9th Class Math Chapter 1 Notes

Q1(i) — Identify as rational or irrational: $2.353535$ (repeating)

Identify $2.\overline{35}$ as rational or irrational.

1. A decimal that repeats forever in a fixed block (here "35" repeats) can always be written as a fraction $p/q$.

Answer: Rational

Q1(ii) — Identify: $0.\overline{6}$

Identify $0.\overline{6}$ as rational or irrational.

1. A repeating decimal is always expressible as $p/q$ (here it equals $\frac{2}{3}$).

Answer: Rational

Q1(iii) — Identify: $2.236067\dots$

Identify $2.236067\dots$ as rational or irrational.

1. This is the decimal expansion of $\sqrt{5}$; it neither terminates nor repeats.

Answer: Irrational

Q1(iv) — Identify: $\sqrt{7}$

Identify $\sqrt{7}$ as rational or irrational.

1. $7$ is not a perfect square, so $\sqrt{7}$ cannot be written as $p/q$.

Answer: Irrational

Q1(v) — Identify: $e$

Identify $e$ (Euler's number) as rational or irrational.

1. $e$ is a transcendental number — its decimal expansion never terminates or repeats.

Answer: Irrational

Q1(vi) — Identify: $\pi$

Identify $\pi$ as rational or irrational.

1. $\pi$ is a transcendental number, like $e$.

Answer: Irrational

Q1(vii) — Identify: $5+\sqrt{11}$

Identify $5+\sqrt{11}$ as rational or irrational.

1. A rational number plus an irrational number is always irrational — if $5+\sqrt{11}$ were rational, then $\sqrt{11}=(5+\sqrt{11})-5$ would also be rational, which is false.

Answer: Irrational

Q1(viii) — Identify: $\sqrt{3}+\sqrt{13}$

Identify $\sqrt{3}+\sqrt{13}$ as rational or irrational.

1. Neither $3$ nor $13$ is a perfect square, and their square roots do not cancel when added, so the sum stays irrational.

Answer: Irrational

Q1(ix) — Identify: $\frac{15}{4}$

Identify $\frac{15}{4}$ as rational or irrational.

1. It is already in the form $p/q$ with integers $p=15$, $q=4\ne 0$.

Answer: Rational

Q1(x) — Identify: $(2-\sqrt{2})(2+\sqrt{2})$

Identify $(2-\sqrt{2})(2+\sqrt{2})$ as rational or irrational.

1. This is a difference-of-squares product: $(2-\sqrt{2})(2+\sqrt{2})=2^2-(\sqrt{2}{)}^2=4-2=2$.
2. $2$ is an integer, hence rational.

Answer: Rational (equals $2$)

Q2(i) — Represent on a number line: $\sqrt{2}$

Show $\sqrt{2}$ on a number line.

1. Draw a unit square on the segment from $0$ to $1$ (base $1$, height $1$) standing on the number line at $0$.
2. Its diagonal has length $\sqrt{1^2+1^2}=\sqrt{2}$ by the Pythagorean theorem.
3. With a compass centred at $0$, swing this diagonal length down onto the number line — the point where it lands is $\sqrt{2}$.

Answer: Point at distance $\sqrt{2}\approx 1.41$ from $0$

Q2(ii) — Represent: $\sqrt{3}$

Show $\sqrt{3}$ on a number line.

1. From the point $\sqrt{2}$ already marked, erect a perpendicular of length $1$.
2. The new hypotenuse has length $\sqrt{(\sqrt{2}{)}^2+1^2}=\sqrt{3}$.
3. Swing this length down onto the number line from $0$ with a compass to mark $\sqrt{3}$.

Answer: Point at distance $\sqrt{3}\approx 1.73$ from $0$

Q2(iii) — Represent: $4\frac{1}{3}$

Show $4\frac{1}{3}$ on a number line.

1. $4\frac{1}{3}=\frac{13}{3}$, which lies between $4$ and $5$.
2. Divide the segment from $4$ to $5$ into $3$ equal parts; the first division point (closer to $4$) is $4\frac{1}{3}$.

