1. Write the following in the form $a\sqrt{b}$, where $b$ is a prime number:
$\sqrt{300}-\sqrt{48}$
$\sqrt{37.5}$
$6\sqrt{3}$
$\frac{5}{2}\sqrt{6}$
2. Write the following in the form $a+b\sqrt{c}$:
$(7+\sqrt{5})(3-\sqrt{5})$
$\dfrac{\sqrt{2}}{1-\sqrt{2}}$
$16-4\sqrt{5}$
$-\sqrt{2}-2$
3. Simplify
$\dfrac{\sqrt{48}-\sqrt{600}}{\sqrt{12}}$
$\dfrac{\sqrt{2}}{4+3\sqrt{2}}$
$2-5\sqrt{2}$
$3-2\sqrt{2}$
4. Simplify
$\sqrt{48}-\dfrac{6}{\sqrt{3}}$
$\dfrac{12\sqrt{3}}{\sqrt{50}-\sqrt{18}}$
$2\sqrt{3}$
$3\sqrt{6}$
5. Simplify
$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}$
$\dfrac{(32x^3)^{\frac{2}{5}}}{2x\sqrt{x}}$
$\dfrac{2}{5x^2}$
$2x^{-\frac{3}{10}}$
6. Solve $32\sqrt{2}=2^x$.
$x = \dfrac{11}{2}$
7. Given that $(2+\sqrt{5})x=6-\sqrt{5}$, find $x$ in the form $a+b\sqrt{5}$.
$8\sqrt{5}-17$
8. Solve:
$10+x\sqrt{8}=\dfrac{6x}{\sqrt{2}}$
$5^x\times5^{x+4}=25$
$x = 5\sqrt{2}$
$x = -1$
9. Given that $y = \dfrac{1}{64}x^3$, find $4y^{-1}$ in the form $kx^n$.
$256x^{-3}$
10. Write:
$2^6\times2^2$ in the form $8^n$
$5\times4^{\frac{2}{3}}+3\times16^{\frac{1}{3}}$ in the form $2^n$
$8^{\frac{8}{3}}$
$2^{\frac{13}{3}}$
11. Find the integer $n$ such that $$n<5\sqrt{3}<n+1$$
$n=8$
12. The area of a rectangle of width $3\sqrt{2}-3$ is 6. Find the height of the rectangle.
$2\sqrt{2}+2$
13. Show that $\left(\dfrac{4}{3}\right)^{\frac{1}{2}}+\left(\dfrac{1}{3}\right)^{-\frac{1}{2}}$ can be written in the form $\dfrac{a}{b}\sqrt{c}$.
$\dfrac{5}{3}\sqrt{3}$
14. Given that $y=2^x$, express the following in terms of $y$:
$2^{2x-1}$
$8^x-4^{-x}$
$\dfrac{y^2}{2}$
$y^3 - \dfrac{1}{y^2}$
15. Write the following in the form $2^n$:
$\Big(\big(2^3\big)^2\Big)^3$
$\big(2^3\big)^{\big(2^3\big)}$
$2^{\Big(\big(3^2\big)^3\Big)}$
$2^{\Big(3^{\big(2^3\big)}\Big)}$
18
24
729
6561
16. A triangle $ABC$ has $AB=BC=4+\sqrt{3}$ and $AC=4+4\sqrt{3}$. $M$ is the midpoint of $AC$. Find the exact length of $BM$ and hence find the area of the triangle.
$BM=\sqrt{3}$, area $=6+2\sqrt{3}$
17. By first expanding $(1-\sqrt{5})^2$, solve $y^2=3-\sqrt{5}$.
$y = \pm \dfrac{\sqrt{2}-\sqrt{10}}{2}$
18. A triangle $ABC$ has $AB = 2\sqrt{3}-1$, $BC = \sqrt{3}+2$, and $\angle ABC = 90^{\circ}$.
