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(d) 4.492

\(\sqrt\frac23+3\times\sqrt\frac32 = \frac{\sqrt2}{\sqrt3}+3\times\frac{\sqrt3}{\sqrt2}\)

\(\frac{\sqrt2\times\sqrt3}{\sqrt3\times\sqrt3} + 3\times\frac{\sqrt3\times\sqrt2}{\sqrt2\times\sqrt2}\)

(Rationalising the denominator)

= \(\frac{\sqrt6}{3}+\frac{3\times\sqrt6}{2}=\frac{2\sqrt6+9\sqrt6}{6}\)

= \(\frac{11\sqrt6}{6}=\frac{11\times2.45}{6}=\frac{26.95}{6}\) = 4.492

(d) simplifies to 6. (Take positive root only)

\(\sqrt{14+6\sqrt5} = \sqrt{9+5+2\times3\times\sqrt5}\)

= \(\sqrt{(3)^2+(\sqrt5)^2+2\times3\sqrt5}\)

= \(\sqrt{(3+\sqrt5)^2}\) = 3 + \(\sqrt5\)

Similarly, \(\sqrt{14+6\sqrt5} = ​3 - \sqrt5 ​\)

∴ \(\sqrt{14+6\sqrt5} + \sqrt{14-6\sqrt5} ​\) =  3 + \(\sqrt5\) +  3 - \(\sqrt5\) = 6.

∴ (d) is the correct answer.

(c) 5

Let x = \(\sqrt{5\sqrt{5\sqrt{5}}}......is\)

⇒ x = \(\sqrt{5x}\) ⇒ x² = 5x

⇒ x² - 5x = 0 ⇒ x (x - 5) = 0 ⇒ 0 or 5

Neglecting x = 0, we have x = 5.

Irrational numbers of the type \(\sqrt{2},\) \(\sqrt{3},\) \(\sqrt{5},\) \(\sqrt{17},\) ...... which are square roots of positive rational ____ numbers that cannot be expressed as squares of any rational numbers are called surds.

Similarly, \(\sqrt[3]{2}\), \(\sqrt[3]{4}\), \(\sqrt[3]{19}\) etc., are numbers which are the cube roots of positive rational numbers that cannot be expressed as the cubes of any rational numbers and so are surds.

Definition : If x is a positive rational number and n is a positive integer such that x^1/n , i.e., \(\sqrt[n]{x}\) is irrational, then

\(\sqrt[n]{x}\) is called a surd or radical of order n.

\(\sqrt[5]{25}\) is a surd of order 5.

(a) If a and b are surds of the same order, say n, then \(\sqrt[n]{a}\) > \(\sqrt[n]{b}\) if a > b. 

 For example,  \(\sqrt[5]{27}\) > \(\sqrt[5]{20}\) as 27 > 20. 

(b) If the given surds are not of the same order, then first convert them to surds of the same order and then compare. For example, to compare \(\sqrt[3]{2}\) and \(\sqrt[4]{3}\), we take the LCM of the orders, i.e., 3 and 4, i.e., 12.

Now, \(\sqrt[3]{2}\)  = \(\sqrt[12]{2^4}\) = \(\sqrt[12]{16}\)             (∴ \(\sqrt[3]{2}\)  = \(2^\frac{1}{3}\) = \(2^\frac{4}{12}\))

  \(\sqrt[4]{3}\)  = \(\sqrt[12]{3^3}\) = \(\sqrt[12]{27}\)                   (∴  \(\sqrt[4]{3}\)  = \(3^\frac{1}{4}\) = \(3^\frac{3}{12}\))

Since, 27 > 16, therefore, \(\sqrt[12]{27}\) > \(\sqrt[12]{16}\), i.e, \(\sqrt[4]{3}\) > \(\sqrt[3]{2}\).

Rationalisation : When surds occur in the denominator of a fraction, it is customary to rid the denominator of the radicals. The surd in the denominator is multiplied by an appropriate expression, such that the product is a rational number. The given surd and the expression by which it is multiplied are called rationalising factors of each other. 

For example,

(i) \(a^{1-\frac{1}{n}}\) is the rationalising factor of \(a^\frac{1}{n}\),as \(a^\frac{1}{n}\) \(a^{1-\frac{1}{n}}\) = \(a^{\frac{1}{n}+1-\frac{1}{n}}\) = a¹ = a, which is a rational number. Hence, the rationalising factor of \(6^\frac{1}{5}\) is \(6^{1-\frac{1}{5}}\) = \(6^\frac{1}{5}\).

(ii) (a + √b) is the rationalising factor of (a- √b) , as, (a + √b)(a - √b) = a² – b, which is a rational. Hence, (3 + √2) is the rationalising factor of (3 - √2) .

(3 + √2) (3 - √2) = 9 – 2 = 7

(iii) (√a + √b) is the rationalising factor of (√a + √b) as (√a + √b)(√a - √b) = (√a)² - (√b)² = a - b, which is a rational number.

Note: Such binominal surds as (√a + √b) and (√a - √b)which differ only in the sign connecting their terms are said to be conjugate surds.

The product of conjugate surds is always a rational.

Thus, the process of multiplication of a surd by its rationalising factor is called rationalisation.

LCM of (3, 6, 4) = 12, Raising each given number to power 12,

∴ \(\sqrt[3]{4}=\) (4)^{\(\frac{1}{3}\)} =  (4^{\(\frac{1}{3}\)} )¹² = 4⁴ = 256

  \(\sqrt{2}=\) (2)^{\(\frac{1}{2}\)} =  (2^{\(\frac{1}{2}\)} )¹² = 4⁶ = 64

  \(\sqrt[6]{13}=\) (13)^{\(\frac{1}{6}\)} =  (13^{\(\frac{1}{6}\)} )¹² = 13² = 169

  \(\sqrt[4]{5}=\) (5)^{\(\frac{1}{4}\)} =  (4^{\(\frac{1}{4}\)} )¹² = 5³ = 125

⇒ \(\sqrt[3]{4}\) is the greatest.

\(\sqrt{A}=\sqrt{5+2\sqrt6} = \sqrt{3+2+2\sqrt{3\times2}}\)

= \(\sqrt{(\sqrt3)^2 + (\sqrt2)^2 + 2\sqrt3\sqrt2}\)

= \(\sqrt{(\sqrt3) + (\sqrt2)^2} = \sqrt3+\sqrt2\)

∴ \(\sqrt{A} + \frac{1}{\sqrt{A}}=(\sqrt3+\sqrt2) + \frac{1}{(\sqrt3+\sqrt2)}\)

= \((\sqrt3+\sqrt2) + \frac{\sqrt3-\sqrt2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\)

= (\(\sqrt3+\sqrt2\)) + \(\frac{\sqrt3-\sqrt2}{3-2} = \sqrt3+\sqrt2+\sqrt3-\sqrt2= 2\sqrt3\).