📘 Class 9 (Maharashtra Board) — Chapter 2: Real Numbers
Mobile-friendly • Comic Sans MS • MathJax
Beautiful, copy-paste ready notes. Split into parts to keep your website menu bar unchanged.
Part A • 1-Mark Questions (20) — Red Q / Green A
1) Write \(N,W,\mathbb{Z},\mathbb{Q},\mathbb{R}\).
\(N=\{1,2,3,\dots\},\ W=\{0,1,2,\dots\},\ \mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\},\ \mathbb{Q}=\{\tfrac{p}{q}\mid p,q\in\mathbb{Z},q\ne0\},\ \mathbb{R}=\) real numbers.
2) State the chain of inclusions.
\(N\subset W\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}\).
3) When are two rational numbers \(\frac pq,\frac rs\) equal?
If \(ps=qr\) with \(q,s>0\).
4) Decimal form of a rational number is…
Either terminating or non-terminating recurring.
5) Write \( \dfrac{2}{5}\) in decimal.
\(0.4\) (terminating).
6) Write \(\dfrac{17}{36}\) in recurring form (short).
\(0.472\overline{2}\).
7) Convert \(0.\overline{7}\) to \(\tfrac pq\).
\(\dfrac{7}{9}\).
8) Is \(\sqrt2\) rational?
No; \(\sqrt2\) is irrational.
9) Give the definition of a surd.
An irrational root of a positive rational number (e.g., \(\sqrt[3]{7}\), \(\sqrt5\)).
10) What is a quadratic surd?
A surd of order \(2\), i.e., \(\sqrt{a}\) with \(a>0\) not a perfect square.
11) Simplest form of \(\sqrt{48}\).
\(4\sqrt3\).
12) Like surds?
Surds with same order and same radicand (e.g., \(3\sqrt5,-\tfrac12\sqrt5\)).
13) Compare: which is larger, \(6\sqrt2\) or \(5\sqrt5\)?
\(5\sqrt5\) (since \(72<125\)).
14) Write the counting formula for union in \(\mathbb{R}\).
\(n(A\cup B)=n(A)+n(B)-n(A\cap B)\).
15) Conjugate of \( \sqrt5+\sqrt3\).
\(\sqrt5-\sqrt3\).
16) Value of \((\sqrt a+\sqrt b)(\sqrt a-\sqrt b)\).
\(a-b\).
17) Rationalizing factor of \(\sqrt2\).
\(\sqrt2\) (since \(\sqrt2\cdot\sqrt2=2\)).
18) Absolute value \(|-7|\).
\(7\).
19) Solve \(|x-5|=2\) (set of solutions).
\(x=7\) or \(x=3\).
20) State: square root of a negative real number is…
Not a real number.
Part B • 2-Mark Questions (20) — Red Q / Green A
1) Convert \(0.\overline{3}\) to \(\dfrac pq\).
Let \(x=0.\overline{3}\Rightarrow 10x=3.\overline{3}\). Subtract: \(9x=3\Rightarrow x=\dfrac{1}{3}\).
2) Convert \(7.\overline{529}\) to \(\dfrac pq\).
\(x=7.\overline{529}\Rightarrow 1000x=7529.\overline{529}\). Subtract: \(999x=7522\Rightarrow x=\dfrac{7522}{999}\).
3) Classify as terminating/recurring: \(\dfrac{29}{16},\ \dfrac{11}{6}\).
\(\dfrac{29}{16}\) terminates (denominator \(2^4\)). \(\dfrac{11}{6}\) recurring (denominator has 3).
4) Show \(\sqrt2\) is irrational (outline).
Assume \(\sqrt2=\tfrac pq\) in lowest terms. Then \(2q^2=p^2\Rightarrow p\) even; write \(p=2t\Rightarrow q\) even. Both even — contradiction. Hence irrational.
5) Decide: \( \sqrt{27}\) surd or not; simplest form.
Yes (irrational root). \(\sqrt{27}=3\sqrt3\).
