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Himpunan penyelesaian dari pertidaksamaan 5 p 2 − 2 p + 4 < ( 2 5 p ) p ⋅ 12 5 2 p adalah ....

Himpunan penyelesaian dari pertidaksamaan  adalah ....

  1. begin mathsize 14px style open curly brackets straight p │ straight p less than 4 minus square root of 20 space atau space straight p greater than 4 plus square root of 20 comma straight p element of straight R close curly brackets end style 

  2. begin mathsize 14px style open curly brackets straight p │ straight p less than negative 4 minus square root of 20 space atau space straight p greater than negative 4 plus square root of 20 comma straight p element of straight R close curly brackets end style 

  3. begin mathsize 14px style open curly brackets straight p │ minus 4 minus square root of 20 less than straight p less than 4 plus square root of 20 comma straight p element of straight R close curly brackets end style 

  4. begin mathsize 14px style open curly brackets straight p │ 4 minus square root of 20 less than straight p less than 4 plus square root of 20 comma straight p element of straight R close curly brackets end style 

  5. begin mathsize 14px style open curly brackets straight p │ minus 4 minus square root of 20 less than straight p less than negative 4 plus square root of 20 comma straight p element of straight R close curly brackets end style 

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Pembahasan

Perhatikan bahwa Karena 5 > 1, maka Didapat pembuat nol yaitu atau . Dengan bantuan garis bilangan, maka didapat daerah sebagai berikut Sehingga didapat penyelesaian atau . Maka, himpunan penyelesaian dari pertidaksamaan adalah

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 5 to the power of straight p squared minus 2 straight p plus 4 end exponent end cell less than cell left parenthesis 25 to the power of straight p right parenthesis to the power of straight p times 125 to the power of 2 straight p end exponent end cell row cell 5 to the power of straight p squared minus 2 straight p plus 4 end exponent end cell less than cell left parenthesis left parenthesis 5 squared right parenthesis to the power of straight p right parenthesis to the power of straight p times left parenthesis 5 cubed right parenthesis squared straight p end cell row cell 5 to the power of straight p squared minus 2 straight p plus 4 end exponent end cell less than cell 5 to the power of 2 straight p squared end exponent times 5 to the power of 6 straight p end exponent end cell row cell 5 to the power of straight p squared minus 2 straight p plus 4 end exponent end cell less than cell 5 to the power of 2 straight p squared plus 6 straight p end exponent end cell row blank blank blank end table end style 

Karena 5 > 1, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight p squared minus 2 straight p plus 4 end cell less than cell 2 straight p squared plus 6 straight p end cell row cell negative straight p squared minus 8 straight p plus 4 end cell less than 0 row cell straight p squared plus 8 straight p minus 4 end cell greater than 0 row cell straight p squared plus 8 straight p plus 16 minus 20 end cell greater than 0 row cell left parenthesis straight p plus 4 right parenthesis squared minus 20 end cell greater than 0 row cell left parenthesis straight p plus 4 right parenthesis squared minus left parenthesis square root of 20 right parenthesis squared end cell greater than 0 row cell left parenthesis left parenthesis straight p plus 4 right parenthesis plus square root of 20 right parenthesis left parenthesis left parenthesis straight p plus 4 right parenthesis minus square root of 20 right parenthesis end cell greater than 0 row cell left parenthesis straight p plus 4 plus square root of 20 right parenthesis left parenthesis straight p plus 4 minus square root of 20 right parenthesis end cell greater than 0 row blank blank blank row blank blank blank row blank blank blank end table end style 

 

Didapat pembuat nol yaitu begin mathsize 14px style space straight p equals negative 4 minus square root of 20 space end root end style  atau begin mathsize 14px style straight p equals negative 4 plus square root of 20 end style  .

Dengan bantuan garis bilangan, maka didapat daerah sebagai berikut

Sehingga didapat penyelesaian begin mathsize 14px style space straight p less than negative 4 minus square root of 20 end style  atau begin mathsize 14px style straight p greater than negative 4 plus square root of 20 end style  .

Maka, himpunan penyelesaian dari pertidaksamaan undefined adalah begin mathsize 14px style open curly brackets straight p │ straight p less than negative 4 minus square root of 20 space end root atau space straight p greater than negative 4 plus square root of 20 comma straight p element of straight R close curly brackets end style 

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