Mathematics Quadratic

QUADRATIC EQUATIONS

Target IIT JEE

1. POLYNOMIAL

The expression of the form 1

0 1 ... 1

n n

n n

a x a x  a x a



    is called polynomial where a0,a1, ...an are real number and n is a non negative integer.

(i) Quadratic Polynomial

The polynomial where n 2 and a 0 0 is called a quadratic polynomial i.e. polynomial with degree 2. such

as 2 2

0 1 2 or

a x a xa ax bxc.

(ii) Quadratic Equations

The general form of quadratic equation is 2

0,

ax bx c where a b c, , are real numbers and a 0.

For example: 2

5 10 0

x  x  is a quadratic equation.

(iii) Zeros of a Quadratic Polynomial

The value(s) of x for which the polynomial reduces to ‘0’ are called zeros of quadratic polynomial. For example:

2

4 4

x  x becomes zero if we put x  2 2

 is a zero of this polynomial.

(iv) Roots of Quadratic Equation

If  , are the zeros of the equatic polynomial, _{p x( )ax}2_{bxc, then  , are the roots of corresponding} quadratic equation i.e. 2

0

ax bx c  or p x ( ) 0.

Solving the quadratic equation means finding the values of ' 'x for which given quadratic equation is satisfied.

For example:

Consider the quadratic equation, 2 5 6 0 x  x  Let us put x 2, 3. 2 2 2   5 2 0 and 3    5 3 6 0 2, 3 x

  are roots of quadratic equation.

2. FINDING SOLUTION OF QUADRATIC EQUATION

(a) Factorisation method:

Let us consider the quadratic equation: 2

0

ax bx c 

To find roots, we factorise it and get:

2 _{( )( ) 0} ax bx c lxm nxp  0 or 0 lxm nxp or m p x x l n     

Hence, these are the roots of the given quadratic equation.

Illustration:

Find the roots of the following equation:

2 2 2 3 2 0 a x  abx b  Sol. 2 2 2 2 2 0 a x  abx abx  b 

Quadratic Equations

( 2 ) ( 2 ) 0

ax ax b b ax b  (ax2 )(b ax b )0

2 / or /

x b a xb a

(b) Method of Completion of Square

The roots of a quadratic equation are found by expressing the quadratic equation in perfect square from and then taking square roots on both sides.

Let us take the quadratic equation 2

0 ax bx c  2 0 b c a x a a          2 2 0 2 b c a x x a a         

On adding and subtracting 2 , 2 b a       we get: 2 2 2 2 0 2 2 2 b b c b a x x a a a a  _{   }                     2 2 2 4 0 2 4 b ac b a x a a _{  }     _{ }        [Using 2 2 2 (xy) x 2xyy ] 2 ₂ 2 4 2 4 b b ac x a a          2 4 2 2 b b ac x a a     _  _    2 ₄ 2 ₄ , 2 2 2 b b ac b b ac x a a a       

Depending on the sign of 2

4 ,

b  ac we make the following cases.

Case I: 2

4 0

b  ac

Then the two roots of quadratic are:

2 ₄ 2 ₄ and 2 2 b b ac b b ac a a  a         Case II: 2 4 0 b  ac

Then the equation does not have any real roots.

Illustration:

Solve the equation 2

6x   x 2 0. Sol. Expressing 6 2 2 6 2 1 6 3 x x   x _x   _   2 2 1 6 0 12 3 x x    _   _   2 2 1 1 1 6 0 12 144 144 3 x x    _     _  

2 1 (1 48) 0 12 144 x     _  _     2 1 49 1 7 12 144 2 12 x x    _  _        1 7 2 1 , 12 12 3 2 x x      

(c) Direct formula Method

From the previous method, we can derive the formula for quadratic equation : 2

0

ax bx c 

(i) If 2

4 0,

b  ac then the equation has real and distinct roots i.e.,

2 ₄ 2 ₄ , 2 2 b b ac b b ac a a         (ii) If 2 4 0,

b  ac then the equation has equal roots

2 b a    (iii) If 2 4 0,

b  ac then the equation has no real roots.

