Ele_mentary Algeb_ra

Chapter-8

Elementary Algebra

1. Algebraic ExpressionIn algebra, we generally come across two types of symbols namelyconstant and variable. A symbol having a fixed numerical valueis called a constant and a symbol which takes various numericalvalues is called a variable. For example, the perimeter P of arectange of sides l and b is given by P = 2(l + b). Here, 2 is aconstant and l and b are variables.A combination of constants and variables connected by the signsof fundamental operations of addition, subtraction, multiplicationand division is called algebraic expression.Various parts of an algebraic expression which are separated bythe signs of + or are called the terms of the expression. Forexample, 2x2 3xy + 5y2 is an algebraic expression consisting ofthree terms, namely 2x2, 3xy, 5y2.

2. Factorization of Algebraic ExpressionsThe process of writing an algebraic expression as the product oftwo or more algebraic expressions is called factorization andthe algebraic expressions that may be multiplied to obtain thegiven algebraic expressions are called factors of the givenexpression. For example, (x + 3) and (x 3) are the factors of(x2 9), because (x + 3) (x 3) = x2 9.Similarly, x, x + 1, x(x +1), (x +1)2 and x(x +1)2 are factors of x2 + 2x2+ x, because x3 + 2x2 + x = x(x +1)2.

3. Algebraic Equations and Algebraic IndentitiesAn equation is a statement that two algebraic expressions areequal. If an equation is satisfied by any value of the variable,then the equation is said to be an identity. These identities areused as expansion formulae and ought to be carefully noted andcommitted to memory.

(1) 2 2 2 2( ) 2 ( ) 4a b a ab b a b ab

(2) 2 2 2 2( ) 2 ( ) 4a b a ab b a b ab

(3) 2 2 2 2( ) ( ) 2( )a b a b a b (4) 2 2( ) ( ) 4a b a b ab

(5) 2 2 2 2( ) 2( )a b c a b c ab bc ca

(6) 2 2 2 2 2( )a b c d a b c d 2 ( ) 2 ( ) 2a b c d b c d cd

70 Magical Book on Arithmetical Formulae

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(7) 2 2( )( )a b a b a b

(8) 2( )( ) ( )x a x b x a b x ab

(9) 3 2( )( )( ) ( )x a x b x c x a b c x ( )ab bc ca x abc

(10) 3 3 3( ) 3 ( )a b a ab a b b

(11) 3 3 3( ) 3 ( )a b a ab a b b

(12) 3 3 3( ) 3 ( )a b a b ab a b 2 2( )( )a b a ab b

(13) 3 3 3( ) 3 ( )a b a b ab a b 2 2( )( )a b a ab b

(14) 3 3 3 3 ( )a b c abc a b c 2 2 2( )a b c ab ac bc

If a + b + c = 0, then 3 3 3 3a b c abc

(15) 1 2 3 2( )( ...n n n n na b a b a a b a b 1)nb for all n.

(16) 1 2 3 2( )( ...n n n n na b a b a a b a b 1)nb if n is even.

(17) 1 2 3 2( )( ...n n n n na b a b a a b a b 1)nb if n is odd.

(18) 4 2 2 4 2 2 2 2( )( )a a b b a ab b a ab b

If we put a = x and b = 1x , in the above mentioned formulae, we

will have the following algebraic identities which are morefrequently used:

(1) 2

22

1 12x x

x x

(2) 2

22

1 12x x

x x

(3) 2 21 1

4x xx x

(4) 2 21 1

4x xx x

(5) 2 21 1 1x x xx x x

(6) 3

33

1 1 13x x x

x x x

(7) 3

33

1 1 13x x x

x x x

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(8) 3 23 21 1 1 1x x xx x x

(9) 3 23 21 1 1 1x x xx x x

4. Remainder TheoremIf an expression f(x) is divided by (x a), then the remainder isf(a).Note:(i) If an expression f(x) is divided by (x + a), then the remainder

is the value of f(x) at x = (a), ie f(a).(ii) If an expression f(x) is divided by (ax b), then the remainder

is the value of f(x) at x = ba , ie

bfa

.