Answer: Point $\frac{1}{3}$ of the way from $4$ to $5$

Q2(iv) — Represent: $-2\frac{1}{7}$

Show $-2\frac{1}{7}$ on a number line.

1. $-2\frac{1}{7}$ lies between $-3$ and $-2$, at $\frac{1}{7}$ of the way from $-2$ towards $-3$.
2. Divide the segment from $-2$ to $-3$ into $7$ equal parts and mark the first division from $-2$.

Answer: Point between $-3$ and $-2$, $\frac{1}{7}$ of the way from $-2$

Q2(v) — Represent: $\frac{5}{8}$

Show $\frac{5}{8}$ on a number line.

1. $\frac{5}{8}$ lies between $0$ and $1$.
2. Divide the unit segment from $0$ to $1$ into $8$ equal parts; the $5$th mark from $0$ is $\frac{5}{8}$.

Answer: Point $\frac{5}{8}$ of the way from $0$ to $1$

Q2(vi) — Represent: $2\frac{3}{4}$

Show $2\frac{3}{4}$ on a number line.

1. $2\frac{3}{4}$ lies between $2$ and $3$.
2. Divide the segment from $2$ to $3$ into $4$ equal parts; the $3$rd mark from $2$ is $2\frac{3}{4}$.

Answer: Point $\frac{3}{4}$ of the way from $2$ to $3$

Q3(i) — Express as $p/q$: $0.\overline{4}$

Write $0.\overline{4}$ as a fraction $\frac{p}{q}$.

1. Let $x=0.\overline{4}=0.4444\dots$
2. Multiply by $10$: $10x=4.4444\dots$
3. Subtract: $10x-x=4.4444\dots -0.4444\dots \Rightarrow 9x=4$
4. So $x=\frac{4}{9}$.

Answer: $\frac{4}{9}$

Q3(ii) — Express as $p/q$: $0.\overline{37}$

Write $0.\overline{37}$ as a fraction $\frac{p}{q}$.

1. Let $x=0.\overline{37}=0.373737\dots$
2. Since $2$ digits repeat, multiply by $100$: $100x=37.373737\dots$
3. Subtract: $100x-x=37\Rightarrow 99x=37$
4. So $x=\frac{37}{99}$.

Answer: $\frac{37}{99}$

Q3(iii) — Express as $p/q$: $0.\overline{21}$

Write $0.\overline{21}$ as a fraction $\frac{p}{q}$.

1. Let $x=0.\overline{21}=0.212121\dots$
2. Multiply by $100$: $100x=21.212121\dots$
3. Subtract: $99x=21\Rightarrow x=\frac{21}{99}=\frac{7}{33}$ (dividing by $3$).

Answer: $\frac{7}{33}$

Q4(i) — Name the property: $(a+4)+b=a+(4+b)$

Which property justifies $(a+4)+b=a+(4+b)$?

1. Only the grouping of the terms changes; the order stays the same.

Answer: Associative property of addition

Q4(ii) — Name the property: $\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}$

Which property justifies $\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}$?

1. The order of the two addends is swapped.

Answer: Commutative property of addition

Q4(iii) — Name the property: $x-x=0$

Which property justifies $x-x=0$?

1. Every real number $x$ has an additive inverse $-x$ such that $x+(-x)=0$.

Answer: Existence of additive inverse

Q4(iv) — Name the property: $a(b+c)=ab+ac$

Which property justifies $a(b+c)=ab+ac$?

1. Multiplication is distributed over each term inside the bracket.

Answer: Distributive property of multiplication over addition

Q4(v) — Name the property: $16+0=16$

Which property justifies $16+0=16$?

1. Adding $0$ never changes a number.

Answer: Additive identity property

Q4(vi) — Name the property: $100\times 1=100$

Which property justifies $100\times 1=100$?

1. Multiplying by $1$ never changes a number.

Answer: Multiplicative identity property

Q4(vii) — Name the property: $4\times (5\times 8)=(4\times 5)\times 8$

Which property justifies $4\times (5\times 8)=(4\times 5)\times 8$?

1. Only the grouping changes, not the order.

Answer: Associative property of multiplication

Q4(viii) — Name the property: $ab=ba$

Which property justifies $ab=ba$?

