Question

11
Given that
f(x) = (4 – 6k)x² + (8 – 2k)x + 4,
where k is a negative number,
(i) explain whether the coefficient of
x? is positive or negative.
(ii) find the range of values of k for
which f(x) = 0 has no real roots.
please help ‍♀️‍♀️​

Answer 1

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.

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