Solve for x: 1.1.1 3x² + 10x + 6 = 0 (correct to TWO decimal places) 1.1.2 √(6x² - 15) = x + 1 1.1.3 x² + 2x - 24 ≥ 0 - English General - NSC Mathematics - Question 1 - 2017 - Paper 1

First, square both sides to eliminate the square root:

$6x^2 - 15 = (x + 1)^2$

Expanding the right side: $6x^2 - 15 = x^2 + 2x + 1$

Rearranging to bring all terms to one side: $6x^2 - x^2 - 2x - 1 - 15 = 0$ $5x^2 - 2x - 16 = 0$

Now, apply the quadratic formula again:

• a = 5, b = -2, c = -16

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-16)}}{2 \cdot 5}$

Calculating the discriminant: $4 + 320 = 324$

So, $x = \frac{2 \pm 18}{10}$

This gives:

1. $x = 2$ (valid since we squared both sides)
2. $x = -1.6$ (valid as both satisfy the original equation)

First, let's find the roots of the quadratic equation:

$x^2 + 2x - 24 = 0$

Using the quadratic formula again:

• a = 1, b = 2, c = -24

$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1}$

Calculating: $4 + 96 = 100$

Thus, $x = \frac{-2 \pm 10}{2}$

The roots are:

1. $x = 4$
2. $x = -6$

Next, we analyze the intervals created by these roots: (-∞, -6), (-6, 4), (4, ∞). By testing points in each interval, we find:

• The quadratic is greater than or equal to zero in the intervals: $(-\infty, -6]$ and $[4, \infty)$.