Exercise 1.2

Question 1: Rationalize the Denominator

To rationalize a denominator containing a square root, multiply the numerator and denominator by the conjugate of the denominator.

Formula

$$(a+b)(a-b)=a^2-b^2$$

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(i)

Given

$$\frac{13}{4+\sqrt3}$$

Solution

Multiply by the conjugate (4-\sqrt3).

$$\frac{13}{4+\sqrt3}\times\frac{4-\sqrt3}{4-\sqrt3}$$

$$= \frac{13(4-\sqrt3)} {(4+\sqrt3)(4-\sqrt3)}$$

Using

$$(a+b)(a-b)=a^2-b^2$$

$$= \frac{13(4-\sqrt3)} {16-3}$$

$$= \frac{13(4-\sqrt3)} {13}$$

$$=4-\sqrt3$$

Final Answer

$$\boxed{4-\sqrt3}$$

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(ii)

Given

$$\frac{\sqrt2+\sqrt5}{\sqrt3}$$

Solution

Multiply numerator and denominator by (\sqrt3).

$$\frac{\sqrt2+\sqrt5}{\sqrt3} \times \frac{\sqrt3}{\sqrt3}$$

$$= \frac{\sqrt6+\sqrt15}{3}$$

Final Answer

$$\boxed{\frac{\sqrt6+\sqrt15}{3}}$$

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(iii)

Given

$$\frac{\sqrt2-1}{\sqrt5}$$

Solution

Multiply numerator and denominator by (\sqrt5).

$$\frac{\sqrt2-1}{\sqrt5} \times \frac{\sqrt5}{\sqrt5}$$

$$= \frac{\sqrt10-\sqrt5}{5}$$

Final Answer

$$\boxed{\frac{\sqrt10-\sqrt5}{5}}$$

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(iv)

Given

$$\frac{6-4\sqrt2}{6+4\sqrt2}$$

Solution

Multiply by the conjugate (6-4\sqrt2).

$$\frac{6-4\sqrt2}{6+4\sqrt2} \times \frac{6-4\sqrt2}{6-4\sqrt2}$$

$$= \frac{(6-4\sqrt2)^2} {6^2-(4\sqrt2)^2}$$

Expand the numerator.

$$(6-4\sqrt2)^2 = 36-48\sqrt2+32 = 68-48\sqrt2$$

Simplify the denominator.

$$36-32=4$$

$$= \frac{68-48\sqrt2}{4}$$

$$=17-12\sqrt2$$

Final Answer

$$\boxed{17-12\sqrt2}$$

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(v)

Given

$$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$$

Solution

Multiply by the conjugate (\sqrt3-\sqrt2).

$$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2} \times \frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$$

$$= \frac{(\sqrt3-\sqrt2)^2} {3-2}$$

Expand.

$$(\sqrt3-\sqrt2)^2 = 3+2-2\sqrt6$$

$$= 5-2\sqrt6$$

Final Answer

$$\boxed{5-2\sqrt6}$$

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(vi)

Given

$$\frac{4\sqrt3}{\sqrt7+\sqrt5}$$

Solution

Multiply by the conjugate (\sqrt7-\sqrt5).

$$\frac{4\sqrt3}{\sqrt7+\sqrt5} \times \frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5}$$

$$= \frac{4\sqrt3(\sqrt7-\sqrt5)} {7-5}$$

$$= \frac{4(\sqrt21-\sqrt15)} {2}$$

$$= 2(\sqrt21-\sqrt15)$$

Final Answer

$$\boxed{2(\sqrt21-\sqrt15)}$$

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Question 2: Simplify the Following

Use the laws of exponents to simplify each expression.

Important Formulas

$$a^m \times a^n=a^{m+n}$$

$$\frac{a^m}{a^n}=a^{m-n}$$

$$(a^m)^n=a^{mn}$$

$$a^{-n}=\frac1{a^n}$$

$$a^{\frac1n}=\sqrt[n]{a}$$

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(i)

Given

$$\left(\frac{81}{16}\right)^{-\frac34}$$

Solution

Invert the fraction because of the negative power.

$$= \left(\frac{16}{81}\right)^{\frac34}$$

Write each number as a power.

$$16=2^4,\qquad81=3^4$$

$$= \left(\frac{2^4}{3^4}\right)^{\frac34}$$

$$= \frac{2^3}{3^3}$$

$$= \frac{8}{27}$$

Final Answer

$$\boxed{\frac{8}{27}}$$

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(ii)

