Solve for $x$: 1.1.1 $3x^2 + 10x + 6 = 0$ (correct to TWO decimal places) 1.1.2 $ ewline(6x^2 - 15 = x + 1)$ 1.1.3 $x^2 + 2x - 24 ewline 0$ - NSC Mathematics - Question 1 - 2017 - Paper 1

To solve the equation $\sqrt{6x^2 - 15} = x + 1$, we square both sides:

1. Squaring both sides gives: $6x^2 - 15 = (x + 1)^2$

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

3. Rearranging the equation: $5x^2 - 2x - 16 = 0$

4. Applying the quadratic formula again with $a = 5$, $b = -2$, and $c = -16$: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(-16)}}{2(5)}$

5. Calculate the discriminant: $4 + 320 = 324$

6. Hence: $x = \frac{2 \pm 18}{10}$

   • First value: $x = 2$ (valid as $eq -1$)
   • Second value: $x = -1.6$ (valid)

So the solutions are $x = 2$ and $x \approx -1.6$.

To solve the inequality $x^2 + 2x - 24 \geq 0$, we first find the roots of the boundary equation:

1. Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   Here, $a = 1$, $b = 2$, and $c = -24$:

2. Calculate the discriminant: $2^2 - 4(1)(-24) = 4 + 96 = 100$

3. Apply the formula: $x = \frac{-2 \pm 10}{2}$

4. The roots are:

   • First root: $x = 4$
   • Second root: $x = -6$

5. The critical points $-6$ and $4$ divide the number line into intervals. Test intervals:

   • For $x < -6$, let $x = -7$: $(-7)^2 + 2(-7) - 24 = 49 - 14 - 24 > 0$ (valid)
   • For $-6 < x < 4$, let $x = 0$: $0^2 + 2(0) - 24 < 0$ (invalid)
   • For $x > 4$, let $x = 5$: $5^2 + 2(5) - 24 > 0$ (valid)

6. Therefore, the solution for the inequality is: $x \leq -6 \text{ or } x \geq 4$