6) Are \(\sqrt{45}\) and \(\sqrt{80}\) like surds?
\(\sqrt{45}=3\sqrt5,\ \sqrt{80}=4\sqrt5\) → like surds (both \(\sqrt5\)).
7) Compare \(8\sqrt3\) and \(\sqrt{192}\).
\((8\sqrt3)^2=192,\ (\sqrt{192})^2=192\Rightarrow\) equal.
8) Simplify \(\sqrt{50}+\sqrt{18}\).
\(5\sqrt2+3\sqrt2=8\sqrt2\).
9) Multiply \(\sqrt7\cdot\sqrt{42}\).
\(\sqrt{7\cdot42}=\sqrt{294}=7\sqrt6\).
10) Divide \(\dfrac{\sqrt{125}}{\sqrt5}\).
\(\sqrt{\tfrac{125}{5}}=\sqrt{25}=5\).
11) Rationalize \(\dfrac{1}{\sqrt5}\).
\(\dfrac{1}{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{\sqrt5}{5}\).
12) Rationalize \(\dfrac{3}{\sqrt2+\sqrt7}\).
\(\dfrac{3}{\sqrt2+\sqrt7}\cdot\dfrac{\sqrt7-\sqrt2}{\sqrt7-\sqrt2}=\dfrac{3(\sqrt7-\sqrt2)}{7-2}=\dfrac{3(\sqrt7-\sqrt2)}{5}\).
13) Compute \((\sqrt5+\sqrt3)(\sqrt5-\sqrt3)\).
\(5-3=2\).
14) Find three rationals between \(0.3\) and \(-0.5\).
E.g., \(-0.1,-0.2,-0.4\).
15) Solve \(|3x-5|=1\).
\(3x-5=\pm1\Rightarrow x=2\) or \(x=\dfrac{4}{3}\).
16) Write the order and radicand of \( \sqrt[4]{10}\).
Order \(4\); radicand \(10\).
17) Simplify \( \sqrt{112}\).
\(4\sqrt7\).
18) Compare \(7\sqrt2\) and \(5\sqrt3\).
Squares: \(98>75\Rightarrow 7\sqrt2>5\sqrt3\).
19) Show that \(\sqrt5+\sqrt7\) is irrational.
If rational, then \(\sqrt7=(\sqrt5+\sqrt7)-\sqrt5\) is difference of rationals ⇒ \(\sqrt7\) rational — contradiction.
20) Find simplest rationalizing factor of \(\sqrt{27}\).
\(\sqrt{27}=3\sqrt3\). A simplest RF is \(\sqrt3\) (since \(\sqrt3\cdot\sqrt{27}=9\)).
Part C • 3-Mark Questions (20) — Red Q / Green A
1) Decide & justify terminating/recurring: \(\dfrac{127}{200},\ \dfrac{25}{99},\ \dfrac{23}{7}\).
\(\frac{127}{200}\) terminates (denom \(2^3\cdot5^2\)); \(\frac{25}{99}\) recurring; \(\frac{23}{7}\) recurring.
2) Write \(0.37\) and \(2.514\) in \(\tfrac pq\) form.
\(0.37=\dfrac{37}{100}\). \(2.514=\dfrac{2514}{1000}=\dfrac{1257}{500}\).
3) Prove: sum/product of non-zero rational with irrational is irrational (state clearly).
Let \(r\in\mathbb{Q}\setminus\{0\},\ \alpha\notin\mathbb{Q}\). If \(r+\alpha\in\mathbb{Q}\Rightarrow \alpha=(r+\alpha)-r\in\mathbb{Q}\) contradiction. If \(r\alpha\in\mathbb{Q}\Rightarrow \alpha=\dfrac{r\alpha}{r}\in\mathbb{Q}\) contradiction.
4) Show that \(\sqrt{48}-\sqrt{27}+2\sqrt{12}\) simplifies to \(?\)
\(\;4\sqrt3-3\sqrt3+2\cdot2\sqrt3=(4-3+4)\sqrt3=5\sqrt3\).
