Question

$\textsf{ Solve this inequality : }$


$\sf \implies 12k^{2} \: + \: 4k \: - \: 8 \: \geq \: 0$

Answer 1

Hey!!


$\sf \implies 12k^{2} \: + \: 4k \: - \: 8 \: \geq \: 0$


Divide both sides of the inequality by 4


$\sf \implies 3k^{2} \: + \: k \: - \: 2 \: \geq \: 0$


Write k as a difference

$\sf \implies 3k^{2} \: + \: 3k \: - \: 2k \: - \: 2 \: \geq \: 0$


$= > 3k(k + 1) - 2(k + 1) \geq0 \\ \\ = > (k + 1)(3k - 2) \geq0 \\ \\ = > k + 1 \geq0 \: \: or \: \: 3k - 2 \geq0 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \\ \\ = > k + 1 \leq0 \: \: or \: \: 3k - 2 \leq0 \\ \\ solve \: \: the \: \: inequality \: \: for \: \: k \\ \\ = > k \geq - 1 \: \: or \: \: 3k - 2 \geq0 \\ \\ or \: \: \: \\ \\ = > k + 1 \leq0 \: \: or \: \: 3k + 2 \leq0$


$= > k \geq - 1 \: \: or \: \: k \geq \frac{2}{3} \\ \\or \\ \\ = > k \leq - 1 \: \: or \: \: k \leq \frac{2}{3}$

Find the intersection



See in the attachment ⬆⬆⬆






Hope it helps!

Answer 2

view the attachments.