Jika α dan β adalah akar-akar persamaan kuadrat: (m−1)x2−(m+2)x−1=0, maka log (1 + (1 - ) + ) ada nilainya untuk ....

Pertanyaan

Jika begin mathsize 14px style alpha end style dan begin mathsize 14px style beta end style adalah akar-akar persamaan kuadrat: begin mathsize 14px style open parentheses straight m minus 1 close parentheses straight x squared minus open parentheses straight m plus 2 close parentheses straight x minus 1 equals 0 end style, maka log (1 + (1 - begin mathsize 14px style alpha end style) begin mathsize 14px style beta end style + begin mathsize 14px style alpha end style) ada nilainya untuk ....

  1. m > -1

  2. m < 1

  3. -1 < m < 1

  4. m < -1 atau m > 1

  5. m < begin mathsize 14px style begin inline style short dash 2 over 3 end style end style atau m > begin mathsize 14px style begin inline style 2 over 3 end style end style

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style open parentheses straight m minus 1 close parentheses straight x squared minus open parentheses straight m plus 2 close parentheses straight x minus 1 equals 0  straight alpha plus straight beta equals negative straight b over straight a equals fraction numerator straight m plus 2 over denominator straight m minus 1 end fraction  αβ equals straight c over straight a equals negative fraction numerator 1 over denominator straight m minus 1 end fraction    log invisible function application left parenthesis 1 plus open parentheses 1 minus straight alpha close parentheses straight beta plus straight alpha right parenthesis  equals log invisible function application left parenthesis 1 plus straight beta minus αβ plus straight alpha right parenthesis  equals log invisible function application open parentheses straight alpha plus straight beta minus αβ plus 1 close parentheses  equals log invisible function application open parentheses fraction numerator straight m plus 2 over denominator straight m minus 1 end fraction plus fraction numerator 1 over denominator straight m minus 1 end fraction plus 1 close parentheses  equals log invisible function application open parentheses fraction numerator straight m plus 2 plus 1 plus straight m minus 1 over denominator straight m minus 1 end fraction close parentheses  equals log invisible function application open parentheses fraction numerator 2 straight m plus 2 over denominator straight m minus 1 end fraction close parentheses    Bentuk space di space atas space akan space ada space nilainya space jika colon  open parentheses fraction numerator 2 straight m plus 2 over denominator straight m minus 1 end fraction close parentheses greater than 0  Pembuat space nolnya space adalah colon  2 straight m plus 2 equals 0 space dan space straight m minus 1 equals 0  straight m equals negative 1 space dan space straight m equals 1    Dibuat space garis space bilangan comma space lalu space uji space titik comma space sehingga space diperoleh colon end style1

begin mathsize 14px style Jadi comma space bentuk space di space atas space ada space nilainya space untuk colon  straight m less than negative 1 space atau space straight m greater than 1 end style

70

5.0 (3 rating)

Pertanyaan serupa

Diketahui persamaan −x2+4x−2=0 mempunyai akar-akar α dan β. Nilai βα​+αβ​ adalah. .. .

265

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia