Polynomials and Exponents

nth root of a:- If a is any real number and n is a positive integer, then the nth root of a is the real number x such xna =

The nth root of a denoted by 1/an or n?a

Thus, n?a = x ? 1/an = xn ? xn = a

(i) ?a = 1/a2 is called the square-root of a;

(ii) 3?a = 1/a3 is called the cube-root of a;

(iii) 4?a = 1/a4 is called the 4th-root of a and so on.

Remark: For positive value of a, the value of 1/an will always e taken as positive.

Example:

(i) ?9 = 1/92 = 3, since 32 = 9;

(ii) 3?8 = 1/83 = 2, since 23 = 8;

(iii) 4?81 = (81)1/4 = 3; since 34 = 81

Remark: If a is a negative real number and n is an even positive integer, then 1/an is not defined.

Thus, (– 4)1/2 is not defined

Similarly, (– 9)1/4, (– 64)1/6 are not defined

For any rational number p/q, we define:

p/aq = (1/aq)p = (ap)1/q = q?ap

Example:

(i) 3/42 = (43)1/2 = (64)1/2 = 8

(ii) (– 8) 2/3 = [ (– 8)2 ]1/3 = (64)1/3 = 4

(iii) (– 4)3/2 = [ (– 4)3 ]1/2 = (– 64)1/2, which is not defined


Examples on Rational and Fractional

Example1:- Simplify: 5n + 3 6 × 5n + 1/ 9 × 5n 22 × 5n

Solution. We have

5n + 3 6 × 5n + 1/ 9 × 5n 22 × 5n = 52 × 5n + 1 6 × 5n + 1/ 9 × 5 4 × 5n = 25 × 5(n + 1) 6 × 5(n + 1) 6 × 5n 4 × 5n

= 5(n + 1) × (25 6)/ 5n × (9 4) = 19 × 5n + 1/ 5 × 5n = 19 × 5n + 1/ 5n + 1= 19

Example2:- Simplify

(i) (xb/ xc)a ·(xc/ xa)b · (xa/ xb)c (ii) (xa/ xb)(a2 + ab + b2) · (xb/ xc)(b2 + bc + c2) · (xc/ xa)(c2 + ca + a2)

Solution. We have

(i) (xb/ xc)a ·(xc/ xa)b · (xa/ xb)c

= (xbc)a ·(xca)b · (xab)c = xa (bc) · xb(c – a) · xc (ab)

= xa (b – c) + b (ca) + c (ab) = x0 = 1

(ii) (xa/ xb)(a2 +ab + b2) · (xb/ xc)(b2 + bc + c2) ·(xc/xa)(c2 + ca + a2)

= (xab)(a2 + ab + b2) · (xbc)(b2 + bc + c2) · (xca)(c2 + ca + a2)

= x(ab) (a2 + ab + b2) · x(bc) (b2 + bc + c2) · x(ca) (c2 + ca + a2)

= x(a3 – b3) · x(b3 – c3) · x(c3 – a3) · = x(a3 – b3) + (b3 – c3) + (c3 – a3) = x0 = 1

Solve the Problem

1. 52(n + 6) × (25) – 7 + 2n/(125)2n

2. 3 × (27)n + 1 + 9 × 3(3n – 1)/ 8 × 33n – 5 × (27)n

3. If abc = 1, prove that: 1/1 + a + b – 1 + 1/1 + b + c – 1 + 1/1 + c + a – 1 = 1

4. If a, b, c are positive real numbers, show that:

?a – 1b · ?b – 1c · ?c – 1a = 1


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