Solve for $x$: 1.1.1 $(x-3)(x+1)=0$ 1.1.2 $orall x ext{ such that } \\ \\sqrt{x} = 512$ 1.1.3 $x(x-4) < 0$ Given: $f(x) = x^2 - 5x + 2$ 1.2.1 Solve for $x$ if $f(x)=0$ 1.2.2 For which values of $c$ will $f(x) = c$ have no real roots? Solve for $x$ and $y$: 1.3 $x = 2y + 2$ $x^2 - 2xy + 3y^2 = 4$ 1.4 Calculate the maximum value of $S$ if $S = \frac{6}{x^2 + 2}$ - NSC Mathematics - Question 1 - 2017 - Paper 1

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

$ext{Square: } \\ x = 512^2 \\ \ ext{Thus } x = 262144.$

Alternatively, expressing $512$ as $8^3$, we have: $x = (8^3)^{2} = 8^{6} = 64.$

Both evaluations provide $x = 64$.

To find the interval for which $x(x-4) < 0$, we first identify the critical points by setting $x(x-4) = 0$:

$x=0$ 2. $x=4$

Next, test the intervals: $(- ightarrow 0)$, $(0 ightarrow 4)$, and $(4 ightarrow ightarrow ext{Infinity})$.

Results:

• For $x < 0$: product positive,
• For $0 < x < 4$: product negative,
• For $x > 4$: product positive.

Hence, the solution interval is $(0, 4)$.

To find the roots of $f(x) = 0$, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -5$, and $c = 2$.

1. Start by calculating the discriminant: $D = b^2 - 4ac = (-5)^2 - 4(1)(2) = 25 - 8 = 17$

2. Next, substituting into the quadratic formula: $x = \frac{5 \pm \sqrt{17}}{2}$

This yields the roots $x = 2.5 + 0.5 ext{√17}$ and $x = 2.5 - 0.5 ext{√17}$.

The condition for $f(x) = c$ to have no real roots relies on the discriminant being negative:

$b^2 - 4a(c) < 0$ where $a=1$ and $b=-5$ leads to:

$(-5)^2 - 4(1)(c) < 0 \\ \Rightarrow 25 - 4c < 0 \\ \Rightarrow c > \frac{25}{4} = 6.25$.

Thus, for $c > 6.25$, the equation will have no real roots.

We substitute $x$ in the second equation:

$x^2 - 2xy + 3y^2 = 4$ Substituting $x = 2y + 2$:

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

Expanding and simplifying leads to: egin{align*} 4y^2 + 8y + 4 - (4y^2 + 4y) + 3y^2 &= 4 \ \ 3y^2 + 4y + 4 - 4 &= 0 \\ \ 3y^2 + 4y = 0\ \ \Rightarrow y(3y + 4) = 0 \ \ y = 0 ext{ or } y= -\frac{4}{3} \ \ ext{Substituting back: } \ If\ y=0,\ \Rightarrow x=2\ \ ext{If } y=-\frac{4}{3}, \ x = -\frac{2}{3}. \ \ ext{Final values: } (x, y) = (2, 0)\ ext{and } (-\frac{2}{3}, -\frac{4}{3}). \ \ ext{Thus, the solutions are: } (x, y) = (2, 0) \text{or } (-\frac{2}{3}, -\frac{4}{3}).

To maximize $S = \frac{6}{x^2 + 2}$, we note:

1. Calculate the function minimum than $x^2 + 2$ which occurs at $x = 0$.
2. Thus: $S = \frac{6}{0^2 + 2} = \frac{6}{2} = 3$.

Negative infinity confirms $x^2 + 2 > 0$; Thus, the maximum value of $S = 3$.