Question

3X²-12≥0
X² ≤ 4x+ 12
16X²-64>0
6X²-5<x
X²-2x-4≥3​

Answer 1

3X²-12≥0

$3 {x}^{2} = 12 \\ {x}^{2} = 4 \\ x = \sqrt{4 } \\ x = 2$

X² ≤ 4x+ 12

${x}^{2} \leqslant 4x + 12 \\ {x}^{2} - 4x - 12 \\ (x - 6)(x + 2) \leqslant 0 \\1) x - 6 \leqslant 0 \\ x \leqslant 6 \\ 2)x + 2 \leqslant 0 \\ x + \leqslant - 2$

16X²-64>0

$16 {x}^{2} - 64 > 0 \\ {x}^{2} = 4 \\ x = 2$

6X²-5<x

$6 {x}^{2} -x - 5 < 0 \\ (6x -5)(x + 1) < 0 \\ 1)6x - 5 < 0 \\ x = \frac{5}{6} \\ 2)x + 1 < 0 \\ x < -1$

X²-2x-4≥3

${x}^{2} - 2x - 7 \geqslant 0$