Los op vir x: 1.1.1 $x^2 + 9x + 14 = 0$ 1.1.2 $4x^2 + 9x - 3 = 0$ (korrek tot TWEE decimal plekke) 1.1.3 $ ext{√}x^2 - 5 = 2 ext{√}x$ Los op vir x en y indien: 1.2 $3x - y = 4$ en $x^2 + 2xy - y^2 = -2$ Gegee: $f(x) = x^2 + 8x + 16$ 1.3.1 Los op vir x indien $f(x) > 0$ - NSC Mathematics - Question 1 - 2017 - Paper 1

For this quadratic equation, we can use the quadratic formula:

1. Apply the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   where $a = 4$, $b = 9$, and $c = -3$.

2. Substitute into the formula:

   $x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}$

   Simplifying:

   $x = \frac{-9 \pm \sqrt{81 + 48}}{8} = \frac{-9 \pm \sqrt{129}}{8}$

3. Calculate the values:

   One solution will be approximately $x = 0.29$ and the other $x = -2.54$.

To solve this equation, follow these steps:

1. Rearrange the equation to isolate square roots:

   $ext{√}x^2 - 2 ext{√}x - 5 = 0$

2. Let $y = ext{√}x$, then:

   $y^2 - 2y - 5 = 0$

3. Use the quadratic formula on this new equation:

   $y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 20}}{2} = \frac{2 \pm \sqrt{24}}{2}$

4. Therefore, $y = 1 + \sqrt{6}$ or $y = 1 - \sqrt{6}$ (only non-negative values matter for $ext{√}x$).

5. Since $y = ext{√}x$, we find:

   • Valid solution: $x = (1 + \sqrt{6})^2$
   • Invalid solution because it results in negative values when squaring: $y = 1 - \sqrt{6} < 0$.

To solve this system of equations:

1. From the first equation, express $y$ in terms of $x$:

   $y = 3x - 4$

2. Substitute this into the second equation:

   $x^2 + 2x(3x - 4) - (3x - 4)^2 = -2$

3. Simplifying:

   $x^2 + 6x^2 - 8x - (9x^2 - 24x + 16) = -2$ $-2x^2 + 16x - 18 = 0$

4. Factor or use the quadratic formula to solve for $x$, then substitute back to find $y$.

Given the function $f(x) = x^2 + 8x + 16$, we first rewrite it in a usable form:

1. Factor the quadratic:

   $f(x) = (x + 4)^2$

2. Set $(x + 4)^2 > 0$. The expression is positive when:

   • Either $x + 4 > 0$ or $x + 4 < 0$ but squared leads to only positive scenarios.

3. Hence, the solution is for all $x$ except $x = -4$.

To find the conditions for $f(y) = p$ to have two unequal negative roots:

1. The discriminant must be positive and $p < 16$:

   $64 - 4(16 - p) > 0$

2. Rearranging gives:

   $0 < p < 16$

   Therefore, the values of $p$ are in the range $(0, 16)$ that lead to two unequal negative roots.