Given

$$\left(\frac34\right)^{-2} \div \left(\frac49\right)^3 \times \frac{16}{27}$$

Solution

$$\left(\frac34\right)^{-2} = \left(\frac43\right)^2 = \frac{16}{9}$$

$$\left(\frac49\right)^3 = \frac{64}{729}$$

$$= \frac{16}{9} \times \frac{729}{64} \times \frac{16}{27}$$

Cancel common factors.

$$=12$$

Final Answer

$$\boxed{12}$$

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(iii)

Given

$$(0.027)^{-\frac13}$$

Solution

$$0.027=\frac{27}{1000} =\left(\frac3{10}\right)^3$$

$$= \left(\frac3{10}\right)^{-1}$$

$$= \frac{10}{3}$$

Final Answer

$$\boxed{\frac{10}{3}}$$

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(iv)

Given

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

Solution

Simplify the powers first.

$$= \sqrt[7]{x^{14}y^7z^{28}}$$

Apply the seventh root.

$$=x^2yz^4$$

Final Answer

$$\boxed{x^2yz^4}$$

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(v)

Given

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

Solution

Use

$$25=5^2$$

Numerator:

$$5^{2n+3}-5^{2n+2} = 5^{2n+2}(5-1) = 4\cdot5^{2n+2}$$

Denominator:

$$5^{2n+4}-5^{2n+2} = 5^{2n+2}(25-1) = 24\cdot5^{2n+2}$$

$$= \frac4{24} = \frac16$$

Final Answer

$$\boxed{\frac16}$$

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(vi)

Given

$$\frac{16^{x+1}+20(4^{2x})} {2^{x-3}\times8^{x+2}}$$

Solution

Convert everything to base 2.

$$16=2^4,\qquad4=2^2,\qquad8=2^3$$

Numerator:

$$2^{4x+4}+20\cdot2^{4x} = 2^{4x}(16+20) = 36\cdot2^{4x}$$

Denominator:

$$2^{x-3}\times2^{3x+6} = 2^{4x+3}$$

$$= 36\times2^{-3}$$

$$=\frac{36}{8} =\frac92$$

Final Answer

$$\boxed{\frac92}$$

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(vii)

Given

$$(64)^{-\frac23}\div(9)^{-\frac32}$$

Solution

$$64=2^6,\qquad9=3^2$$

$$(64)^{-\frac23} = 2^{-4} = \frac1{16}$$

$$(9)^{-\frac32} = 3^{-3} = \frac1{27}$$

$$= \frac1{16} \div \frac1{27} = \frac{27}{16}$$

Final Answer

$$\boxed{\frac{27}{16}}$$

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(viii)

Given

$$\frac{3^n\times9^{n+1}} {3^{n-1}\times9^{n-1}}$$

Solution

Use

$$9=3^2$$

$$= \frac{3^n\times3^{2n+2}} {3^{n-1}\times3^{2n-2}}$$

$$= 3^{3n+2-(3n-3)}$$

$$=3^5$$

$$=243$$

Final Answer

$$\boxed{243}$$

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(ix)

Given

$$\frac{5^{n+3}-6\cdot5^{n+1}} {9\cdot5^n-4\cdot5^n}$$

Solution

Factor out common powers.

Numerator:

$$5^{n+1}(25-6) = 19\cdot5^{n+1}$$

Denominator:

$$5^n(9-4) = 5^{n+1}$$

$$= 19$$

Final Answer

$$\boxed{19}$$

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Question 3: If $x = 3+\sqrt8$, Find the Following

Given

$$x=3+\sqrt8$$

Since

$$\sqrt8=2\sqrt2$$

we have

$$x=3+2\sqrt2$$

First, find the reciprocal of (x).

$$\frac1x=\frac1{3+2\sqrt2}$$

Multiply by the conjugate.

$$\frac1x = \frac{3-2\sqrt2}{(3+2\sqrt2)(3-2\sqrt2)}$$

$$= \frac{3-2\sqrt2}{9-8}$$

$$= 3-2\sqrt2$$

Now use

$$x=3+2\sqrt2$$

and

$$\frac1x=3-2\sqrt2$$

to solve each part.

