Solve for $x$: 1.1.1 $(x-3)(x+1)=0$ 1.1.2 $\, \, \, \, \, \sqrt{x} = 512$ 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? 1.3 Solve for $x$ and $y$: $x = 2y + 2$ $x^2 - 2xy + 3y^2 = 4$ 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 $x$, we square both sides:

$\sqrt{x} = 512$

Squaring both sides gives: $x = (512)^2 = 262144$

Therefore, the solution is: $x = 262144$

To solve the inequality $x(x-4) < 0$, we first find the critical points by setting each factor to zero:

1. $x = 0$
2. $x - 4 = 0 \Rightarrow x = 4$

Next, we test intervals created by these points:

• For $x < 0$, say $x = -1$: $(-1)(-1 - 4) = 5$ (not a solution)
• For $0 < x < 4$, say $x = 2$: $(2)(2 - 4) = -4$ (solution)
• For $x > 4$, say $x = 5$: $(5)(5 - 4) = 5$ (not a solution)

Thus, the solution set is: $0 < x < 4$

To solve the quadratic equation $x^2 - 5x + 2 = 0$, we apply the quadratic formula:

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

where $a = 1$, $b = -5$, and $c = 2$.

Calculating the discriminant:

$b^2 - 4ac = (-5)^2 - 4(1)(2) = 25 - 8 = 17$

Now using the quadratic formula:

$x = \frac{5 \pm \sqrt{17}}{2}$

So, the solutions are: $x = \frac{5 + \sqrt{17}}{2} \, \text{and} \, x = \frac{5 - \sqrt{17}}{2}$

To identify the values of $c$ for which $f(x) = c$ has no real roots, we need the condition for the discriminant to be negative:

$b^2 - 4ac < 0$

Substituting for $a = 1$, $b = -5$, and $c = c - 2$ leads us to:

$(-5)^2 - 4(1)(2 - c) < 0$

Calculating the left side:

$25 - 8 + 4c < 0 \Rightarrow 4c < -17 \Rightarrow c < -\frac{17}{4}$

So, the values of $c$ that yield no real roots are: $c < -\frac{17}{4}$

1. Substitute $x$ into the second equation: $x^2 - 2xy + 3y^2 = 4$ Substituting $x = 2y + 2$: $(2y + 2)^2 - 2(2y + 2)y + 3y^2 = 4$

Simplifying: $4y^2 + 8y + 4 - (4y^2 + 4y) + 3y^2 = 4$ Combining like terms: $3y^2 + 4y + 4 - 4 = 0 \Rightarrow 3y^2 + 4y = 0$

Factoring: $y(3y + 4) = 0$

Thus, $y = 0$ or $y = -\frac{4}{3}$. For each $y$ value:

• If $y = 0$: $x = 2(0) + 2 = 2$
• If $y = -\frac{4}{3}$: $x = 2(-\frac{4}{3}) + 2 = -\frac{2}{3}$

Therefore, the pairs $(x, y)$ are: $(2, 0) \, \text{and} \, (-\frac{2}{3}, -\frac{4}{3})$

To find the maximum value of $S$, we first rewrite it in terms of $x$:

A maximum occurs when $x^2 + 2$ is minimized. The minimum value of $x^2$ is $0$, so:

$x^2 + 2 \geq 2$

Thus: $S \leq \frac{6}{2} = 3$

To achieve this, we substitute $x = 0$ into $S$:

$S = \frac{6}{0^2 + 2} = 3$

Therefore, the maximum value of $S$ is: $S = 3$