5) Multiply & simplify: \(\sqrt{12}\cdot\sqrt{18}\).
\(\sqrt{216}=\sqrt{36\cdot6}=6\sqrt6\).
6) Divide \(\dfrac{\sqrt{98}}{\sqrt2}\) and write in simplest surd form.
\(\sqrt{49}=\;7\).
7) Rationalize: \(\dfrac{1}{\sqrt5-\sqrt3}\).
\(\dfrac{\sqrt5+\sqrt3}{(\sqrt5)^2-(\sqrt3)^2}=\dfrac{\sqrt5+\sqrt3}{2}\).
8) Rationalize: \(\dfrac{8}{\sqrt3+2\sqrt5}\).
\(\dfrac{8(\sqrt3-2\sqrt5)}{3-20}=\dfrac{8(2\sqrt5-\sqrt3)}{17}\).
9) Evaluate \((\sqrt3-2)(2\sqrt3-3\sqrt2)\).
\(6-3\sqrt6-4\sqrt3+3\sqrt2\).
10) Put three rationals between \(5.2\) and \(5.3\).
\(5.21,\,5.25,\,5.29\).
11) Compare \(4\sqrt2\) and \(9\).
Squares: \(32<81\Rightarrow 4\sqrt2<9\).
12) Simplify \( \sqrt{168}\).
\(\sqrt{16\cdot10.5}\) not neat; use \(168=24\cdot7=4\cdot42\Rightarrow \sqrt{168}=2\sqrt{42}=2\sqrt{6\cdot7}=2\sqrt{42}\) (no further square factor) → also \(= 2\sqrt{42}\).
13) Show that if \(a,b>0\) and \(a
Square both sides (monotone for non-negatives): \(a
14) Simplify \(7\sqrt{48}-\sqrt{27}-3\sqrt{12}\).
\(7\cdot4\sqrt3-3\sqrt3-3\cdot2\sqrt3=(28-3-6)\sqrt3=19\sqrt3\).
15) Solve \(|7-2x|=5\).
\(7-2x=\pm5\Rightarrow x=1\) or \(x=6\).
16) Write radicals in simplest form: \(\sqrt{250},\ \sqrt{112}\).
\(5\sqrt{10},\ 4\sqrt7\).
17) Show \((\sqrt a+\sqrt b)^2=a+b+2\sqrt{ab}\).
Expand using \((u+v)^2=u^2+v^2+2uv\).
18) If \(x^5=2\), write \(x\) and state whether rational.
\(x=\sqrt[5]{2}\), irrational.
19) Simplify \(\dfrac{6}{3\sqrt3}\).
\(=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\).
20) Compute \((\sqrt5+\sqrt2)(\sqrt5+\sqrt2)\).
\(7+2\sqrt{10}\).
Part D • Textbook Exercises — Perfect Solutions (MathJax)
Practice Set 2.1 — Solutions
1) Classify as terminating / non-terminating recurring: (i) \(\dfrac{13}{5}\) (ii) \(\dfrac{2}{11}\) (iii) \(\dfrac{29}{16}\) (iv) \(\dfrac{17}{125}\) (v) \(\dfrac{11}{6}\).
(i) Terminating (\(=2.6\)). (ii) Recurring (11 in denom). (iii) Terminating (\(2^4\)). (iv) Terminating (\(5^3\)). (v) Recurring (factor 3).
2) Write decimals: (i) \(\dfrac{127}{200}\) (ii) \(\dfrac{25}{99}\) (iii) \(\dfrac{23}{7}\) (iv) \(\dfrac{4}{5}\) (v) \(\dfrac{17}{8}\).
(i) \(0.635\). (ii) \(0.\overline{25}\). (iii) \(3.\overline{285714}\). (iv) \(0.8\). (v) \(2.125\).
