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Tanya [424]
3 years ago
12

11

Mathematics
1 answer:
enot [183]3 years ago
6 0

Step-by-step explanation:

  1. Since, k is negative, Hence, coefficient of x will always be positive.
  2. Equating (4 – 6k)x² + (8 – 2k)x + 4=0 with ax² + bx + c =0, we find: a = (4-6k), b= (8-2k) & c = 4
  3. f(x) = 0 has no real roots.

\therefore \:  {b}^{2}  - 4ac < 0 \\  \\  \therefore \:  {(8 - 2k)}^{2}  - 4(4 - 6k) \times 4 < 0 \\  \\  \therefore \: 64 + 4 {k}^{2} - 32k  - 64 + 96k< 0 \\  \\ \therefore \: 64 + 4 {k}^{2} - 32k  - 64   \\  \\\therefore \:  4 {k}^{2} + 64k < 0 \\  \\ \therefore \:  4 k(k + 16) < 0 \\  \\ \therefore \:   k(k + 16) < 0  \\  \\  \therefore \:   k < 0 \:  \: or \: k + 16< 0  \\  \\ \therefore \:   k < 0 \:  \: or \: k <  - 16\\  \\ \therefore \:  k <  - 16 \\  \\ \therefore \: k \in  \{ -  \infty,  \:  \:  - 16 \}

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