Solve for x: 1.1.1 $x^2 - x - 20 = 0$ 1.1.2 $3x^3 - 2x - 6 = 0$ (correct to TWO decimal places) 1.1.3 $(x - 1)^2 > 9$ 1.1.4 $2 ext{√}x + 6 + 2 = x$ 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

To solve the cubic equation $3x^3 - 2x - 6 = 0$, we can use numerical methods or synthetic division. Upon applying the method and using a calculator, we find the roots:

1. $x \approx 1.12$
2. $x \approx 1.79$.

Thus, the roots are approximately $x = 1.12$ and $x = 1.79$.

To solve the inequality $(x - 1)^2 > 9$, we first find the critical values by taking the square root:

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

Combining these results, we have:

$x < -2 \text{ or } x > 4$.

Rearranging the equation yields:

$2\text{√}x + 8 = x$

To isolate the term with the square root, we have:

$2\text{√}x = x - 8$

Dividing by 2:

$\text{√}x = \frac{x - 8}{2}$

Squaring both sides produces:

$x = \left(\frac{x - 8}{2}\right)^2$

Expanding and solving will eventually lead to:

1. $x = 10$.

Thus, the solution is $x = 10$.

To solve the equations:

1. $4x + y = 2$
2. $4x + y^2 = 8$,

We can substitute $y = 2 - 4x$ from the first equation into the second:

$4x + (2 - 4x)^2 = 8$

Expanding the equation:

$(2 - 4x)^2 = 4 - 16x + 16x^2\Rightarrow 4x + 4 - 16x + 16x^2 = 8$

Combining like terms:

$16x^2 - 12x - 4 = 0$

We can use the quadratic formula to find $x$ values:

$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 16 \cdot (-4)}}{2 \cdot 16} = 1\text{ or } -\frac{1}{4}.$

Substituting these back into either equation gives corresponding y-values as:

1. $y = 1$ when $x = 1$
2. $y = 3$ when $x = -\frac{1}{4}$.

Given:

$2^{x} * 3^{y} = 24^{x} = (2^3 * 3^1)^{x} = 2^{3x} * 3^{x}.$

Equating the exponents of the base $2$ and the base $3$ gives us:

1. From $2^{x} = 2^{3x} \Rightarrow x = 3x \Rightarrow 2x = 0 \Rightarrow x = 0$.
2. From $3^{y} = 3^{x} \Rightarrow y = x = 0$.

Thus, the values are $x = 0$ and $y = 0$, leading to:

$x - y = 0 - 0 = 0.$