(iii) If an expression f(x) is divided by (ax + b), then the remainder

is the value of f(x) at x = ba , ie

bfa

.

(iv) If an expression f(x) is divided by (b ax), then the remainder

is equal to the value of f(x) at x = ba , ie

bfa

.

5. Factor TheoremIf f(x) is completely divisible by (x a), then f(a) = 0. Thus (x a) isa factor of f(x) f(a) = 0.Note:(i) (x + a) is a factor of an expression f(x), if f(a) = 0.

(ii) (ax b) is a factor of an expression f(x), if bfa

= 0.

(iii) (ax + b) is a factor of an expression f(x), if bfa

= 0.

(iv) (x a) (x b) is a factor of an expression f(x), if f(a) = 0 andf(b) = 0.

6. Important Results Regarding Remainder and FactorTheorem(i) (xn an) is divisible by (x a) for all values of n.(ii) (xn + an) is divisible by (x + a) only when n is odd.(iii) (xn an) is divisible by (x + a) only for even values of n.(iv) (xn + an) is never divisible by (x a).


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7. PolynomialsAn expression of the form p(x) = a0 + a1x + a2x

2 + ... + anxn, where a0,

a1, a2, ..., an are real numbers and n is non-negative integer iscalled a polynomial of degree n, where an 0.For example, (7x3 5x2 + 2x + 1) is polynomial of degree 3. But

(6x2 + 3x + x + 2) is not a polynomial, since in a polynomial,every power of x must be a non-negative integer.The above expression (7x3 5x2 + 2x + 1) is a polynomial in onevariable x. Consider another example.

The expression 2 2 253 2 4 8

3x x y y xy is a polynomial in

two variables x and y.

(a) Degree of a Polynomial in One VariableIn a polynomial in one variable, the highest power of the

variable is called its degree. For example, 4 2 5 13 7

2 3xx x

is a polynomial in x of degree 4.

(b) Degree of a Polynomial in Two VariablesIn a polynomial in more than one variable, the sum of thepowers of the variables in each term so obtained is called thedegree of the polynomial. For example (3x4 2x3y2 + 7xy3 9x+ 5y + 4) is a polynomial in x and y of degree 5.

(c) Type of Polynomial According to its Degree

Name ofPolynomial Degree Example

1. Linear Polynomial 1 (2x + 1), (5y + 4)

2. Quadratic Polynomial 2 (2x

2 3x + 4), (2 x + x2)

3. Cubic Polynomial 3 (x

3 7x2 + 2x 3)

4. Biquadratic Polynomial 4 (3x

4 7x3 + x2 x + 9)

Note:A polynomial consists of a constant term only is called aconstant polynomial. The degree of constant polynomial iszero.


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8. MonomialAn algebraic expression containing only one term is called a

monomial. For example, 5, 3x, 7xy, 2 223

x y , 2 253

a b etc are all

monomials.

9. BinomialAn algebraic expression containing two terms is called binomial.For example (2x 3), (3x + 2y), (xyz 5) etc are binomials. Notethat (3x + 7x) is not a binomial, because (3x + 7x =) 10x, which isa monomial.

10. TrinomialAn algebraic expression containing three terms is called atrinomial. For example, (a b + 2), (x2 y2 + xy), (x3 2y3 3x2y2z)etc are trinomials.

11. FactorsEach term in algebraic expression is a product of one or morenumber(s) and/or literal(s). These number(s) and/or literals(s)are known as the factors of that term.A constant factor is called numerical factor, while a variablefactor is known as a literal factor.

12. Co-efficientIn a term of an algebraic expression any of the factors with thesign of the term is called the co-efficient of the product of theother factors. Consider the following examples:(a) In 5xy, the co-efficient of x is 5y; the co-efficient of y is 5x

and co-efficient of xy is 5.(b) In x, the co-efficient of x is 1.(c) In 3a2bc, the co-efficient of a2 is 3bc, the co-efficient of b is

3a2c and the co-efficient of c is 3a2b.

13. Constant TermA term of the expression having no literal factors is called aconstant term. For example, in the algebraic expressionx2 xy + yz 4, the constant term is 4.