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(i)

Find

$$x+\frac1x$$

Solution

$$=(3+2\sqrt2)+(3-2\sqrt2)$$

$$=6$$

Final Answer

$$\boxed{6}$$

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(ii)

Find

$$x-\frac1x$$

Solution

$$=(3+2\sqrt2)-(3-2\sqrt2)$$

$$=4\sqrt2$$

Final Answer

$$\boxed{4\sqrt2}$$

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(iii)

Find

$$x^2+\frac1{x^2}$$

Formula

$$\left(x+\frac1x\right)^2 = x^2+\frac1{x^2}+2$$

Solution

$$x^2+\frac1{x^2} = 6^2-2$$

$$=36-2$$

$$=34$$

Final Answer

$$\boxed{34}$$

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(iv)

Find

$$x^2-\frac1{x^2}$$

Formula

$$x^2-\frac1{x^2} = \left(x-\frac1x\right) \left(x+\frac1x\right)$$

Solution

$$=(4\sqrt2)(6)$$

$$=24\sqrt2$$

Final Answer

$$\boxed{24\sqrt2}$$

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(v)

Find

$$x^4+\frac1{x^4}$$

Formula

$$\left(x^2+\frac1{x^2}\right)^2 = x^4+\frac1{x^4}+2$$

Solution

$$=34^2-2$$

$$=1156-2$$

$$=1154$$

Final Answer

$$\boxed{1154}$$

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(vi)

Find

$$\left(x-\frac1x\right)^2$$

Solution

$$=(4\sqrt2)^2$$

$$=16\times2$$

$$=32$$

Final Answer

$$\boxed{32}$$

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Question 4

Prove that

$$\sqrt5+\sqrt6>\sqrt{11}$$

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Formula

For positive numbers,

$$(a+b)^2=a^2+b^2+2ab$$

If

$$A^2>B^2$$

and both (A) and (B) are positive, then

$$A>B$$

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Solution

Let

$$A=\sqrt5+\sqrt6$$

Square both sides.

$$A^2=(\sqrt5+\sqrt6)^2$$

Expand the expression.

$$=5+6+2\sqrt{30}$$

$$=11+2\sqrt{30}$$

Since

$$2\sqrt{30}>0$$

we have

$$11+2\sqrt{30}>11$$

Therefore,

$$A^2>11$$

But

$$11=(\sqrt{11})^2$$

Hence,

$$A^2>(\sqrt{11})^2$$

Since both numbers are positive,

$$A>\sqrt{11}$$

Substitute the value of (A).

$$\boxed{\sqrt5+\sqrt6>\sqrt{11}}$$

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Final Answer

$$\boxed{\sqrt5+\sqrt6>\sqrt{11}}$$

Question 5

Simplify the Following

Important Formula

$$(a+b)^2=a^2+2ab+b^2$$

$$(a-b)^2=a^2-2ab+b^2$$

$$(a+b)(a-b)=a^2-b^2$$

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(i)

Given

$$(\sqrt5+\sqrt3)^2$$

Solution

Apply the identity

$$(a+b)^2=a^2+2ab+b^2$$

$$=(\sqrt5)^2+2(\sqrt5)(\sqrt3)+(\sqrt3)^2$$

$$=5+2\sqrt{15}+3$$

$$=8+2\sqrt{15}$$

Final Answer

$$\boxed{8+2\sqrt{15}}$$

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(ii)

Given

$$(\sqrt7-\sqrt2)^2$$

Solution

Apply

$$(a-b)^2=a^2-2ab+b^2$$

$$=(\sqrt7)^2-2(\sqrt7)(\sqrt2)+(\sqrt2)^2$$

$$=7-2\sqrt{14}+2$$

$$=9-2\sqrt{14}$$

Final Answer

$$\boxed{9-2\sqrt{14}}$$

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(iii)

Given

$$(\sqrt6+\sqrt5)(\sqrt6-\sqrt5)$$

Solution

Use the identity

$$(a+b)(a-b)=a^2-b^2$$

$$=(\sqrt6)^2-(\sqrt5)^2$$

$$=6-5$$

$$=1$$

Final Answer

$$\boxed{1}$$

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(iv)

Given

$$(\sqrt8+\sqrt2)(\sqrt8-\sqrt2)$$

Solution

Apply

$$(a+b)(a-b)=a^2-b^2$$

$$=(\sqrt8)^2-(\sqrt2)^2$$

$$=8-2$$

$$=6$$

Final Answer

$$\boxed{6}$$

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