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 $ ewline ext{ } ext{√}x^2 - 5 = 2 ext{√}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

We apply the quadratic formula, where:

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

For the equation $4x^2 + 9x - 3 = 0$, we have $a = 4$, $b = 9$, $c = -3$:

\begin{align*} \Delta & = b^2 - 4ac = 9^2 - 4(4)(-3) = 81 + 48 = 129\\ x & = \frac{-9 \pm \sqrt{129}}{2(4)} = \frac{-9 \pm \sqrt{129}}{8}\end{align*}

Calculating:

$\sqrt{129} ≈ 11.36$, hence:

\begin{align*} x & = \frac{-9 + 11.36}{8} ≈ 0.295\\ x & = \frac{-9 - 11.36}{8} ≈ -2.795\end{align*}

Rounded to two decimal places: $x ≈ 0.29, -2.54$.

To solve the equation, we square both sides:

\begin{align*} x^2 - 5 & = 4x \\ \Rightarrow x^2 - 4x - 5 & = 0\end{align*}

Factoring gives:

\begin{align*}(x - 5)(x + 1) = 0\end{align*}

Thus:

\begin{align*}x - 5 = 0 & \Rightarrow x = 5\ x + 1 = 0 & \Rightarrow x = -1\end{align*}

Final solutions: $x = 5$ or $x = -1$.

From the first equation, we express $y$:
$y = 3x - 4$.
Substituting into the second equation:

\begin{align*} x^2 + 2x(3x - 4) - (3x - 4)^2 = -2\\ x^2 + 6x^2 - 8x - (9x^2 - 24x + 16) + 2 = 0\end{align*}

Combining terms:

$-2x^2 + 16x - 14 = 0$ or equivalently:

\begin{align*}x^2 - 8x + 7 = 0\end{align*}

Factoring gives:

\begin{align*}(x - 7)(x - 1) = 0\end{align*}

Thus, the solutions are $x = 7$ or $x = 1$. Substituting back: For $x = 7$:
$y = 3(7) - 4 = 17$. For $x = 1$:
$y = 3(1) - 4 = -1$.
Final ordered pairs: $(1, -1)$ and $(7, 17)$.

To solve the inequality $f(x) = x^2 + 8x + 16 > 0$, we first note:

\begin{align*}f(x) = (x + 4)^2\end{align*}

As a perfect square, $f(x) ext{ is non-negative.}$ Hence:

\begin{align*}f(x) > 0 & \Rightarrow (x + 4)^2 > 0\\x + 4 & \neq 0\\x & \neq -4\end{align*}

Thus, solutions are all real numbers except $x = -4$.

We need the quadratic $x^2 + 8x + (16 - p) = 0$ to have two distinct negative roots. For the roots to be:

1. Distinct, the discriminant should be positive: $\Delta = 64 - 4(16 - p) > 0$
   Simplifying gives: $p < 16$.
2. Negative: The vertex $\left(-\frac{b}{2a}\right) = -4 < 0$ which is satisfied always.
   For two negative roots, we also require the leading coefficient to be positive. Therefore, the values of ${p}$ must satisfy: $0 < p < 16$.