14. Like and Unlike TermsThe terms having the same literal factors are called like or similarterms, otherwise, they are called unlike terms. For example, inthe algebraic expression 2a2b + 3ab2 7ab 4ba2, we have 2a2band 4ba2 are like terms, whereas 3ab2 and 7ab are unlike terms.


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15. HCF of PolynomialsA polynomial h(x) is called the HCF or GCD of two or more givenpolynomials, if h(x) is a polynomial of highest degree dividingeach one of the given polynomials. The coefficient of highestdegree term in HCF is always taken as positive.We can have a rule to find the HCF of two or more givenpolynomials. See the stepwise procedure given below:Step I: Express each polynomial as a product of powers of

irreducible factors. (This requires that the numericalfactor be also expressed as product of powers of primes.)

Step II: If there is no common factor, the HCF is 1. If there arecommon irreducible factors, find the smallest (least)exponents of these irreducible factors in the factorizedform of the given polynomials.

Step III: Raise the common irreducible factors to the smallestexponents found in Step II and multiply to get theHCF.

Ex. 1: Find the HCF of the polynomials 230(x 3x + 2) and250(x 2x +1) .

Soln: Let 2( ) 30( 3 2)f x x x , 2g x = 50 x 2x +1 .

Step I: Writing f(x) and g(x) as a product of powers ofirreducible factors,

2 3 5 1 2f x x x , 222 5 1g x x .

Step II: Common irreducible factors and the smallest(least) exponents of these in f(x) and g(x) takentogether are:Common Irreducible Least

Factor Exponent2 15 1

x 1 1Step III: HCF = 21 51 (x 1)1 = 10(x 1)

With practice, the reader will be able to writethe HCF directly after factorizing thepolynomials into irreducible factors.

16. LCM of PolynomialsA polynomial p(x) is called LCM of two or more given polynomials,if it is a polynomial of smallest degree which is divided by eachone of the given polynomials.We can also have a rule to find the LCM of two or more givenpolynomials.


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We may use the following three-step procedure to calculate theLCM of two (or more) polynomials:Step I: Express each polynomial as a product of powers of

irreducible factors.Step II: List all the irreducible factors (once only) occurring in

the given polynomials. For each of these factors, findthe greatest exponent in the factorized form of the givenpolynomials.

Step III: Raise each irreducible factor to the greatest exponentfound in Step II, and multiply to get the LCM.

Ex. 2: Find the LCM of the polynomials

2 2f x = 4 x 1 x + 6x + 8 and 2g x = 10 x 1 x + 2 x +7x +10 .

Soln: Step I: Writing the polynomials as a product of powersof irreducible factors, we get

222 1 2 4f x x x x ,

2 5 1 2 5 2g x x x x x

or, 22 5 1 2 5g x x x x Step II:

Irreducible GreatestFactor Exponent

2 2 5 1

x 1 2x + 2 2x + 4 1x + 5 1

Step III:

LCM = 2 22 12 5 1 2x x 1 14 5x x

= 2 220 1 2 4 5x x x x

17. Relation Between HCF and LCM of Two PolynomialsIf the HCF and LCM of the two polynomials f(x) and g(x) are h(x)and l(x) respectively, then h(x) l(x) = f(x) g(x)

18. Linear EquationsAn equation involving linear polynomials (ie a polynomial ofdegree one) is called a linear equation. For example,3 4 2 32

x x is a linear equation.


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There are two types of linear equations.(i) Linear Equations in One Variable.(ii) Linear Equations in Two Variables.

19. Linear Equations in One VariableA linear equation in one variable is an equation of the form ax +b = 0 or ax = c; where a, b and c are real numbers, a 0 and x isa variable. For example,

2x + 3 = 0 and 3 4 2 32

x x are linear equations in one

variable.

20. Solution of a Linear Equation in One VariableA value of the variable which satisfies the given linear equationis known as its solution. A solution of an equation is also knownas root.

If ax + b = 0 is a linear equation, then bxa

is its root.

