Solve for x: x²+x−6=0

We need two numbers that multiply to -6 and add to 1. These numbers are 3 and -2.

The possible values for $x$ are $\boxed{x=-3 \text{ or } x=2}$.

Question 1.1.2
Solve for $x$:

(correct to TWO decimal places)

Here, $a=3$, $b=-2$, $c=-10$.

The values for $x$ are $\boxed{x \approx 2.19 \text{ or } x \approx -1.52}$.

Question 1.1.3
Solve for $x$:

$81 = 3^4$.

So, $\frac{16}{81} = \left(\frac{2}{3}\right)^4$.

The value for $x$ is $\boxed{x=4}$.

Question 1.1.4
Solve for $x$:

We need two numbers that multiply to -12 and add to -4. These numbers are -6 and 2.

The domain requires $x+6 \ge 0 \implies x \ge -6$ and $10-x \ge 0 \implies x \le 10$. Both $x=6$ and $x=-2$ are within this domain $[-6, 10]$.
For $x=6$: $\sqrt{6+6}\sqrt{10-6} = \sqrt{12}\sqrt{4} = 2\sqrt{3} \cdot 2 = 4\sqrt{3}$. This is correct.
For $x=-2$: $\sqrt{-2+6}\sqrt{10-(-2)} = \sqrt{4}\sqrt{12} = 2 \cdot 2\sqrt{3} = 4\sqrt{3}$. This is correct.
The possible values for $x$ are $\boxed{x=6 \text{ or } x=-2}$.

Question 1.2
Solve simultaneously for $x$ and $y$ if:

For $x=0$:

For $x=-\frac{8}{7}$:

The solutions are $\boxed{x=0, y=2 \text{ or } x=-\frac{8}{7}, y=-\frac{2}{7}}$.

Question 1.3
The product of two numbers is less than 15. The second number is 2 more than the first number. Calculate ALL the possible values of the first number.

Let the first number be $x$.
The second number is $x+2$.

The product of the two numbers is less than 15.

Factor the quadratic: $(x+5)(x-3) = 0$.
The critical values are $x=-5$ and $x=3$.

We can test values or use the parabola's shape (opens upwards).
If $x < -5$, e.g., $x=-6$: $(-6+5)(-6-3) = (-1)(-9) = 9$, which is not $<0$.
If $-5 < x < 3$, e.g., $x=0$: $(0+5)(0-3) = (5)(-3) = -15$, which is $<0$.
If $x > 3$, e.g., $x=4$: $(4+5)(4-3) = (9)(1) = 9$, which is not $<0$.
The inequality $x^2 + 2x - 15 < 0$ is true when $x$ is between -5 and 3.
The possible values of the first number are $\boxed{-5 < x < 3}$.

Question 2.1.1
Consider the arithmetic sequence: $-1; -\frac{2}{3}; -\frac{1}{3}; ...; 27$.
Write down the constant difference of the sequence.

The constant difference is $\boxed{d=\frac{1}{3}}$.

Question 2.1.2
How many terms are there in this sequence?

We have $a = -1$, $d = \frac{1}{3}$, and the last term $T_n = 27$.

There are $\boxed{85}$ terms in this sequence.

Question 2.1.3
Without calculating the answer, express the sum of the last 65 terms of this sequence in sigma notation.

The sequence has 85 terms. The last 65 terms start from term $85 - 65 + 1 = 21$. So we need to sum from $T_{21}$ to $T_{85}$.

Substitute $a=-1$ and $d=\frac{1}{3}$.

The sum of the last 65 terms in sigma notation is $\boxed{\sum_{k=21}^{85} \left(-1 + \frac{1}{3}(k-1)\right)}$.

Question 2.2.1
The $n^{th}$ term of a geometric series is given by $T_n = (4)^{-n+5}$. Write down the first TWO terms of this series.

The first two terms are $\boxed{T_1=256, T_2=64}$.

Question 2.2.2
Which term of the series is equal to $\frac{1}{1024}$? Show ALL your working.

$4 = 2^2$ and $1024 = 2^{10}$.

The 10th term of the series is equal to $\frac{1}{1024}$.
The term is $\boxed{T_{10}}$.

Question 2.2.3
Determine $S_\infty$ of this series.

From 2.2.1, $a = T_1 = 256$.
The common ratio $r = \frac{T_2}{T_1} = \frac{64}{256}$.

Alternatively, from $T_n = (4)^{-n+5} = 4^5 \cdot 4^{-n} = 4^5 \cdot (4^{-1})^n = 1024 \cdot \left(\frac{1}{4}\right)^n$.
This is not in the standard form $ar^{n-1}$. Let's rewrite $T_n = 4^5 \cdot 4^{-n} = 4^5 \cdot 4^{-1} \cdot 4^{-(n-1)} = 4^4 \cdot \left(\frac{1}{4}\right)^{n-1}$.
So $a=4^4=256$ and $r=\frac{1}{4}$.

Here, $r=\frac{1}{4}$, so $|r|<1$ is satisfied.

The sum to infinity of this series is $\boxed{S_\infty = \frac{1024}{3}}$.