Between which two integers does sqrt(17) lies?

1.2

Step 1: Identify integers $n$ where $n^2 < 17 < (n+1)^2$.
$4^2 = 16$
$5^2 = 25$
Since $16 < 17 < 25$, $\sqrt{17}$ lies between $4$ and $5$.

4 \text{ and 5}

1.3.1
Step 1: Recognise perfect square trinomial $x^2 - 2x + 1$.
Formula: $(x - a)^2 = x^2 - 2ax + a^2$.
Here, $2a = 2$ so $a = 1$.
$(x - 1)^2 = x^2 - 2x + 1.$
Thus, $x^2 - 2x + 1 = (x - 1)^2$.

(x-1)^2

1.3.2
Step 1: Factor quadratic $y^2 - y - 1$.
Discriminant: $b^2 - 4ac = (-1)^2 - 4(1)(-1) = 1 + 4 = 5$.
Roots: $y = \frac{1 \pm \sqrt{5}}{2}.$
Thus,
$y^2 - y - 1 = \left(y - \frac{1 + \sqrt{5}}{2}\right)\left(y - \frac{1 - \sqrt{5}}{2}\right).$

\left(y - \dfrac{1 + \sqrt{5}{2}\right)\left(y - \dfrac{1 - \sqrt{5}}{2}\right)}

1.3.3
Step 1: Rewrite $9x - x^2 = -(x^2 - 9x)$.
Factor $x^2 - 9x = x(x - 9)$.
So $-x(x - 9) = x(9 - x)$.
Verify: $x(9 - x) = 9x - x^2.$

x(9 - x)

1.3.4
Step 1: Recognise sum of cubes $x^3 + 27 = x^3 + 3^3$.
Formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Substitute $a = x$, $b = 3$:
$(x + 3)(x^2 - 3x + 9).$
Step 2: Verify expansion.
$x(x^2 - 3x + 9) = x^3 - 3x^2 + 9x$
$3(x^2 - 3x + 9) = 3x^2 - 9x + 27$
Add: $x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 = x^3 + 27.$

(x + 3)(x^2 - 3x + 9)

1.4.1
Step 1: Expand $(x - 3)(x^2 - 3x + 9)$.
Distribute $x$: $x \cdot x^2 = x^3, \quad x \cdot (-3x) = -3x^2, \quad x \cdot 9 = 9x.$
Distribute $-3$: $-3 \cdot x^2 = -3x^2, \quad -3 \cdot (-3x) = 9x, \quad -3 \cdot 9 = -27.$
Step 2: Combine:
$x^3 + (-3x^2 - 3x^2) + (9x + 9x) - 27 = x^3 - 6x^2 + 18x - 27.$

x^3 - 6x^2 + 18x - 27

1.4.2
Step 1: Convert exponents.
$25^{-1} = \frac{1}{25}, \quad 5^{-1} = \frac{1}{5}.$
Expression: $\frac{\frac{1}{25} - 1}{\frac{1}{5} + 1}.$
Step 2: Numerator: $\frac{1 - 25}{25} = -\frac{24}{25}.$
Denominator: $\frac{1 + 5}{5} = \frac{6}{5}.$
Step 3: $\frac{-24/25}{6/5} = -\frac{24}{25} \times \frac{5}{6} = -\frac{24 \cdot 5}{25 \cdot 6} = -\frac{120}{150}.$
Step 4: Simplify $-\frac{120 \div 30}{150 \div 30} = -\frac{4}{5}$.

-\frac{4{5}}