1.1 Solve for x: 1.1.1 $x^2 - x - 20 = 0$ 1.1.2 $3x^2 - 2x - 6 = 0$ (correct to TWO decimal places) 1.1.3 $(x - 1)^2 > 9$ 1.1.4 $2\\sqrt{x + 6} + 2 = x$ 1.2 Solve simultaneously for x and y: $4x + y = 2$ and $4x + y^2 = 8$ 1.3 If it is given that $2^{x} * 3^{y} = 24^{x}$, determine the numerical value of $x - y$. - NSC Mathematics - Question 1 - 2021 - Paper 1

Using the quadratic formula:

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

Here, $a = 3$, $b = -2$, and $c = -6$:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-6)}}{2(3)}$ $x = \frac{2 \pm \sqrt{4 + 72}}{6}$ $x = \frac{2 \pm \sqrt{76}}{6}$ $x = \frac{2 \pm 2\sqrt{19}}{6}$ $x = \frac{1 \pm \frac{\sqrt{19}}{3}}{3}$

Calculating the two decimal places yields: $x \approx 1.12, \: -1.79$

First, take the square root of both sides:

$|x - 1| > 3$

This leads to two cases:

1. $x - 1 > 3$ \Rightarrow $x > 4$
2. $x - 1 < -3$ \Rightarrow $x < -2$

Thus, the solution sets combine to: $x < -2 \text{ or } x > 4$

First, isolate the surd:

$2\sqrt{x + 6} = x - 2$

Next, square both sides: $4(x + 6) = (x - 2)^2$

Expanding both sides gives: $4x + 24 = x^2 - 4x + 4$

Rearranging leads to: $x^2 - 8x - 20 = 0$

Factoring or using the quadratic formula, we find: $x = 10 \text{ or } x = -2$

From the first equation, rearranging gives: $y = 2 - 4x$

Substituting this into the second equation results in: $4x + (2 - 4x)^2 = 8$

Thus, expanding gives: $4x + 4 - 16x + 16x^2 = 8$ $16x^2 - 12x - 4 = 0$

Using the quadratic formula: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(16)(-4)}}{2(16)}$ Calculating yields: $x = 1 \text{ or } x = -\frac{1}{4}$

Then substituting back gives the corresponding y-values: For $x = 1$, $y = -2$ and for $x = -\frac{1}{4}$, $y = 3$.

Rewriting $24^{x}$ gives: $24 = 2^3 * 3^1 \Rightarrow 24^x = 2^{3x} * 3^{x}$

Thus, we have: $2^x * 3^y = 2^{3x} * 3^{x}$

Setting the bases equal gives:

1. $x = 3x \Rightarrow x = 0$
2. $y = x \Rightarrow y = 0$

Therefore, $x - y = 0 - 0 = 0$.