3) Write in \(\tfrac pq\): (i) \(0.6\) (ii) \(0.37\) (iii) \(3.17\) (iv) \(15.89\) (v) \(2.514\).
(i) \(\dfrac{3}{5}\). (ii) \(\dfrac{37}{100}\). (iii) \(\dfrac{317}{100}\). (iv) \(\dfrac{1589}{100}\). (v) \(\dfrac{1257}{500}\) (reduced).
Practice Set 2.2 — Solutions
1) Show \(4\sqrt2\) is irrational.
If \(4\sqrt2\) were rational, dividing by \(4\ne0\) gives \(\sqrt2\) rational — contradiction. Hence irrational.
2) Prove \(\sqrt3+\sqrt5\) is irrational.
If rational, subtract \(\sqrt3\) (irrational) to get \(\sqrt5\) rational, impossible.
3) Represent \(\sqrt5\) and \(\sqrt{10}\) on number line (method).
Use right triangle construction (Pythagoras): segment of length \(\sqrt5\) from 0 via 2–1 triangle; similarly extend to get \(\sqrt{10}\). (Geometric marking on the line).
4) Write any three rationals between: (i) \(0.3\) and \(-0.5\) (ii) \(-2.3\) and \(-2.33\) (iii) \(5.2\) and \(5.3\) (iv) \(-4.5\) and \(-4.6\).
(i) \(-0.1,-0.2,-0.4\). (ii) \(-2.305,-2.31,-2.32\). (iii) \(5.21,5.25,5.29\). (iv) \(-4.51,-4.55,-4.59\).
Practice Set 2.3 — Solutions
1) State order of surds: (i) \(\sqrt[3]{7}\) (ii) \(\sqrt{12}\) (iii) \(\sqrt[4]{10}\) (iv) \(\sqrt[9]{3}\) (v) \(\sqrt[3]{18}\).
(i) 3 (ii) 2 (iii) 4 (iv) 9 (v) 3.
2) Which are surds? justify: (i) \(\sqrt[3]{51}\) (ii) \(\sqrt[4]{16}\) (iii) \(\sqrt[5]{81}\) (iv) \(\sqrt{256}\) (v) \(\sqrt[3]{64}\) (vi) \(\dfrac{22}{7}\).
(i) Surd (irrational). (ii) Not a surd (=2). (iii) Surd (irrational). (iv) Not (=\(16\)). (v) Not (=\(4\)). (vi) Not a surd (rational, no radical).
3) Like or unlike: (i) \(\sqrt{52},\ \sqrt{5\cdot13}\) (ii) \(\sqrt{68},\ \sqrt{5}\sqrt3\) (iii) \(4\sqrt{18},\ 7\sqrt2\) (iv) \(19\sqrt{12},\ 6\sqrt3\) (v) \(5\sqrt{22},\ 7\sqrt{33}\) (vi) \(5\sqrt5,\ \sqrt{75}\).
Reduce: \(\sqrt{52}=2\sqrt{13}\), \(\sqrt{5\cdot13}=\sqrt{65}\) → unlike. (ii) \(\sqrt{68}=2\sqrt{17}\), \(\sqrt{15}\) → unlike. (iii) \(4\sqrt{18}=12\sqrt2\) → like with \(7\sqrt2\). (iv) \(19\cdot2\sqrt3=38\sqrt3\) → like with \(6\sqrt3\). (v) radicands differ → unlike. (vi) \(\sqrt{75}=5\sqrt3\) → unlike with \(5\sqrt5\).
4) Simplify: (i) \(\sqrt{27}\) (ii) \(\sqrt{50}\) (iii) \(\sqrt{250}\) (iv) \(\sqrt{112}\) (v) \(\sqrt{168}\).
(i) \(3\sqrt3\). (ii) \(5\sqrt2\). (iii) \(5\sqrt{10}\). (iv) \(4\sqrt7\). (v) \(2\sqrt{42}\).
