Pertanyaan

A ka r − aka r p ers amaan x 2 + ( a − 1 ) x + 2 = 0 a d a l ah α d an β . J ika α = 2 β d an a > 0 maka ni l ai a = ...

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F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

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Pembahasan

x squared plus left parenthesis a minus 1 right parenthesis x plus 2 equals 0 rightwards arrow A equals 1 comma B equals a minus 1 comma C equals 2  • P e n j u m l a h a n space a k a r minus a k a r  alpha plus beta equals negative B over A  alpha plus beta equals negative fraction numerator left parenthesis a minus 1 right parenthesis over denominator 1 end fraction  alpha plus beta equals 1 minus a horizontal ellipsis horizontal ellipsis horizontal ellipsis left parenthesis 1 right parenthesis  • P e r k a l i a n space a k a r minus a k a r  alpha beta equals C over A  alpha beta equals 2 over 1  alpha beta equals 2 horizontal ellipsis horizontal ellipsis horizontal ellipsis left parenthesis 2 right parenthesis    S u b s t i t u s i space alpha equals 2 beta space k e space p e r s a m a a n space left parenthesis 2 right parenthesis space m a k a space d i p e r o l e h comma  2 beta squared equals 2  beta squared equals 1 rightwards arrow beta equals plus-or-minus 1  M a k a space k i t a space d a p a t space m e n g e t a h u i space alpha space m e l a l u i space alpha equals 2 beta rightwards arrow alpha equals 2 left parenthesis plus-or-minus 1 right parenthesis equals plus-or-minus 2 comma space s e h i n g g a  alpha plus beta equals 1 minus a  2 plus 1 equals 1 minus a  3 equals 1 minus a  minus a equals 3 minus 1  a equals negative 2  A t a u space  alpha plus beta equals 1 minus a  minus 2 minus 1 equals 1 minus a  1 minus a equals negative 3  minus a equals negative 4  a equals 4  K a r e n a space a greater than 0 space m a k a space y a n g space m e m e n u h i space a d a l a h space a equals 4.

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