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A k a r minus a k a r space p e r s a m a a n space x squared plus open parentheses a minus 1 close parentheses x plus 2 equals 0 space a d a l a h space alpha space d a n space beta. space J i k a space alpha equals 2 beta space d a n space a greater than 0 space m a k a space n i l a i space a equals...

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D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

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i right parenthesis C a r i space h u b u n g a n space a k a r minus a k a r space p e r s a m a a n space k u a d r a t  U n t u k space p e r s a m a a n space k u a d r a t space A x squared plus b x plus c equals 0 space y a n g space m e m i l i k i space a k a r minus a k a r space alpha space d a n space beta comma space m a k a  alpha plus beta equals b over A  alpha beta equals c over A    box enclose P e r s a m a a n space k u a d r a t space x squared plus left parenthesis a minus 1 right parenthesis x plus 2 equals 0 space d e n g a n space A equals 1 comma space b equals left parenthesis a minus 1 right parenthesis comma space d a n space c equals 2 space  m e m i l i k i space a k a r minus a k a r space alpha space d a n space beta space d i m a n a space alpha equals 2 beta comma space m a k a  alpha beta equals 2 over 1  2 beta cross times beta equals 2  2 beta squared equals 2  beta squared equals 1  beta equals 1  alpha plus beta equals b over A  2 beta plus beta equals fraction numerator open parentheses a minus 1 close parentheses over denominator 1 end fraction  3 beta equals left parenthesis a minus 1 right parenthesis end enclose    i i right parenthesis S u b t i t u s i space n i l a i space beta  box enclose 3 beta equals left parenthesis a minus 1 right parenthesis space  3 equals a minus 1  space a equals 4 end enclose

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Diketahui ( − 1 , − 2 ) merupakan salah satu solusi real dari sistem persamaan berikut. { 2 a x 2 − 7 x y − 2 a y 2 = − 20 − a x 2 + 4 x y + a y 2 = 11 ​ Carilah Solusi lainnya!

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