5) Compare: (i) \(7\sqrt2,\,5\sqrt3\) (ii) \(\sqrt{247},\,\sqrt{274}\) (iii) \(2\sqrt7,\,\sqrt{28}\) (iv) \(5\sqrt5,\,7\sqrt2\) (v) \(4\sqrt{42},\,9\sqrt2\) (vi) \(5\sqrt3,\,9\) (vii) \(7,\,2\sqrt5\).
(i) \(7\sqrt2>5\sqrt3\) (98>75). (ii) \(\sqrt{247}<\sqrt{274}\). (iii) Equal (both \(\sqrt{28}\)). (iv) \(5\sqrt5>7\sqrt2\) (125>98). (v) Square: \(16\cdot42=672\) vs \(81\cdot2=162\) ⇒ \(4\sqrt{42}>\,9\sqrt2\). (vi) \(5\sqrt3<9\) (75<81). (vii) \(7>2\sqrt5\) (49>20).
6) Simplify: (i) \(5\sqrt3+8\sqrt3\) (ii) \(9\sqrt5-4\sqrt5+\sqrt{125}\) (iii) \(7\sqrt{48}-\sqrt{27}-3\sqrt{}\! \text{(typo→}3\sqrt{?})\) (iv) \(\dfrac{7}{\sqrt5}-\dfrac{3}{\sqrt7}+2\sqrt7\).
(i) \(13\sqrt3\). (ii) \(5\sqrt5+5\sqrt5=10\sqrt5\). (iii) Interpret as \(7\sqrt{48}-\sqrt{27}-3\sqrt{12}=28\sqrt3-3\sqrt3-6\sqrt3=19\sqrt3\). (iv) \(\dfrac{7\sqrt5}{5}-\dfrac{3\sqrt7}{7}+2\sqrt7=\dfrac{7\sqrt5}{5}+\dfrac{11\sqrt7}{7}\).
7) Multiply & simplify: (i) \(\sqrt{3}\sqrt{12}\cdot\sqrt{18}\) (ii) \(\sqrt3\sqrt{12}\cdot 7\sqrt{15}\) (iii) \(\sqrt3\sqrt8\cdot\sqrt5\) (iv) \(5\sqrt8\cdot2\sqrt8\).
(i) \(\sqrt{648}=18\sqrt2\). (ii) \(7\sqrt{540}=7\cdot6\sqrt{15}=42\sqrt{15}\). (iii) \(\sqrt{120}=2\sqrt{30}\). (iv) \(10\cdot 8=80\).
8) Divide (simplest form): (i) \(\dfrac{\sqrt{98}}{\sqrt2}\) (ii) \(\dfrac{\sqrt{125}}{\sqrt{50}}\) (iii) \(\dfrac{\sqrt{54}}{\sqrt{27}}\) (iv) \(\dfrac{\sqrt{310}}{\sqrt5}\).
(i) \(7\). (ii) \(\sqrt{\tfrac{125}{50}}=\sqrt{\tfrac52}=\dfrac{\sqrt{10}}{2}\). (iii) \(\sqrt2\). (iv) \(\sqrt{62}\).
9) Rationalize denominator: (i) \(\dfrac{3}{\sqrt5}\) (ii) \(\dfrac{1}{\sqrt{14}}\) (iii) \(\dfrac{5}{\sqrt7}\) (iv) \(\dfrac{6}{9\sqrt3}\) (v) \(\dfrac{11}{\sqrt3}\).
(i) \(\dfrac{3\sqrt5}{5}\). (ii) \(\dfrac{\sqrt{14}}{14}\). (iii) \(\dfrac{5\sqrt7}{7}\). (iv) \(\dfrac{2}{3\sqrt3}=\dfrac{2\sqrt3}{9}\). (v) \(\dfrac{11\sqrt3}{3}\).
Practice Set 2.4 — Solutions
1) Multiply: (i) \(\sqrt3(\sqrt7-\sqrt3)\) (ii) \((\sqrt5-\sqrt7)^2\) (iii) \((3\sqrt2-\sqrt3)(4\sqrt3-2)\).