21. Linear Equations in Two VariablesAn equation of the form ax + by + c = 0 or ax + by = c, where a, b,c are real numbers, where 0a , 0b and x, y are variables, iscalled a linear equation in two variables.For example, x + 2y = 3 and x + 3y = 4 are linear equations intwo variables.

22. Solution of the Linear Equations in Two VariablesLet ax + by + c = 0, where a, b, c are real numbers such that 0aand 0b . Then, any pair of values of x and y which satisfiesthe equation ax + by + c = 0, is called a solution of it.For example, x = 3, y = 2 is a solution of 3x 2y = 5 because,when x = 3, y = 2, we have LHS = 3 3 2 2 = 5 = RHS. But,x = 3, y = 2 is not its solution, because, 3 3 2 (2) 5ie, LHS RHS; when x = 3 and y = 2.

23 Infinitely Many Solutions of a Linear Equations in TwoVariablesLet us take an example.Show that (x = 1, y = 1) and (x = 2, y = 5) is a solution of 4x y 3 = 0. To examine the above solution, we put x = 1 and y = 1 inthe given equation. Now, we have,LHS = 4 1 1 3 = 0 = RHSTherefore, x = 1, y = 1 is a solution of 4x y 3 = 0.Again, if we put x = 2, y = 5 in the equation 4x y 3 = 0, wehave,


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LHS = 4 2 5 3 = 0 = RHSHence, x = 2, y = 5 is a solution of 4x y 3 = 0.In the above illustration, we have seen that the linear equation4x y 3 = 0 has two solutions. Question now arises. Does ithave more solutions? In fact, we can obtain as many solutionsas we wish in the following way:Substitute a value of your choice for x (say x = 0), in 4x y 3 =0. The equation reduces to 4 0 y 3 = 0 y = 3 y = 3.Therefore, (0, 3) is another solution of 4x y 3 = 0Similarly, substitute x = 1, we get,4 1 y 3 = 0 7 y = 0 y = 7Continuing in this manner, we can obtain any number of solutionsof 4x y 3 = 0. Thus, a linear equation in two variables hasinfinitely many solutions.

24. Simultaneous Linear EquationsA pair of linear equations in two variables is said to form asystem of simultaneous linear equations.A pair of values of x and y satisfying each one of the equationsin a given system of two simultaneous equations in x and y iscalled a solution of the system.

25. Consistency of Linear EquationsA system consisting of two simultaneous linear equations intwo variables is said to be consistent, if it has at least onesolution.A system consisting of two simultaneous linear equations intwo variables is said to be inconsistent, if it has no solution.For example, consider the system of equations: x + y = 2 and2x + 2y = 5. Clearly, there are no values of x and y which maysimultaneously satisfy the given equations. Hence, the systermgiven above is inconsistent.

26. Conditions For Consistency (Solvability) of System ofSimultaneous Linear EquationsThe system of equations,(i) a1x + b1y + c1 = 0(ii) a2x + b2y + c2 = 0

(a) is consistent with unique solution, if 1 12 2

a ba b

, ie lines

represented by equations (i) and (ii) are not parallel.

(b) is consistent with infinitely many solutions, if 1 1 12 2 2

a b ca b c

,

ie lines represented by equations (i) and (ii) are coincident.


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(c) is inconsistent, if 1 1 12 2 2

a b ca b c

, ie lines represented by equa-

tions (i) and (ii) are parallel and non-coincident.

27. Method of Solving Simultaneous Linear Equations(i) Method of Elimination by Substitution

In this method, we express one of the variables in terms ofthe other variable from either of the two equations and thenthis expression is put in the other equation to obtain anequation in one variable as explained in the followingprocedure:Step I: Obtain the two equations. Let the equations be

1 1 1 0a x b y c ... (i) and2 2 2 0a x b y c ... (ii)

Step II: Choose either of the two equations, say (i), andfind the value of one variable, say y, in terms ofthe other, ie x.

Step III:Substitute the value of y, obtained in step II, inthe other equation ie (ii) to get an equation in x.

Step IV: Solve the equation obtained in step III to get thevalue of x.

Step V: Substitute the value of x obtained in step IV inthe expression for y in terms of x obtained instep II to get the value of y.