1. The order of the two factors is swapped.

Answer: Commutative property of multiplication

Q4(ix) — Name the property: $-3<-1\Rightarrow 0<2$

Which property justifies going from $-3<-1$ to $0<2$?

1. Adding $3$ to both sides of $-3<-1$ gives $0<2$; the inequality direction is preserved.

Answer: Additive property of order (if $a<b$ then $a+c<b+c$)

Q4(x) — Name the property: if $a<b$ then $\frac{1}{a}>\frac{1}{b}$

Which property justifies: if $a<b$ then $\frac{1}{a}>\frac{1}{b}$ (for $a,b$ of the same sign)?

1. Taking reciprocals of two positive (or two negative) numbers reverses the inequality.

Answer: Property of reciprocals for inequalities

Q4(xi) — Name the property: if $a<b$ then $a+c<b+c$

Which property justifies: if $a<b$ then $a+c<b+c$?

1. Adding the same number to both sides of an inequality keeps its direction.

Answer: Additive property of order

Q4(xii) — Name the property: if $ac<bc$, $c>0$ then $a<b$

Which property justifies: if $ac<bc$ and $c>0$ then $a<b$?

1. Dividing both sides of an inequality by the same positive number keeps its direction.

Answer: Multiplicative property of order (positive multiplier)

Q4(xiii) — Name the property: if $ac<bc$, $c<0$ then $a>b$

Which property justifies: if $ac<bc$ and $c<0$ then $a>b$?

1. Dividing both sides of an inequality by the same negative number reverses its direction.

Answer: Multiplicative property of order (negative multiplier reverses the inequality)

Q4(xiv) — Name the property: $a>b$ or $a=b$ or $a<b$

Which property justifies that exactly one of $a>b$, $a=b$, $a<b$ must hold?

1. For any two real numbers, exactly one of these three relations is always true.

Answer: Trichotomy property

Q5(i) — Two rational numbers between $\frac{1}{4}$ and $\frac{1}{3}$

Find two rational numbers between $\frac{1}{4}$ and $\frac{1}{3}$.

1. $\frac{1}{4}=0.25$ and $\frac{1}{3}\approx 0.333$.
2. Pick any two decimals in between, e.g. $0.27=\frac{27}{100}$ and $0.3=\frac{3}{10}$.

Answer: $\frac{27}{100}$ and $\frac{3}{10}$ (many other answers are possible)

Q5(ii) — Two rational numbers between $3$ and $4$

Find two rational numbers between $3$ and $4$.

1. Any two decimals between $3$ and $4$ work, e.g. $3.25=\frac{13}{4}$ and $3.5=\frac{7}{2}$.

Answer: $\frac{13}{4}$ and $\frac{7}{2}$

Q5(iii) — Two rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$

Find two rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.

1. $\frac{3}{5}=0.6$ and $\frac{4}{5}=0.8$.
2. Pick $0.65=\frac{13}{20}$ and $0.7=\frac{7}{10}$.

Answer: $\frac{13}{20}$ and $\frac{7}{10}$

Q5(iv) — Two rational numbers between $4$ and $5$

Find two rational numbers between $4$ and $5$.

1. Pick any two decimals in between, e.g. $4.25=\frac{17}{4}$ and $4.5=\frac{9}{2}$.

Answer: $\frac{17}{4}$ and $\frac{9}{2}$

Q6(i) — Rationalize: $\frac{13}{4+\sqrt{3}}$

Rationalize the denominator of $\frac{13}{4+\sqrt{3}}$.

1. Multiply top and bottom by the conjugate $4-\sqrt{3}$: $\frac{13(4-\sqrt{3})}{(4+\sqrt{3})(4-\sqrt{3})}$
2. Denominator: $4^2-(\sqrt{3}{)}^2=16-3=13$.
3. So the fraction becomes $\frac{13(4-\sqrt{3})}{13}=4-\sqrt{3}$.

Answer: $4-\sqrt{3}$

Q6(ii) — Rationalize: $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$

Rationalize the denominator of $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$.

