Q.1. Expand;

  1.     {(x+3)}^2=x^2+6x+9      
  2.     {\left(2a+7\right)}^2={4a}^2+28a+49      
  3.     {(8+3p)}^2={64+48p+9p}^2      
  4.     {(\sqrt{3}x+2)}^2={3x}^2+4\sqrt{3}x+4      
  5.     {(4+\sqrt{5}y)}^2={16+8\sqrt{5}y+5y}^2      
  6.     {(6x+11y)}^2={36x^2+132xy+121y}^2      
  7.     {(\frac{x}{2}+\frac{y}{3})}^2={\frac{x^2}{4}+\frac{1}{3}xy+\frac{y}{9}}^2      
  8.     {(\frac{3a}{5}+\frac{5b}{3})}^2=\frac{9a^2}{25}+2ab+\frac{25b^2}{9}      

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Q.2. Expand

  1.     {(x-9)}^2=x^2-18x+81     
  2.     {(6-y)}^2=36-127+y^2     
  3.     {(3a-2)}^2=9a^2-12a+4     
  4.     {(8y-5z)}^2={64y}^2-80yz+25z^2     
  5.     {\left(\frac{x}{2}-\frac{y}{2}\right)}^2={\frac{x}{4}}^2-\frac{1}{2}xy+\frac{y^2}{4}     
  6.     {\left(2a-\frac{5}{2}\right)}^2={4a}^2-2a+\frac{25}{4}     
  7.     \left(\frac{2}{a}-\frac{3}{b}\right)=\frac{4}{a^2}-\frac{12}{ab}+\frac{9}{b^2}     

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Q.3. Using special expressions to find the value ofː

  1.     ({53)}^2=({50+3)}^2=2500+300+9=2809    
  2.     ({84)}^2=({100-16)}^2=1000-3200+256=7056    
  3.     ({1011)}^2=({1000+11)}^2=1000000+22000+121=1022121    
  4.     ({988)}^2=({1000-12)}^2=1000000-24000+144=976144    

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Q.4. Using special expressions find the value ofː

  1.     ({67)}^2=({70-3)}^2=4900-420+9=4489    
  2.     ({795)}^2=({800-5)}^2=640000-8000+25=632025    
  3.     ({10.9)}^2=({11-0.1)}^2=121-2.2+0.01=118.81    
  4.     ({9.2)}^2=({10-0.8)}^2=100-16+0.64=84.64    

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Q.5. If      \ \ \left(x+\frac{1}{x}\right)=4   , find the value of:

  •     x-\frac{1}{x}   
  •     \left(x^2+\frac{1}{x^2}\right)   
  •     \left(x^4+\frac{1}{x^4}\right)   

Answer:

    \left(x+\frac{1}{x}\right)=4   

    {\left(x+\frac{1}{x}\right)}^2=16   

    x^2+\frac{1}{x^2}+2=16   

    x^2+\frac{1}{x^2}=14   

    {Now\ \left(x-\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}-2   

    {\left(x-\frac{1}{x}\right)}^2=14-2=12   

    Therefore\ \left(x-\frac{1}{x}\right)=\sqrt{12}=\pm{}2\sqrt{3}   

    \left(x^4+\frac{1}{x^4}\right)={\left(x^2+\frac{1}{x^2}\right)}^2-2={14}^2-2=194   

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Q.6. If     \left(z-\frac{1}{z}\right)=6   find the value of

  •     \left(z+\frac{1}{z}\right)  
  •     \left(z^2+\frac{1}{z^2}\right)  
  •     \left(z^4+\frac{1}{z^4}\right)  