(i) \(\sqrt{21}-3\). (ii) \(5+7-2\sqrt{35}=12-2\sqrt{35}\). (iii) \(12\sqrt6-6-4\sqrt{10}+2\sqrt3\).
2) Rationalize denominator: (i) \(\dfrac{1}{\sqrt7+\sqrt2}\) (ii) \(\dfrac{3}{\sqrt5(\,3\sqrt2\,)}\) (iii) \(\dfrac{4}{\sqrt7+\sqrt3}\) (iv) \(\dfrac{\sqrt5}{\sqrt5+\sqrt3}\).
(i) \(\dfrac{\sqrt7-\sqrt2}{7-2}=\dfrac{\sqrt7-\sqrt2}{5}\). (ii) \(\dfrac{3}{3\sqrt{10}}=\dfrac{1}{\sqrt{10}}=\dfrac{\sqrt{10}}{10}\). (iii) \(\dfrac{4(\sqrt7-\sqrt3)}{7-3}=\dfrac{\sqrt7-\sqrt3}{1}\cdot 1\) → \(=\sqrt7-\sqrt3\). (iv) \(\dfrac{\sqrt5(\sqrt5-\sqrt3)}{5-3}=\dfrac{5-\sqrt{15}}{2}\).
Practice Set 2.5 — Solutions (Absolute Value)
1) Evaluate: (i) \(|15-2|\) (ii) \(|4-9|\) (iii) \(|7|\cdot|-4|\).
(i) \(13\). (ii) \(5\). (iii) \(28\).
2) Solve: (i) \(|3x-5|=1\) (ii) \(|7-2x|=5\) (iii) \(\left|\dfrac{8}{2}x\right|=5\) (iv) \(\left|5+\dfrac{x}{4}\right|=5\).
(i) \(x=2,\ \dfrac{4}{3}\). (ii) \(x=1,\ 6\). (iii) \(|4x|=5\Rightarrow x=\pm\dfrac{5}{4}\). (iv) \(5+\dfrac{x}{4}=5\) or \(-5\Rightarrow x=0\) or \(-40\).
Problem Set 2 — Solutions
1) MCQs: (i) irrational? (A) \(\tfrac{16}{25}\) (B) \(\sqrt5\) (C) \(\tfrac{3}{9}\) (D) \(\sqrt{196}\)
(B).
1(ii) Irrational?
(D) \(0.101001000\ldots\) (non-terminating, non-recurring).
1(iii) Non-terminating recurring?
(C) \(\tfrac{3}{11}\).
1(iv) Every point on number line represents…
(D) Real numbers.
1(v) \(0.\overline{4}\) in \(\tfrac pq\) form.
\(\dfrac{4}{9}\).
1(vi) \(\sqrt n\) when \(n\) not a perfect square is…
(C) Irrational.
1(vii) Not a surd?
(A) \(7\) (no radical/irrational root).
1(viii) Order of \(5\sqrt3\).
2 (coefficient \(5\) doesn’t change order).
1(ix) Conjugate pair of \(\sqrt2+\sqrt3\).
(C) \(\sqrt2-\sqrt3\).
1(x) Value of \(|12-(13+7)\cdot4|\).
\(|12-80|=68\) → (B).
2) Write in \(\tfrac pq\): (i) \(0.\overline{5}\) (ii) \(29.568\) (iii) \(9.315315\ldots\) (iv) \(357.\overline{417}\) (v) \(30.219\).
(i) \(\dfrac{5}{9}\). (ii) \(\dfrac{29568}{1000}=\dfrac{3696}{125}\). (iii) Let \(x=9.\overline{315}\Rightarrow 1000x- x=9315-9=9306\Rightarrow x=\dfrac{9306}{999}=\dfrac{1034}{111}\). (iv) \(x=357.\overline{417}\Rightarrow 1000x-x=357417-357=357060\Rightarrow x=\dfrac{357060}{999}=\dfrac{3964}{11}\). (v) \(\dfrac{30219}{1000}\).