1. Multiply top and bottom by $\sqrt{3}$: $\frac{(\sqrt{2}+\sqrt{5})\sqrt{3}}{3}=\frac{\sqrt{6}+\sqrt{15}}{3}$.

Answer: $\frac{\sqrt{6}+\sqrt{15}}{3}$

Q6(iii) — Rationalize: $\frac{\sqrt{2}-1}{\sqrt{5}}$

Rationalize the denominator of $\frac{\sqrt{2}-1}{\sqrt{5}}$.

1. Multiply top and bottom by $\sqrt{5}$: $\frac{(\sqrt{2}-1)\sqrt{5}}{5}=\frac{\sqrt{10}-\sqrt{5}}{5}$.

Answer: $\frac{\sqrt{10}-\sqrt{5}}{5}$

Q6(iv) — Rationalize: $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$

Rationalize the denominator of $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$.

1. Multiply top and bottom by the conjugate $6-4\sqrt{2}$.
2. Denominator: $6^2-(4\sqrt{2}{)}^2=36-32=4$.
3. Numerator: $(6-4\sqrt{2}{)}^2=36-48\sqrt{2}+32=68-48\sqrt{2}$.
4. Fraction: $\frac{68-48\sqrt{2}}{4}=17-12\sqrt{2}$.

Answer: $17-12\sqrt{2}$

Q6(v) — Rationalize: $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Rationalize the denominator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.

1. Multiply top and bottom by the conjugate $\sqrt{3}-\sqrt{2}$.
2. Denominator: $3-2=1$.
3. Numerator: $(\sqrt{3}-\sqrt{2}{)}^2=3-2\sqrt{6}+2=5-2\sqrt{6}$.

Answer: $5-2\sqrt{6}$

Q6(vi) — Rationalize: $\frac{4\sqrt{3}}{\sqrt{7}+\sqrt{5}}$

Rationalize the denominator of $\frac{4\sqrt{3}}{\sqrt{7}+\sqrt{5}}$.

1. Multiply top and bottom by the conjugate $\sqrt{7}-\sqrt{5}$.
2. Denominator: $7-5=2$.
3. Numerator: $4\sqrt{3}(\sqrt{7}-\sqrt{5})=4\sqrt{21}-4\sqrt{15}$.
4. Fraction: $\frac{4\sqrt{21}-4\sqrt{15}}{2}=2\sqrt{21}-2\sqrt{15}$.

Answer: $2\sqrt{21}-2\sqrt{15}$

Q7(i) — Simplify: ${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}$

Simplify ${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}$.

1. A negative exponent flips the fraction: ${\left(\frac{16}{81}\right)}^{\frac{3}{4}}$.
2. $16^{1/4}=2$ and $81^{1/4}=3$, so ${\left(\frac{16}{81}\right)}^{1/4}=\frac{2}{3}$.
3. Cube it: ${\left(\frac{2}{3}\right)}^3=\frac{8}{27}$.

Answer: $\frac{8}{27}$

Q7(ii) — Simplify: ${\left(\frac{3}{4}\right)}^{-2}÷{\left(\frac{4}{9}\right)}^3\times \frac{16}{27}$

Simplify ${\left(\frac{3}{4}\right)}^{-2}÷{\left(\frac{4}{9}\right)}^3\times \frac{16}{27}$.

1. ${\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^2=\frac{16}{9}$.
2. ${\left(\frac{4}{9}\right)}^3=\frac{64}{729}$.
3. Divide: $\frac{16}{9}÷\frac{64}{729}=\frac{16}{9}\times \frac{729}{64}=\frac{81}{4}$.
4. Multiply by $\frac{16}{27}$: $\frac{81}{4}\times \frac{16}{27}=12$.

Answer: $12$

Q7(iii) — Simplify: $(0.027{)}^{-\frac{1}{3}}$

Simplify $(0.027{)}^{-\frac{1}{3}}$.

1. $0.027=\frac{27}{1000}={\left(\frac{3}{10}\right)}^3$, so $(0.027{)}^{1/3}=\frac{3}{10}$.
2. A negative exponent means reciprocal: $(0.027{)}^{-1/3}=\frac{10}{3}$.

