Solve for x: 1.1.1 $x^2 + 9x + 14 = 0$ 1.1.2 $4x^2 + 9x - 3 = 0$ (correct to TWO decimal places) 1.1.3 $orall x, ext{ } ext{ if } ext{ } ext{ } ext{ } ext{ } \sqrt{x^2 - 5} = 2 ext{ } Solve for x and y if: 3x - y = 4 $x^2 + 2xy - y^2 = -2$ Given: $f(x) = x^2 + 8x + 16$ 1.3.1 Solve for x if $f(x) > 0$ 1.3.2 For which values of p will $f(x) = p$ have TWO unequal negative roots? - NSC Mathematics - Question 1 - 2017 - Paper 1

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

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

Setting $a = 4$, $b = 9$, and $c = -3$, the discriminant is:

$D = 9^2 - 4 \cdot 4 \cdot (-3) = 81 + 48 = 129$

Now applying the formula: $x = \frac{-9 \pm \sqrt{129}}{8}$

Calculating the two values:

1. $x = \frac{-9 + \sqrt{129}}{8} \approx 0.29$
2. $x = \frac{-9 - \sqrt{129}}{8} \approx -2.54$

Hence, the solutions are approximately $x \approx 0.29$ and $x \approx -2.54$.

To solve this equation, we start by squaring both sides:

$x^2 - 5 = 4x$

Rearranging gives:

$x^2 - 4x - 5 = 0$

Now we can factor this as: $(x - 5)(x + 1) = 0$

Thus, the solutions are:

1. $x = 5$
2. $x = -1$

From the first equation, we can express $y$ in terms of $x$:

$y = 3x - 4$

Now substituting this expression for $y$ into the second equation:

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

Expanding gives:

$x^2 + 6x^2 - 8x - (9x^2 - 24x + 16) = -2$ $-2x^2 + 16x - 14 = 0$

Dividing everything by -2 leads to: $x^2 - 8x + 7 = 0$

Factoring: $(x - 7)(x - 1) = 0$

Thus, $x = 7$ or $x = 1$.

For $x = 7$: $y = 3(7) - 4 = 17$.
For $x = 1$: $y = 3(1) - 4 = -1$.

Therefore, the two pairs are $(7, 17)$ and $(1, -1)$.

The function $f(x) = x^2 + 8x + 16$ can be rewritten as:

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

This expression is always non-negative because a squared term is non-negative. Thus, $f(x) > 0$ when: $x + 4 \neq 0$
or
$x \neq -4$

Therefore, $f(x) > 0$ for: $x \in \mathbb{R}, x \neq -4$.

Setting $f(x) = p$, we get: $x^2 + 8x + 16 - p = 0$

For this quadratic to have two unequal negative roots, the following conditions must hold:

1. The discriminant must be positive:
   $D > 0 \Rightarrow 8^2 - 4(1)(16 - p) > 0$
   Simplifying gives: $64 - 64 + 4p > 0 \Rightarrow p > 0$

2. The vertex of the parabola, given by $x = -\frac{b}{2a} = -\frac{8}{2} = -4$, must be less than 0, which is always satisfied.

3. In order for both roots to be negative: The condition also requires: $4p < 64 \Rightarrow p < 16$

Thus, we have: $0 < p < 16$

This means that $p$ must be within the interval $(0, 16)$.