3) Write decimals: (i) \(-\dfrac{5}{7}\) (ii) \(\dfrac{9}{11}\) (iii) \(\sqrt5\) (iv) \(\dfrac{121}{13}\) (v) \(\dfrac{29}{8}\).
(i) \(-0.\overline{714285}\). (ii) \(0.\overline{81}\). (iii) \(2.236\ldots\) (irrational). (iv) \(9.\overline{307692}\). (v) \(3.625\).
4) Show \( \sqrt5+\sqrt7\) is irrational.
As in 2.2(2): otherwise \(\sqrt7=(\sqrt5+\sqrt7)-\sqrt5\) rational ⇒ contradiction.
5) Simplest form: (i) \(\dfrac{3}{\sqrt{48}}\) (ii) \(-\dfrac{5}{\sqrt{9\cdot45}}\).
(i) \(\dfrac{3}{4\sqrt3}=\dfrac{3\sqrt3}{12}=\dfrac{\sqrt3}{4}\). (ii) Denominator \(\sqrt{405}=9\sqrt5\): value \(=-\dfrac{5}{9\sqrt5}=-\dfrac{5\sqrt5}{45}=-\dfrac{\sqrt5}{9}\).
6) Simplest rationalizing factor for: (i) \(\sqrt{32}\) (ii) \(\sqrt{50}\) (iii) \(\sqrt{27}\) (iv) \(3\sqrt{5\cdot10}\) (v) \(3\sqrt{72}\) (vi) \(4\sqrt{11}\).
(i) \(\sqrt2\) (since \(\sqrt{32}\cdot\sqrt2=8\)). (ii) \(\sqrt2\). (iii) \(\sqrt3\). (iv) \(\sqrt{10}\) (or \(\tfrac{1}{3}\sqrt{10}\) scaled). (v) \(\sqrt2\) (as \(3\sqrt{72}\cdot\sqrt2=6\cdot6=36\)). (vi) \(\sqrt{11}\).
7) Simplify: (i) \(4\sqrt7+ \sqrt{147}+ \dfrac{3\sqrt8}{\sqrt{192}}+\dfrac{\sqrt{75}}{5}\) (ii) \(5\sqrt3+2\sqrt{27}+\dfrac{1}{\sqrt3}\) (iii) \(\sqrt{216}-\dfrac{5}{\sqrt6}+ \dfrac{6}{\sqrt{294}}\).
(i) \(\sqrt{147}=7\sqrt3\) (oops → check: \(147=49\cdot3\Rightarrow 7\sqrt3\) but terms are of \(\sqrt7\) family); better regroup carefully:
• \(4\sqrt7\) stays. • \(\sqrt{147}=7\sqrt3\). • \(\dfrac{3\sqrt8}{\sqrt{192}}=\dfrac{3\cdot2\sqrt2}{8\sqrt3}=\dfrac{3\sqrt2}{4\sqrt3}=\dfrac{3\sqrt6}{12}=\dfrac{\sqrt6}{4}\). • \(\dfrac{\sqrt{75}}{5}=\dfrac{5\sqrt3}{5}=\sqrt3\). Final: \(4\sqrt7 + (7\sqrt3+\sqrt3) + \dfrac{\sqrt6}{4}=4\sqrt7+8\sqrt3+\dfrac{\sqrt6}{4}\).
(ii) \(2\sqrt{27}=6\sqrt3\). So \(5\sqrt3+6\sqrt3+\dfrac{\sqrt3}{3}=11\sqrt3+\dfrac{\sqrt3}{3}=\dfrac{34\sqrt3}{3}\).