Answer: $\frac{10}{3}$

Q7(iv) — Simplify: $\sqrt[7]{\frac{x^{14}{y}^{21}{z}^{35}}{y^{14}{z}^7}}$

Simplify $\sqrt[7]{\frac{x^{14}{y}^{21}{z}^{35}}{y^{14}{z}^7}}$.

1. Simplify inside first: $\frac{x^{14}{y}^{21}{z}^{35}}{y^{14}{z}^7}=x^{14}{y}^7{z}^{28}$.
2. Take the 7th root by dividing each exponent by $7$: $x^2{y}^1{z}^4$.

Answer: $x^{2}y{z}^4$

Q7(v) — Simplify: $\frac{5(25{)}^{n+1}-25(5{)}^{2n}}{5(5{)}^{2n+3}-(25{)}^{n+1}}$

Simplify $\frac{5(25{)}^{n+1}-25(5{)}^{2n}}{5(5{)}^{2n+3}-(25{)}^{n+1}}$.

1. Write everything as a power of $5$: $25^{n+1}=5^{2n+2}$.
2. Numerator: $5\cdot 5^{2n+2}-5^2\cdot 5^{2n}=5^{2n+3}-5^{2n+2}=5^{2n+2}(5-1)=4\cdot 5^{2n+2}$.
3. Denominator: $5\cdot 5^{2n+3}-5^{2n+2}=5^{2n+4}-5^{2n+2}=5^{2n+2}(25-1)=24\cdot 5^{2n+2}$.
4. Ratio: $\frac{4\cdot 5^{2n+2}}{24\cdot 5^{2n+2}}=\frac{4}{24}=\frac{1}{6}$.

Answer: $\frac{1}{6}$

Q7(vi) — Simplify: $\frac{(16{)}^{x+1}+20(4^{2x})}{2^{x-3}\times 8^{x+2}}$

Simplify $\frac{(16{)}^{x+1}+20(4^{2x})}{2^{x-3}\times 8^{x+2}}$.

1. Write all terms as powers of $2$: $16^{x+1}=2^{4x+4}$, $4^{2x}=2^{4x}$, $8^{x+2}=2^{3x+6}$.
2. Numerator: $2^{4x+4}+20\cdot 2^{4x}=2^{4x}(16+20)=36\cdot 2^{4x}$.
3. Denominator: $2^{x-3}\times 2^{3x+6}=2^{4x+3}$.
4. Ratio: $\frac{36\cdot 2^{4x}}{2^{4x+3}}=\frac{36}{2^3}=\frac{36}{8}=\frac{9}{2}$.

Answer: $\frac{9}{2}$

Q7(vii) — Simplify: $(64{)}^{-\frac{2}{3}}÷(9{)}^{-\frac{3}{2}}$

Simplify $(64{)}^{-\frac{2}{3}}÷(9{)}^{-\frac{3}{2}}$.

1. $64^{-2/3}=\frac{1}{64^{2/3}}=\frac{1}{4^2}=\frac{1}{16}$ (since $64^{1/3}=4$).
2. $9^{-3/2}=\frac{1}{9^{3/2}}=\frac{1}{27}$ (since $9^{1/2}=3$, cubed $=27$).
3. Divide: $\frac{1}{16}÷\frac{1}{27}=\frac{27}{16}$.

Answer: $\frac{27}{16}$

Q7(viii) — Simplify: $\frac{3^n\times 9^{n+1}}{3^{n-1}\times 9^{n-1}}$

Simplify $\frac{3^n\times 9^{n+1}}{3^{n-1}\times 9^{n-1}}$.

1. Write $9$ as $3^2$: $9^{n+1}=3^{2n+2}$, $9^{n-1}=3^{2n-2}$.
2. Numerator: $3^n\times 3^{2n+2}=3^{3n+2}$.
3. Denominator: $3^{n-1}\times 3^{2n-2}=3^{3n-3}$.
4. Ratio: $3^{(3n+2)-(3n-3)}=3^5=243$.

Answer: $243$

Q7(ix) — Simplify: $\frac{5^{n+3}-6\cdot 5^{n+1}}{9\times 5^n-4\times 5^n}$

Simplify $\frac{5^{n+3}-6\cdot 5^{n+1}}{9\times 5^n-4\times 5^n}$.

