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Soal Mandiri PTN TPA dan TPSA Tahun 2013 No. 4 – buku-on-line.com

Soal Mandiri PTN TPA dan TPSA Tahun 2013 No. 4

SOAL

4.       Agar parabol y = 2×2 – 2x + k (k=konstanta) dan garis lurus y = 6x – 10 berpotongan di dua titik berbeda, maka konstanta k harus memenuhi

a.       k > -2

b.      k > 2

c.       k < -2

d.      0 < k < 2

e.      -2 < k < 0

JAWABAN DAN PEMBAHASAN

4.       Parabola y = 2×2 – 2x + k

Persamaan parabola ax2 + bx + c = mx + n

2×2 – 2x + k = 6x – 10

2×2 – 2x – 6x + k + 10 = 0

2×2 – (-8)x + (k + 10) = 0

2×2 + 8x + (k + 10) = 0

Karena berpotongan di dua titik berbeda D > 0

D > b2 – 4 . a . c

>  b2 – 4 . 2 . (k + 10)

> 64 – 8k – 80

> -16 – 8k

8k > -16

k > -2

jawaban A

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