Step VI: The values of x and y obtained in steps IV and Vrespectively constitute the solution of the givensystem of two linear equations.

Ex. 3: Solve the following system of equations by usingthe method of substitution:

(a) 3x 5y = 1 x y = 1

(b) x + 2y = 1 2x 3y = 12

Soln:(a) The given system of equations is3x 5y = 1 .... (i)x y = 1 .... (ii)From (ii), we get: y = x + 1.Substituting y = x + 1 in (i), we get3x 5 (x + 1) = 1or, 2x 5 = 1or, 2x = 4or, x = 2Putting x = 2 in y = x + 1,we get, y = 1.Hence, the solution is x = 2, y = 1.


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(b) The given system of equations isx + 2y = 1 ..... (i)2x 3y = 12 .... (ii)From (i), we get x = 1 2ySubstituting x = 1 2y in (ii), we get2 (1 2y) 3y = 12or, 2 4y 3y = 12or, 7y = 14or, y = 2Putting y = 2 in x = 1 2y,we get, x = 1 2 (2) = 3Hence, the solution is x = 3, y = 2.

(ii) Method of Elimination by Equating the CoefficientIn this method we eliminate one of the two variables toobtain an equation in one variable which can easily be solved.Putting the value of this variable in any one of the givenequations, the value of the other variable can be obtained.See the following procedure that will illustrate our points:Step I: Obtain the two equations.Step II: Multiply the equations so as to make the

coefficients of the variable to be eliminated equal.Step III: Add or subtract the equations obtained in step II

according as the terms having the same coefficientsare of opposite or of the same sign.

Step IV: Solve the equation in one variable obtained in stepIII.

Step V: Substitute the value found in step IV in any one ofthe given equations and find the value of the othervariable.The values of the variables in step IV and Vconstitute the solutions of the given system ofequations.

Ex. 4: Solve the following system of linear equationsby using the method of elimination by equatingthe coefficients: (a) 3x + 2y = 11 2x + 3y = 4(b) 8x + 5y = 9 3x + 2y = 4

Soln: (a)The given system of equations is3x + 2y = 11 .... (i)2x + 3y = 4 .... (ii)Let us eliminate y from the given equations. Thecoefficients of y in the given equations are 2 and 3respectively. The LCM of 2 and 3 is 6. So, we makethe coefficient of y equal to 6 in the two equations.


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Multiplying (i) by 3 and (ii) by 2, we get9x + 6y = 33 .... (iii)4x + 6y = 8 .... (iv)Subtracting (iv) from (iii), we get 5x = 25 or, x = 5Substituting x = 5 in (i), we get 15 + 2y = 11 or, 2y = 4 or, y = 2Hence, the solution is x = 5, y = 2

(b) The given system of equations is8x + 5y = 9 .... (i)3x + 2y = 4 .... (ii)Let us eliminate x from the given equations. Thecoefficients of x in the given equations are 8 and 3respectively. The LCM of 8 and 3 is 24. So, we makeboth the coefficients equal to 24.Multiplying (i) by 3 and (ii) by 8, we get24x + 15y = 27 .... (iii)24x + 16y = 32 .... (iv)Subtracting (iv) from (iii), we gety = 5 or, y = 5Putting y = 5 in (i), we get8x + 25 = 9 or, 8x = 16 or, x = 2.Hence, the solution is x = 2, y = 5.

(iii) Method of Cross-MultiplicationLet 1 1 1 0a x b y c and 2 2 2 0a x b y c be a system ofsimultaneous linear equations in two variables x and y

such that 1 1

2 2

a ba b

ie 1 2 2 1 0.a b a b Then the system has

a unique solution given by

x = 1 2 2 1

1 2 2 1

( )( )b c b ca b a b

and y =

1 2 2 1

1 2 2 1

( )( )c a c aa b a b

Note:(a) The above solution is generally written as

1 2 2 1 1 2 2 1 1 2 2 1

1x yb c b c c a c a a b a b

or, 1 2 2 1 1 2 2 1 1 2 2 1

1x yb c b c a c a c a b a b

(b) The following procedure is very helpful in determining

the solution without remembering the above formula:Step I: Obtain the two equations.Step II: Shift all terms on LHS in the two equations to

introduce zeros on RHS ie write the two equationsin the following form:


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1 1 1 0a x b y c

2 2 2 0a x b y c Step III: In the above system of equations there are three

columns viz column containing x, ie 1

2

aa

, column

containing y, ie 1

2

bb

and column containing

constant terms, ie 1

2

cc

. To obtain the solution,

write x, y and 1 separated by equality signs asshown below:

In the denominator of xy, leave column containingx and write remaining two columns in the sameorder. In the denominator of y, leave columncontaining y and write the remaining two columns.Similarly, in the denominator of one write columnscontaining x and y. Mark crossed-arrows pointingdownward from top to bottom and pointing upwardfrom bottom to top as shown above.The arrows between two numbers indicate thatthe numbers are to be multiplied.

Step IV: To obtain the denominators of x, y and 1, multiplythe numbers with downward arrow and from theirproduct subtract the product of the numbers withupward arrow.Applying this, we get

1 2 2 1 1 2 2 1 1 2 2 1

1x yb c b c a c a c a b a b

Step V: Obtain the value of x by equating first and third

expressions in step IV. The value of y is obtainedby equating second and third expressions instep IV.

Ex. 5: Solve each of the following system of equationsby using the method of cross-multiplication.(a) x + y = 7 (b) 2x + 3y = 17

5x + 12y = 7 3x 2y = 6Soln: (a)The given system of equations is

x + y 7 = 05x + 12y 7 = 0By cross-multiplication, we get


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or, 1 7 12 7 1 7 5 7x y

11 12 5 1

or, 1

7 84 7 35 12 5x y

or, 1

77 28 7x y

or, x = 777 and y =

287

or, x = 11 and y = 4Hence, the solution is x = 11, y = 4.

(b) The given system of equations is2x + 3y 17 = 03x - 2y 6 = 0By cross-multiplication, we have

or 3 6 ( 2) 17 2 6 3 17x y

=

12 2 3 3

or, 1

18 34 12 51 4 9x y

or, 1

52 39 13x y

or, x = 5213

and y =

3913

or, x = 4 and y = 3.Hence, x = 4, y = 3 is the solution.

28. Quadratic EquationsLet p(x) be a quadratic polynomial. Then, the equation p(x) = 0 iscalled quadratic equation. The values of x satisfying p(x) = 0 arecalled its roots or zeros. For example, 25x2 30x + 9 = 0 is a

quadratic equation. And the value of 35

x is the solution of

the given equation. Since, if we put 35

x in 25x2 30x + 9 = 0,


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we have LHS = 23 3

25 30 95 5

= 9 18 + 9 = 0 = RHS.

The general form of a quardratic equation is ax2 + bx + c = 0;where a, b and c are real numbers and 0a .

29. Roots of a Quadratic Equation(i) If and are the two roots of ax2 + bx + c = 0, then

2 42

b b aca

and

2 42

b b aca

(ii) Sum of the roots ( )ba

(iii)Product of the roots ( )ca

(iv) A quardratic equation whose roots are and is given by2 ( ) 0x x , ie x2 (sum of roots)x + product

of roots = 0(v) In ax2 + bx + c = 0 the expression D = b2 4ac is called its

discriminant.

30. Nature of Roots of ax2 + bx + c = 0Let D = b2 4ac be the discriminant of the given equation. Thenroots of the equation ax2 + bx + c = 0 are(i) real and equal, if D = 0.(ii) real, unequal and rational, when D > 0 and D is a perfect

square.(iii) real, unequal and irrational, when D > 0 and D is not a

perfect square.(iv) imaginary, if D < 0.(v) integers, when a = 1, b and c are integers and the roots are

rational.