1. Numerator: $5^{n+3}-6\cdot 5^{n+1}=5^{n+1}(5^2-6)=5^{n+1}(19)$.
2. Denominator: $9\cdot 5^n-4\cdot 5^n=5^n(5)$.
3. Ratio: $\frac{19\cdot 5^{n+1}}{5\cdot 5^n}=19\cdot \frac{5^{n+1}}{5^{n+1}}=19$.

Answer: $19$

Q7(x) — Simplify: $\frac{(25{)}^{\frac{3}{2}}\times (243{)}^{\frac{3}{5}}}{(16{)}^{\frac{5}{4}}\times (8{)}^{\frac{4}{3}}}$

Simplify $\frac{(25{)}^{\frac{3}{2}}\times (243{)}^{\frac{3}{5}}}{(16{)}^{\frac{5}{4}}\times (8{)}^{\frac{4}{3}}}$.

1. $25^{3/2}=(5^2{)}^{3/2}=5^3=125$.
2. $243^{3/5}=(3^5{)}^{3/5}=3^3=27$.
3. $16^{5/4}=(2^4{)}^{5/4}=2^5=32$.
4. $8^{4/3}=(2^3{)}^{4/3}=2^4=16$.
5. Ratio: $\frac{125\times 27}{32\times 16}=\frac{3375}{512}$.

Answer: $\frac{3375}{512}$

Q7(xi) — Simplify: $\frac{54\times \sqrt[3]{(27{)}^{2x}}}{9^{x+1}+216(3^{2x-1})}$

Simplify $\frac{54\times \sqrt[3]{(27{)}^{2x}}}{9^{x+1}+216(3^{2x-1})}$.

1. $\sqrt[3]{27^{2x}}=27^{2x/3}=3^{2x}$.
2. $9^{x+1}=3^{2x+2}$; $216\cdot 3^{2x-1}=\frac{216}{3}\cdot 3^{2x}=72\cdot 3^{2x}$.
3. Denominator: $3^{2x+2}+72\cdot 3^{2x}=3^{2x}(9+72)=81\cdot 3^{2x}$.
4. Ratio: $\frac{54\cdot 3^{2x}}{81\cdot 3^{2x}}=\frac{54}{81}=\frac{2}{3}$.

Answer: $\frac{2}{3}$

Q7(xii) — Simplify: $\sqrt{\frac{(216{)}^{\frac{2}{3}}\times (25{)}^{\frac{1}{2}}}{(0.04{)}^{-\frac{3}{2}}}}$

Simplify $\sqrt{\frac{(216{)}^{\frac{2}{3}}\times (25{)}^{\frac{1}{2}}}{(0.04{)}^{-\frac{3}{2}}}}$.

1. $216^{2/3}=(6^3{)}^{2/3}=6^2=36$.
2. $25^{1/2}=5$.
3. $0.04=\frac{1}{25}=(\frac{1}{5}{)}^2$, so $(0.04{)}^{-3/2}=25^{3/2}=5^3=125$.
4. Inside the root: $\frac{36\times 5}{125}=\frac{180}{125}=\frac{36}{25}$.
5. Square root: $\sqrt{\frac{36}{25}}=\frac{6}{5}$.

Answer: $\frac{6}{5}$

Q7(xiii) — Simplify: $\left(a^{\frac{1}{3}}+b^{\frac{2}{3}}\right)\left(a^{\frac{2}{3}}-a^{\frac{1}{3}}{b}^{\frac{2}{3}}+b^{\frac{4}{3}}\right)$

Simplify $\left(a^{\frac{1}{3}}+b^{\frac{2}{3}}\right)\left(a^{\frac{2}{3}}-a^{\frac{1}{3}}{b}^{\frac{2}{3}}+b^{\frac{4}{3}}\right)$.

1. Let $X=a^{1/3}$, $Y=b^{2/3}$. Then $X^2=a^{2/3}$, $XY=a^{1/3}{b}^{2/3}$, $Y^2=b^{4/3}$.
2. The expression is exactly $(X+Y)(X^2-XY+Y^2)$, the sum-of-cubes identity, which equals $X^3+Y^3$.
3. $X^3=(a^{1/3}{)}^3=a$ and $Y^3=(b^{2/3}{)}^3=b^2$.