(iii) \(\sqrt{216}=6\sqrt6\). Also \(\dfrac{6}{\sqrt{294}}=\dfrac{6\sqrt{294}}{294}=\dfrac{\sqrt{294}}{49}=\dfrac{7\sqrt6}{49}=\dfrac{\sqrt6}{7}\). Hence \(6\sqrt6-\dfrac{5}{\sqrt6}+\dfrac{\sqrt6}{7}=\left(6+\dfrac{1}{7}\right)\sqrt6-\dfrac{5}{\sqrt6}=\dfrac{43}{7}\sqrt6-\dfrac{5\sqrt6}{6}=\left(\dfrac{258-35}{42}\right)\sqrt6=\dfrac{223}{42}\sqrt6\).
• \(4\sqrt7\) stays. • \(\sqrt{147}=7\sqrt3\). • \(\dfrac{3\sqrt8}{\sqrt{192}}=\dfrac{3\cdot2\sqrt2}{8\sqrt3}=\dfrac{3\sqrt2}{4\sqrt3}=\dfrac{3\sqrt6}{12}=\dfrac{\sqrt6}{4}\). • \(\dfrac{\sqrt{75}}{5}=\dfrac{5\sqrt3}{5}=\sqrt3\). Final: \(4\sqrt7 + (7\sqrt3+\sqrt3) + \dfrac{\sqrt6}{4}=4\sqrt7+8\sqrt3+\dfrac{\sqrt6}{4}\).
(ii) \(2\sqrt{27}=6\sqrt3\). So \(5\sqrt3+6\sqrt3+\dfrac{\sqrt3}{3}=11\sqrt3+\dfrac{\sqrt3}{3}=\dfrac{34\sqrt3}{3}\).
(iii) \(\sqrt{216}=6\sqrt6\). Also \(\dfrac{6}{\sqrt{294}}=\dfrac{6\sqrt{294}}{294}=\dfrac{\sqrt{294}}{49}=\dfrac{7\sqrt6}{49}=\dfrac{\sqrt6}{7}\). Hence \(6\sqrt6-\dfrac{5}{\sqrt6}+\dfrac{\sqrt6}{7}=\left(6+\dfrac{1}{7}\right)\sqrt6-\dfrac{5}{\sqrt6}=\dfrac{43}{7}\sqrt6-\dfrac{5\sqrt6}{6}=\left(\dfrac{258-35}{42}\right)\sqrt6=\dfrac{223}{42}\sqrt6\).
8) Rationalize: (i) \(\dfrac{1}{\sqrt5}\) (ii) \(\dfrac{2}{\sqrt3+\sqrt7}\) (iii) \(\dfrac{1}{\sqrt3-\sqrt2}\) (iv) \(\dfrac{1}{\sqrt3+\dfrac{5}{2}}\) (v) \(\dfrac{12}{\sqrt{4\cdot3\cdot2}}\).
(i) \(\dfrac{\sqrt5}{5}\). (ii) \(\dfrac{2(\sqrt7-\sqrt3)}{7-3}=\dfrac{\sqrt7-\sqrt3}{2}\). (iii) \(\sqrt3+\sqrt2\). (iv) Multiply by \( \left(\sqrt3-\dfrac{5}{2}\right)\): \(\dfrac{\sqrt3-\tfrac{5}{2}}{3-\tfrac{25}{4}}=\dfrac{\sqrt3-\tfrac{5}{2}}{\tfrac{-13}{4}}=-\dfrac{4\sqrt3-10}{13}\). (v) Denominator \(\sqrt{24}=2\sqrt6\): value \(=\dfrac{12}{2\sqrt6}=\dfrac{6}{\sqrt6}=\sqrt6\).
Mini Reference
• Terminating iff denominator in lowest terms has only \(2\)s and \(5\)s.
• Like surds: same order & radicand. Simplify first!
• Conjugates: \((\sqrt a+\sqrt b)\leftrightarrow(\sqrt a-\sqrt b)\), product \(=a-b\).
• Absolute value: \(|x|=\begin{cases}x,&x\ge0\\-x,&x<0\end{cases}\).
Tip: Copy each part (1-mark, 2-mark, 3-mark, Exercises) separately to keep your site’s menu bar intact.