30.2 \right)10+\left\{ +\text{ }0.06 \right\}\]

\[=1000.5\times 100.06\]

\[=1005+0.06\]

= 95.06