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1.Algebra Booster

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2.6 Algebra <strong>Booster</strong><br />

3. To determine the nature of the roots we should remember<br />

the following rules:<br />

(a) If the co-efficients are all positive, the equation<br />

has no positive root. Thus, the equation x 5 + 4x 3 +<br />

2x + 1 = 0 has no positive root.<br />

(b) If the co-efficients of the even powers of x are all<br />

of the same sign and the co-efficients of the odd<br />

powers of x are all of opposite sign, the equation<br />

has no negative root.<br />

For example, the equation x 7 + x 5 – 3x 4 + x 3 – 3x 2 +<br />

2x – 5 = 0 has no negative root.<br />

(c) If the equation contains only even powers of x and<br />

the co-efficients are all of the same sign, the equation<br />

has no real root. For example, the equation x 5<br />

+ 3x 4 + 2x 2 + 1 = 0 cannot have a real root.<br />

(d) If the equation contains only odd powers of x, and<br />

the co-efficients are all of the same sign, the equation<br />

has no real roots other than x = 0. For example,<br />

the equation x 9 + 4x 5 + 5x 3 + 3x = 0 has no real<br />

root other than x = 0.<br />

1.19 DESCARTES RULE OF SIGNS<br />

Consider a polynomial<br />

n n-1 n-2<br />

f( x) = a0x + a1x + a2x + ... + a n<br />

with real co-efficients and a n<br />

π 0, n Œ W.<br />

We have<br />

n n-1 n-2<br />

f(–x) = a0( - x) + a1( - x) + a2( - x) + ..... + a n<br />

By observing the sign variation, in the co-efficients of f(x)<br />

and f(–x), we can predict the following things about the nature<br />

of the roots.<br />

(i) The maximum number of positive roots of f(x) is equal<br />

to the number of sign changes in the co-efficients of<br />

f(x). Let this number be p.<br />

(ii) The maximum number of negative roots of f(x) is equal<br />

to the number of sign changes in the co-efficients of<br />

f(–x). Let this number be q.<br />

(iii) The maximum number of imaginary roots of<br />

f(x) = n – (p + q).<br />

IN-EQUATIONS<br />

2.1 CONCEPT OF SOLVING ALGEBRAIC IN-EQUATION<br />

Solving an algebraic in-equation f(x) > 0, where f(x) is a product<br />

(or quotient or both) of distinct linear factors. We proceed<br />

methodically<br />

(i) Make sure that the coefficients of x in all the linear factors<br />

are positive.<br />

(ii) Equate all factors to zero to get the critical points (the<br />

points where these factors change their signs) c 1<br />

, c 2<br />

, c 3<br />

,<br />

… where c 1<br />

< c 2<br />

< c 3<br />

< …<br />

(iii) Write + and – sign alternately on the number line starting<br />

from the extreme right.<br />

(iv) If f(x) > 0, we shall use the open interval with the critical<br />

points.<br />

(v) If f(x) ≥ 0 or £ 0, we shall use the closed interval with<br />

the critical points.<br />

(vi) If f(x) contains an even power factors, say x 2 , (x – 2) 2 ,<br />

(x – 3) 10 , (x – 4) 100 , we shall not represent them on the<br />

number line, we neglect them.<br />

(vii) If f(x) contains an irreducible quadratic factor (ax 2 + bx<br />

+ c), where a > 0 and D < 0, we discard it.<br />

2.2 EQUATION CONTAINING ABSOLUTE VALUES<br />

Definition<br />

Let for every x in R, the magnitude of x is called its modulus<br />

or absolute value. It is denoted by |x|.<br />

Ï x, x≥<br />

0<br />

Thus , f() x = || x = Ì .<br />

Ó - x, x < 0<br />

X¢<br />

y = x<br />

Properties of Modulus<br />

O<br />

Y<br />

Y¢<br />

y=<br />

x<br />

(i) Geometrically, the modulus of x means, it is the distance<br />

between x and the origin, O.<br />

(ii) ||x|| = |–x| = |x|<br />

(iii) |x| = max (–x, x)<br />

(iv) –|x| = min (–x, x)<br />

(v)<br />

x<br />

2<br />

= || x<br />

(vi) |x + y| = |x| + |y|, when x and y are of the opposite sign.<br />

fi xy £ 0<br />

(vii) |x + y| = |x| + |y|, when x and y are of the same sign.<br />

fi xy ≥ 0<br />

(viii) |xy| = |x| |y|.<br />

x | x| (ix) , y 0.<br />

y = |y|<br />

π<br />

(x) |x| = 1 fi x = ±1<br />

(xi) |x| = –2 fi x = j<br />

(xii) |x| < 1 fi –1 < x < 1<br />

(xiii) |x| > 1 fi x > 1 and x < –1<br />

X

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