Answer: $a+b^2$

Q7(xiv) — Simplify: $\sqrt[5]{\frac{x^{15}{y}^{35}}{z^{20}}}$

Simplify $\sqrt[5]{\frac{x^{15}{y}^{35}}{z^{20}}}$.

1. Divide each exponent by $5$: $x^{15/5}{y}^{35/5}{z}^{-20/5}=x^3{y}^7{z}^{-4}$.

Answer: $\frac{x^3{y}^7}{z^4}$

Q7(xv) — Simplify: $\sqrt[3]{(27{)}^{2x}}$

Simplify $\sqrt[3]{(27{)}^{2x}}$.

1. $27=3^3$, so $27^{2x}=3^{6x}$.
2. Cube root: divide the exponent by $3$: $3^{6x/3}=3^{2x}$.

Answer: $3^{2x}$ (also written $9^x$)

Q7(xvi) — Simplify: $\frac{6(3{)}^{n+2}}{3^{n+1}-3^n}$

Simplify $\frac{6(3{)}^{n+2}}{3^{n+1}-3^n}$.

1. Denominator: $3^{n+1}-3^n=3^n(3-1)=2\cdot 3^n$.
2. Ratio: $\frac{6\cdot 3^{n+2}}{2\cdot 3^n}=3\cdot 3^2=3\times 9=27$.

Answer: $27$

Q8(i) — If $x=3+\sqrt{8}$, find $x+\frac{1}{x}$

Given $x=3+\sqrt{8}=3+2\sqrt{2}$, find $x+\frac{1}{x}$.

1. Rationalize $\frac{1}{x}=\frac{1}{3+2\sqrt{2}}$ by multiplying by $\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$: denominator $=9-8=1$, so $\frac{1}{x}=3-2\sqrt{2}$.
2. $x+\frac{1}{x}=(3+2\sqrt{2})+(3-2\sqrt{2})=6$.

Answer: $6$

Q8(ii) — Find $x-\frac{1}{x}$

Given $x=3+2\sqrt{2}$ and $\frac{1}{x}=3-2\sqrt{2}$, find $x-\frac{1}{x}$.

1. $x-\frac{1}{x}=(3+2\sqrt{2})-(3-2\sqrt{2})=4\sqrt{2}$.

Answer: $4\sqrt{2}$

Q8(iii) — Find $x^2+\frac{1}{x^2}$

Using $x+\frac{1}{x}=6$, find $x^2+\frac{1}{x^2}$.

1. Square the known identity: ${\left(x+\frac{1}{x}\right)}^2=x^2+2+\frac{1}{x^2}$.
2. So $x^2+\frac{1}{x^2}=6^2-2=34$.

Answer: $34$

Q8(iv) — Find $x^2-\frac{1}{x^2}$

Using $x+\frac{1}{x}=6$ and $x-\frac{1}{x}=4\sqrt{2}$, find $x^2-\frac{1}{x^2}$.

1. $x^2-\frac{1}{x^2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)=6\times 4\sqrt{2}=24\sqrt{2}$.

Answer: $24\sqrt{2}$

Q8(v) — Find $x^4+\frac{1}{x^4}$

Using $x^2+\frac{1}{x^2}=34$, find $x^4+\frac{1}{x^4}$.

1. Square again: ${\left(x^2+\frac{1}{x^2}\right)}^2=x^4+2+\frac{1}{x^4}$.
2. So $x^4+\frac{1}{x^4}=34^2-2=1156-2=1154$.

Answer: $1154$

Q8(vi) — Find ${\left(x-\frac{1}{x}\right)}^2$

Using $x-\frac{1}{x}=4\sqrt{2}$, find ${\left(x-\frac{1}{x}\right)}^2$.

1. $(4\sqrt{2}{)}^2=16\times 2=32$.

