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Quadratic Equations and Expressions 2.7<br />

Inequalities with the Absolute Value<br />

Basic rules for the absolute value of the inequality.<br />

1. |x| = 2 fi x = ±2<br />

2. |x| = –3 fi x = f<br />

3. |x| < 4 fi –4 < x < 4<br />

4. |x| £ 5 fi –5 £ x £ 5<br />

5. | x| 2 x 2, x –2 x (– , –2) (2, )<br />

6. | x| 3 x 3, x –3 x ( , 3] [3, )<br />

2.3 IRRATIONAL EQUATIONS<br />

Introduction<br />

In an equation, if the unknown quantities are under radical<br />

sign, it is known as an irrational equation.<br />

For example, x + x - 2 = 1, x - 3 = -2,<br />

x - x + 1=<br />

2,<br />

etc., are irrational equations.<br />

The concept of irrational equation comes from irrational<br />

expression. This expression contains at least one fractional<br />

power of the unknown quantity.<br />

For example,<br />

1. f( x)<br />

= x is an irrational expression.<br />

3<br />

2. f( x)<br />

= x is an irrational expression.<br />

= x is an irrational expression.<br />

Thus, we get the even root as well as odd root of the unknown<br />

quantity of an irrational expression.<br />

If roots of an irrational expression are even, i.e.<br />

4 6 10 100<br />

x, x, x, x, x , ..., it is defined for non-negative real<br />

values of the radicand.<br />

4<br />

If the radicand is negative ( – x, - x, ...) , the roots are<br />

imaginary.<br />

If all the roots are odd, that is, 3 x, 5 x, 9 x, 2013 x...,<br />

it is<br />

defined for all real values of the radicand.<br />

If the odd radicand is negative, its roots are also negative.<br />

2013<br />

3. f( x)<br />

2.4 IRRATIONAL IN-EQUATIONS<br />

Type I: An in-equation is of the form<br />

2n+ 1 2n+<br />

1<br />

f() x < g(),<br />

x nŒN<br />

fi f(x) < g(x)<br />

Type II: An in-equation is of the form<br />

fi<br />

f() x < g(),<br />

x nŒN<br />

2n<br />

2n<br />

Ï f() x ≥ 0<br />

Ô<br />

Ì gx () > 0<br />

Ô<br />

Ógx<br />

() > f()<br />

x<br />

Type III: An in-equation is of the form<br />

fi<br />

2n+<br />

1<br />

f() x < g(), x nŒN.<br />

f(x) < g 2n+1 (x)<br />

Type IV: An in-equation is of the form<br />

2 n f() x < g(),<br />

x nŒN<br />

Type V: An in-equation is of the form<br />

fi<br />

fi<br />

2n+<br />

1<br />

f() x > g(), x nŒN.<br />

f(x) > g 2n+1 (x)<br />

Ï f() x ≥ 0<br />

Ô<br />

Ì gx () > 0<br />

Ô<br />

2n<br />

Ó f() x < g () x<br />

Type VI: An in-equation is of the form 2 n<br />

f() x > g(),<br />

x n ΠN.<br />

ÏÔ gx () ≥ 0 Ïgx<br />

() < 0<br />

fi Ì<br />

and<br />

2n<br />

Ì<br />

ÔÓ f() x > g () x Ó f() x ≥ 0<br />

2.5 EXPONENTIAL EQUATIONS<br />

Type I: An equation is of the form<br />

a f(x) = 1, a > 0, a π 1<br />

fi f(x) = 0<br />

Type II: An equation is of the form f(ax) = 0<br />

fi f(t) = 0, where t = a x .<br />

If t 1<br />

, t 2<br />

, t 3<br />

, … t n<br />

are the roots of f(t) = 0, then ax = t 1<br />

, …,<br />

a x<br />

= t n<br />

.<br />

Type III: An equation is of the form<br />

f () x f () x f () x<br />

a◊ a + b◊ b + g ◊ c = 0 ,<br />

where a, b, g Œ R and a, b, g π 0 and the bases satisfy the<br />

condition b 2 = ac.<br />

2<br />

Êaˆ<br />

fi a◊ t + b◊ t + g = 0 , where t = Á<br />

Ë<br />

˜<br />

b¯<br />

Type IV: An equation is of the form<br />

f () x f () x<br />

f () x<br />

a◊ a + b◊ b + c = 0,<br />

where a, b, c Œ R, a, b, c π 0<br />

and ab = 1 (where a and b are inverse positive numbers).<br />

2<br />

fi a◊ t + c◊ t + b = 0 , where t = a f(x)<br />

Type V: An equation is of the form<br />

a f(x) + b f(x) = c,<br />

where a, b, c ΠR<br />

and a 2 + b 2 = c.<br />

The solution of the given equation is f(x) = 2.<br />

Type VI: An equation is of the form<br />

a f(x) + b f(x) + c f(x) = d,<br />

where a, b, c, d ΠR and a 3 + b 3 + c 3 = d. The required solution<br />

is f(x) = 3.<br />

Type VII: An equation is of the form<br />

a f(x) + a g(x) = c,<br />

where f(x) + g(x) = 1, and<br />

a, c (π 0)<br />

We shall put a f(x) = t<br />

Type VIII: An equation is of the form<br />

a f(x) + a f(x)–1 + a f(x)–2 = b f(x) + b f(x)–1 + b f(x)–2<br />

Then the required solution is<br />

.

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