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
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