31. Methods of Solving Quadratic Equations(i) By Factorization

This can be understood by the examples given below:Ex. 6: Find the solutions of the quadratic equation

2x + 6x + 5 = 0 and check the solutions.Soln: The quadratic polynomial 2 6 5x x can be factorized as

follows:= x2 + 6x + 5 = x2 + 5x + x + 5= ( 5) 1( 5)x x x

= ( 5)( 1)x x


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Therefore the given quadratic equation becomes( 5)( 1) 0x x This gives x = 5 or, x = 1Therefore, x = 1, 5 are the required solutions of thegiven equation.Check: We substitute x = 1 and x = 5 in the givenequation and get(a) (1)2 + 6(1) + 5 = 1 6 + 5 = 0(b) (5)2 + 6(5) + 5 = 25 30 + 5 = 0Therefore, the solutions are correct.

(ii) By Using Method of Completing SquareIt is not always easy to factorise polynomials and solve qua-dratic equations as discussed above. For example, considerthe quadratic equation x2 + 5x + 5 = 0. If we want to factorisethe left-hand side of the equation using the method of split-ting the middle term, we must determine two integral factorsof 5 whose sum is 5. But, the only factors of 5 are 1 and 5 or1 and 5. In both the cases, the sum is not 5. Therefore,using factorisation, we are unable to solve the quadratic equa-tion x2 + 5x + 5 = 0. Here, we shall discuss a method to solvesuch quadratic equations. Let us consider the following ex-ample:Ex. 7:Solve: x2 + 3x + 1 = 0.Soln: We have

x2 + 3x + 1 = 0

We add and subtract (12 coefficient of x)

2 in LHS and get2 2

2 3 33 1 02 2

x x

2 2

2 3 3 32 1 02 2 2

x x

23 5

02 4

x

223 52 2

x

or

3 52 2

x

This gives 3 5

2x

or

3 52

x

Therefore, 3 5 3 5,

2 2x are the solutions of the

given equation.


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(iii) By Using Quadratic Formula If the equation is ax2 + bx + c = 0, then

2 42

b b acxa

or,2 4

2b b acx

a

and 2 4

2b b ac

a

is often referred

to as quadratic formula.

(a) When b2 4ac = 0, ie b2 = 4ac, then 2ba

and

2ba

where and are the two roots of the above

equation ie both the roots are equal.(b) When b2 4ac > 0 ie b2 > 4ac, then the equation has

two distinct real roots , given by2 4

2b b ac

a

and 2 4

2b b ac

a

Ex. 8: Solve the following equation:2x2 + 5x 6 = 0.

Soln: Here, the given equation :2x2 + 5x 6 = 0

ie a = 2, b = 5, c = 6

x = 2 4

2b b ac

a

= 25 (5) 4 2 ( 6)2 2

=

5 25 484

= 5 73

4

= 5 73

4

, 5 73

4

32. Finding Roots of a Quadratic Equation Having RationalRootsSuppose we have to find the roots of 210 21 0x x .Note the following steps: 210 21x x

Step I: 21 10 = 210 (Multiply the coefficient of x2 and the constant term)

Step II: 14 15 (Find the factors of 210 which give coefficient of x ie (1) in any possible way: 14 (15) = 210 and 14 + (15) = 1))


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Step III: 14 710 5

15 3

10 2

(Divide the factors obtained in

step II by coefficient of x2)

Step IV:75

32 (Change the sign of values

obtained in step III)

Hence the roots are 75

and 35

Now see the examples given below:Ex. 9: Find he roots of 12x2 + 25 x 117 = 0Soln:

52 1312 3

27 9

12 4

133

94

Therefore, the roots are 133

and

9.

4

Ex. 10: Find the roots of 35x2 + x 12 = 0Soln: 35x2 + x 12 = 0

35 12

21 20

21 335 5

20 435 7

35

47


and 47


K KUNDAN

Ex. 11: Find the roots of 91x2 + 20x + 1 = 0.Soln: 91x2 + 20x + 1 = 0

91

7 13

7 191 13

13 191 7

1

13 1

7


13

and 1

.7

Ex. 12: I. 6x2 x 35 = 0 II. 6y2 + 41y + 63 = 0

Compare the roots of the quadratic equations in I andII. Which of the following is true?1) x > y2) x < y3) x > y4) x < y5) x = y

Soln: 3;I. 26 35 0x x II. 26 41 63 0y y

35 6 6 63

15 14 27 14

156

146

27 96 2

14 76 3

52

73

92

73

Comparing the pair of values obtained from quadraticequations (I) and (II), we get x > y.