Answer:

   z-\frac{1}{z}=6  

   z^2+\frac{1}{z^2}-2=36  

   z^2+\frac{1}{z^2}=38  

   {\left(z+\frac{1}{z}\right)}^2=z^2+\frac{1}{z^2}+2  

   =38+2=40  

   z+\frac{1}{z}=\pm{}2\sqrt{10}  

   z^4+\frac{1}{z^4}={\left(z^2+\frac{1}{z^2}\right)}^2-2={38}^2-2=1442  

   \left(z+\frac{1}{z}\right)=\pm{}2\sqrt{10}  

   \left(z^2+\frac{1}{z^2}\right)=38  

   \left(z^4+\frac{1}{z^4}\right)=1442  

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Q.7. If     \ \ \left(a^2+\frac{1}{a^2}\right)=23  , find the value of    \left(a+\frac{1}{a}\right)  

   {\left(a+\frac{1}{a}\right)}^2=a^2+\frac{1}{a^2}+2  

   =23+2=25  

   \left(a+\frac{1}{a}\right)=\pm{}5  

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Q.8.If     \ \ \left(x^2+\frac{1}{x^2}\right)=102  , find the value of    \left(x-\frac{1}{x}\right)  

   {\left(x-\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}-2  

   = 102-2=100  

Therefore    \ \left(x-\frac{1}{x}\right)=\pm{}10  

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Q.9. If     \ \ \left(2p+\frac{1}{2p}\right)=5  , find the value of    \left(4p^2+\frac{1}{4p^2}\right)  

   {\left(2p+\frac{1}{2p}\right)}^2=4p^2+\frac{1}{{4p}^2}+2  

   or\ 4p^2+\frac{1}{{4p}^2}=25-2=23  

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Q.10. If   \ \ \left(3c-\frac{1}{3c}\right)=8  , find the value of   \left({9c}^2+\frac{1}{{9c}^2}\right)  

  {\left(3c-\frac{1}{3c}\right)}^2={9c}^2+\frac{1}{9c^2}-2  

  {9c}^2+\frac{1}{9c^2}=8^2+2=64+2=66  

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Q.11. If  \ \ \left(a+b\right)=8  , and  ab=15  , find the value of  a^2+b^2  

 \left(a+{b)}^{2\ }={(a}^2+b^2+2ab\right)  

 a^2+b^2=8^2-2\times{}15=64-30=34  

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Q.12. If   \ \ a+b=11  , and  a^2+b^2=61  , find the value of ab  

 \left(a+{b)}^{2\ }={(a}^2+b^2+2ab\right)  

 2ab={11}^2-61=121-61=60  

 ab=30  

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Q.13. If \ a^2+b^2=13  , and ab=6  , find the value of \left(a+b\right)  

\left(a+{b)}^{2\ }=(a^2+b^2+2ab\right)  

=13+12=25  

Hence\ \left(a+b\right)=\pm{}5  

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Q.14. If  a+b=15  , and  \ ab=56  , find  \left(a^2+b^2\right)  

 \left(a+{b)}^{2\ }={(a}^2+b^2+2ab\right)  

 a^2+b^2={15}^2+2\times{}56=225-112=133  

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Q.15. If  a-b=1\ and\ ab=12,\ find\ \left(\ a^2+b^2\right)  

 = \left(a-{b)}^{2\ }=a^2+b^2-2ab\right)  

 = a^2+b^2=1^2+2\times{}12=25  

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Q.16. If a-b=5\ \&\ a^2+b^2=52\ find\ the\ value\ of\ ab    

Answer:

\left(a-{b)}^{2\ }={(a}^2+b^2-ab\right)    

2ab=a^2+b^2-(a-{b)}^2=53-25=28    

or\ \ ab=14    

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Q.17 If a^2+b^2=52\ \&\ ab=24      Find (a-b)     

{\left(a-b\right)}^2=a^2+b^2-2ab=52-48=4     

\left(a-b\right)=\ \pm{}\ 2     

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Q.18 Find the value of:

36x^2+49y^2+84xy\ \ \ Given\ x=3\ \&\ y=6     

36x^2+49y^2+84xy=\left(6x^2+7y^2\right)=\left(54+36\times{}7\right)=3600     

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25x^2+16y^2-40xy\ \ \ Given\ x=6\ \&\ y=7     

25x^2+16y^2-40xy=(5x-4{y)}^2=(30-{28)}^2=4     

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