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 $\sqrt{x^2 - 5} = 2\sqrt{x}$ 1.2 Solve for x and y if: $3x - y = 4$ and $x^2 + 2xy - y^2 = -2$ 1.3 Given: $f(x) = x^2 + 8x + 16$ 1.3.1 Solve for x if $f(x) > 0$ - NSC Mathematics - Question 1 - 2017 - Paper 1

To find the roots of the equation, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 4$, $b = 9$, and $c = -3$.

Substituting these values into the formula gives:

$x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}$
$= \frac{-9 \pm \sqrt{81 + 48}}{8}$
$= \frac{-9 \pm \sqrt{129}}{8}$

Calculating the approximations: $\sqrt{129} \approx 11.36$
Thus, $x = \frac{-9 + 11.36}{8} \approx 0.295$
$or$
$x = \frac{-9 - 11.36}{8} \approx -2.545$

The solutions are $x \approx 0.30$ and $x \approx -2.54$.

To solve for $x$, first square both sides:

$(\sqrt{x^2 - 5})^2 = (2\sqrt{x})^2$ $x^2 - 5 = 4x$

Rearranging the equation gives:

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

Factoring: $(x - 5)(x + 1) = 0$

Setting each factor to zero:

1. $x - 5 = 0 \Rightarrow x = 5$
2. $x + 1 = 0 \Rightarrow x = -1$

Thus, the solutions are $x = 5$ or $x = -1$.

From the first equation $3x - y = 4$, we can express $y$ in terms of $x$:

$y = 3x - 4$

Substituting this into the second equation:

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

Expanding gives: $x^2 + 6x^2 - 8x - (9x^2 - 24x + 16) = -2$

Combining terms results in: $-2x^2 + 16x - 16 = 0$

Factoring out -2 gives: $2x^2 - 8x + 8 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 2$, $b = -8$, and $c = -8$.

Calculating: $x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 2 \cdot (-8)}}{2 \cdot 2}$
$x = \frac{8 \pm \sqrt{64 + 64}}{4} = \frac{8 \pm 8\sqrt{2}}{4}$
$= 2 \pm 2\sqrt{2}$

Thus, we have two values for $x$, and substituting them back gives us the corresponding $y$ values.
The solutions are $x_1 = 2 + 2\sqrt{2}$, $y_1 = 2 + 2\sqrt{2}$ and $x_2 = 2 - 2\sqrt{2}$, $y_2 = 2 - 2\sqrt{2}$.

The given function is $f(x) = x^2 + 8x + 16$. To find where $f(x) > 0$, we first factor the expression:

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

This function is equal to zero when $x + 4 = 0 \Rightarrow x = -4$. Since $f(x)$ represents a perfect square, it is always non-negative. Thus, $f(x) > 0$ when: $x \in (-\infty, -4) \cup (-4, \infty)$
or in simpler terms, $x \in \mathbb{R}, x \neq -4$.

To determine when $f(x) = p$ has two unequal negative roots, we rewrite the function as:

$x^2 + 8x + 16 = p$

Rearranging gives: $x^2 + 8x + (16 - p) = 0$

For real and unequal roots, the discriminant must be positive:

$\Delta = b^2 - 4ac > 0$
Substituting: $8^2 - 4(1)(16 - p) > 0$
$64 - 64 + 4p > 0\Rightarrow 4p > 0\Rightarrow p > 0$

Additionally, since the roots must also be negative, we consider: $x = \frac{-8 \pm \sqrt{64 - 4(16 - p)}}{2}$ To ensure the roots are negative, we derive through interval tests:

The conditions lead us to: $0 < p < 16$ Thus, the values of $p$ must satisfy: $0 < p < 16$.