33. Rational Expressions

An expression of the form ( )( )

p xq x , where p(x) and q(x) are

polynomials and q(x) 0, is called a rational expression. Everypolynomial is a rational expression but a rational expressionneed not be a polynomial.


K KUNDAN

34. Rational Expression in Lowest TermsIf p(x) and q(x) are polynomials such that HCF of p(x) and q(x) is

1 then )()(

xqxp

is a rational expression in its lowest terms. This is

also called simplification of algebraic fractions.

35. Results on Rational Expressions(i) Sum or difference of two rational expressions is a rational

expression.(ii) Product of two rational expressions is a rational expression.(iii) Addition as well as multiplication on rational expressions

satisfies the cumulative and associative laws.(iv) 0 is the additive identity.

(v) Additive inverse of )()(

xqxp

is )()(

xqxp

.

36. Condition for Common Roots

Let 21 1 1 0a x b x c and 22 2 2 0a x b x c be two quadratic equa-tions such that a1, a1 0 and a1b2 a2b1.Let be the common root of these two equations.

Then, 21 1 1 0a b c

and 22 2 2 0a b c Solving these two equations by cross-multiplication, we get

2

1 2 2 1 1 2 2 1 1 2 2 1

1b c b c c a c a a b a b

2 1 2 2 1

1 2 2 1

b c b ca b a b

and 1 2 2 1

1 2 2 1

c a c aa b a b

Eliminating , we get2

1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

b c b c c a c aa b a b a b a b

21 2 2 1 1 2 2 1 1 2 2 1b c b c a b a b c a c a The above is the required condition for the two quadratic equa-tions to have a common root.The common root is given by

1 2 2 1

1 2 2 1

c a c aa b a b

or 1 2 2 1

1 2 2 1

b c b cc a c a


K KUNDAN

Note: (i) To find the common root of two equations, make thecoefficient of second degree terms in two equationsequal and subtract. The value of x so obtained is therequired common root.

(ii) If the two equations have both roots common, then

1 1 1

2 2 2

a b ca b c

Ex. 13: Find the value of K, so that the equations 2x x 12 0 and 2Kx 10x 3 0 may have one root

common. Also find the common root.Soln: Let be the common root of the two equations.

Hence, 2 12 = 0 andK2 + 10+ 3 = 0Solving the two equations,

2 1117 12 3 10K K

212 3 117 10K K

29 4 1 117 10K K 216 8 1 130 13K K K 216 5 129 0K K

16K2 48K + 43K 129 = 0 16K (K 3) + 43(K 3) = 0 (16K + 43) (K 3) = 0

K = 4316

or 3

= 12 310

KK

= 3 or 4

37. Maximum or Minimum Value of a Quadratic ExpressionAs we have already seen, equation of the type ax2 + bx + c = 0(where, a 0) is called a quadratic equation. An expression ofthe type ax2 + bx + c is called a quadratic expression.The quadratic expression ax2 + bx + c takes different values as xtakes different values.As x varies from to +, the quadratic expression ax2 + bx + c(i) has a minimum value whenever a > 0. The minimum value of

the quadratic expression is 24

4ac b

a and it occurs at

2bxa

.


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(ii) has a miximum value whenever a < 0. The miximum value of

the quadratic expression is 24

4ac b

a and it occurs at

2bxa

.

Ex. 14: Find the maximum or minimum value of5x2 + 20x + 40.

Soln. A quadratic expression of the form ax2 + bx + c, will havea minimum value when a > 0 and maximum value whena < 0. Its maximum or minimum value is given by

244

ac ba

and it occurs at 2bxa

.

Given, a = 5, b = 20 and c = 40Since, a < 0, the expression has a maximum value.

the maximum value = 24( 5)(40) 20

4( 5)

= 60

Ex. 15: In the previous example, find the value ofx for which the maximum value occurs.

Soln. The maximum vlaue of the expression occurs at20 2

2 2( 5)bxa

Ele_mentary Algeb_ra

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