Roboguru

Jika akar-akar dan saling berkebalikan, maka b - a = ....

Pertanyaan

Jika akar-akar 3 x squared space plus space a x space minus space 2 space equals space 0 space dan 2 x squared space plus space 6 x space plus space 3 b space equals space 0 spacesaling berkebalikan, maka b - a = ....

  1. -7

  2. -5 

  3. 6

  4. 7

Pembahasan Soal:

3 straight x squared space plus space ax space minus space 2 space equals space 0 space space open curly brackets table attributes columnalign left end attributes row cell straight x subscript 1 end cell row cell straight x subscript 2 end cell end table close  Didapat space  straight x subscript 1 space plus space straight x subscript 2 space equals space minus straight a over 3 horizontal ellipsis horizontal ellipsis horizontal ellipsis left parenthesis 1 right parenthesis  straight x subscript 1 straight x subscript 2 space equals space minus 2 over 3 horizontal ellipsis horizontal ellipsis horizontal ellipsis left parenthesis 2 right parenthesis  2 straight x squared space plus space 6 straight x space plus space 3 straight b space equals space 0  1 over straight x subscript 1 space plus space 1 over straight x subscript 2 space equals space minus 6 over 2 space equals space minus 3  fraction numerator left parenthesis straight x subscript 1 plus straight x subscript 2 right parenthesis over denominator left parenthesis straight x subscript 1 straight x subscript 2 right parenthesis end fraction space equals space minus 3  Subtitusi space persamaan space left parenthesis 1 right parenthesis space dan space left parenthesis 2 right parenthesis space  fraction numerator negative begin display style straight a over 3 end style over denominator negative begin display style 2 over 3 end style end fraction space equals space minus 3  straight a over 2 equals space minus 3 space  straight a space equals space minus 6  Kemudian  1 over straight x subscript 1 times 1 over straight x subscript 2 space equals space fraction numerator 3 straight b over denominator 2 end fraction  fraction numerator 1 over denominator negative begin display style 2 over 3 end style end fraction space equals space fraction numerator 3 straight b over denominator 2 end fraction  fraction numerator 3 straight b over denominator 2 end fraction space equals space minus 3 over 2  straight b space equals space minus 1  Maka space straight b space minus space straight a space equals space minus 1 minus left parenthesis negative 6 right parenthesis space equals space 5

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 20 Desember 2020

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