Answer: $32$

Q9 — Find rational $p,q$: $\frac{8-3\sqrt{2}}{4+3\sqrt{2}}=p+q\sqrt{2}$

Find rational numbers $p,q$ such that $\frac{8-3\sqrt{2}}{4+3\sqrt{2}}=p+q\sqrt{2}$.

1. Multiply top and bottom by the conjugate $4-3\sqrt{2}$.
2. Denominator: $4^2-(3\sqrt{2}{)}^2=16-18=-2$.
3. Numerator: $(8-3\sqrt{2})(4-3\sqrt{2})=32-24\sqrt{2}-12\sqrt{2}+9(2)=50-36\sqrt{2}$.
4. Fraction: $\frac{50-36\sqrt{2}}{-2}=-25+18\sqrt{2}$.

Answer: $p=-25,q=18$

Q10(i) — Verify $a(b+c)=ab+ac$

For $a=\frac{3}{2}$, $b=\frac{5}{3}$, $c=\frac{7}{5}$, verify $a(b+c)=ab+ac$.

1. $b+c=\frac{5}{3}+\frac{7}{5}=\frac{25+21}{15}=\frac{46}{15}$.
2. $a(b+c)=\frac{3}{2}\times \frac{46}{15}=\frac{138}{30}=\frac{23}{5}$.
3. $ab=\frac{3}{2}\times \frac{5}{3}=\frac{5}{2}$; $ac=\frac{3}{2}\times \frac{7}{5}=\frac{21}{10}$.
4. $ab+ac=\frac{25}{10}+\frac{21}{10}=\frac{46}{10}=\frac{23}{5}$ — matches $a(b+c)$.

Answer: Both sides equal $\frac{23}{5}$ — verified

Q10(ii) — Verify $(a+b)c=ac+bc$

For the same $a,b,c$, verify $(a+b)c=ac+bc$.

1. $a+b=\frac{3}{2}+\frac{5}{3}=\frac{9+10}{6}=\frac{19}{6}$.
2. $(a+b)c=\frac{19}{6}\times \frac{7}{5}=\frac{133}{30}$.
3. $ac=\frac{21}{10}=\frac{63}{30}$; $bc=\frac{5}{3}\times \frac{7}{5}=\frac{7}{3}=\frac{70}{30}$.
4. $ac+bc=\frac{63}{30}+\frac{70}{30}=\frac{133}{30}$ — matches $(a+b)c$.

Answer: Both sides equal $\frac{133}{30}$ — verified

WP1 — Three consecutive integers summing to 42

The sum of three consecutive integers is $42$. Find the three integers.

1. Let the integers be $n-1,n,n+1$.
2. Sum: $(n-1)+n+(n+1)=3n=42$.
3. $n=14$.

Answer: $13,14,15$

WP2 — Right triangle side in surd form

Right-angled triangle $ABC$ has the right angle at $A$, with $AC=\sqrt{3}+\sqrt{5}$ cm as one leg and area $=1+\sqrt{15}$ cm${}^2$. Find $AB$ in the form $(a\sqrt{3}+b\sqrt{5})$ cm.

1. With the right angle at $A$, $AB$ and $AC$ are the two legs, so Area $=\frac{1}{2}\cdot AB\cdot AC$.
2. $AB=\frac{2(1+\sqrt{15})}{\sqrt{3}+\sqrt{5}}$; rationalize by multiplying by $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ (denominator becomes $5-3=2$).
3. Numerator: $2(1+\sqrt{15})(\sqrt{5}-\sqrt{3})=2(\sqrt{5}-\sqrt{3}+\sqrt{75}-\sqrt{45})=2(\sqrt{5}-\sqrt{3}+5\sqrt{3}-3\sqrt{5})=2(4\sqrt{3}-2\sqrt{5})=8\sqrt{3}-4\sqrt{5}$.
4. Divide by $2$: $AB=4\sqrt{3}-2\sqrt{5}$ cm.

Answer: $AB=4\sqrt{3}-2\sqrt{5}$ cm

WP3 — Area of a rectangle in surd form

A rectangle has sides $2+\sqrt{18}$ m and $\left(5-\frac{4}{\sqrt{2}}\right)$ m. Express its area in the form $a+b\sqrt{2}$.