Between which two integers does sqrt(17) lies?

Since $16 < 17 < 25$, $\sqrt{17}$ lies between 4 and 5.

4 and 5

1.3.1

Formula: $(x - a)^2 = x^2 - 2ax + a^2$.
Here, $2a = 2$ so $a = 1$.

Thus, $x^2 - 2x + 1 = (x - 1)^2$.

(x-1)^2

1.3.2

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

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

1.3.3

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

Formula: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Substitute $a = x$, $b = 3$:

Add: $x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 = x^3 + 27.$

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

1.4.1

Distribute $x$: $x \cdot x^2 = x^3, \quad x \cdot (-3x) = -3x^2, \quad x \cdot 9 = 9x.$
Distribute $-3$: $-3 \cdot x^2 = -3x^2, \quad -3 \cdot (-3x) = 9x, \quad -3 \cdot 9 = -27.$

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

1.4.2

Expression: $\frac{\frac{1}{25} - 1}{\frac{1}{5} + 1}.$

Denominator: $\frac{1 + 5}{5} = \frac{6}{5}.$

-\frac{4}{5}