Note: 2

4

b  ac is called the Discriminant of the quadratic equation and is represented by  or D.

Illustration:

Find the roots of the equation 2 2 2

3 2 0 a x  abx b  . Sol. _B2_{4AC ( 3ab)}2_4(2b2_)(a2₎ 2 2 2 2 2 2 9 8 0 D a b b a a b D      

 The equation has two distinct roots i.e.

, 2 2 B D B D A A     2 2 2 2 2 2 3 3 , 2 2 ab a b ab a b a a    2 , b b x a a  

(d) Sum and product of roots of a Quadratic Equation

Let  and be the roots of a quadratic equation: 2

0

ax bx c  where a b c, , R anda0. Using the direct formula method,

2 ₄ 2 ₄ , 2 2 b b ac b b ac a a         Sum of roots 2 2 coeff. of 2 coeff. of b b x a a x           Product of roots =





2 2 2 2 ( ) 4 4 b b ac a      2 2 2 2 4 constant term 4 coeff. of b b ac c a a x    

Illustration:

Find the sum and product of roots of the equation 2

4x 2 3x 5 0. Sol.



2 3



3 4 2     5 4  

(e) Formation of a quadratic equation when roots are given

Let  , be the roots of equation 2 2

0 ax bx c  . & b c a a    Consider 2 0 ax bx c  2 2 0 0 b c b c x x x x a a a a         _{ }     2 _{( ) 0} x   x      

2 _{(sum of roots) product of roots 0}

x x

   

Illustration:

Find the quadratic equation whose roots are 2 and 3/2.

Sum of roots 2 3 7 and Product of roots 2 3 3

2 2 2       2 7 _{3 0 2} 2 _{7 6 0} 2 x x x x        

Note: If one root of a quadratic equation is in the form p q and a b, andc are rational numbers, then other

root must be p q since

2 ₄ 2 b b ac x a    

3. FINDING THE VALUES OF SYMMETRIC EXPRESSION

If  and are the roots of 2

0

ax bx c where a b c, , R and a 0, then the values of symmetrical expressions in  and are obtained by using the sum of roots formula and product of roots formula.

Some important formulae to convert symmetrical expressions in terms of  and 

(i) _2_2 _{()}2_2 (ii) 22 ( )( )()









2



 















24     (iii) 33 ( )



22









 





23   (iv) 3_3 _



_ 





2_2_



_{ }



_



2_4_



 _



2_       (v) 4 4



2 2



2 ₂ 2 2





2 ₂ 2 ₂ 2 2                 (vi) 4_4 _



2_2





_ 



_



_{ }



_



2_2



_







_



2_4   _ _ (vii) 5 5



2 2



3 3



2 3 3 2           





2







2 2



2 2





2                    _      _  





2



 



2 2 2 2 3 ( )                             (viii) _6_6 _



_2_2



_4_4_{ }2 2







2 ₂





2 2



2 ₂



2 ₃ 2 2                _    _  _{  } (ix) 2





2 2 1 1 ( ) a b a b a b a ab b                 or using s 2 2 0 & 0 a b c a b c & c c a b  b         1 1 ( ) a b a b a c c                 (x) 2 2 1 1 ( ) 2 ( ) b c b c b c b bc c                 or using _a2_{b c 0 & a}2_{b c 0} 2 2 b  c a b  c 



2 2



2 2 2 2 2 2 2 ( ) 2 1 1 1 a a a a                          Illustration:

Let  , be the roots of 2

0.

ax bx c  Find the values of:

(1) 1 2 2 2 2 2 2 2 3 3 3( ) 4 2 2 2( ) 5                     2 3( ) 2 ( ) 2 5               (2) 2 2 2 2     4 4 2 2 2 2 2 2 2 2 2 ( ) 2                2 2 2 2 2 2 ( ) 2 2            2 2 2 2 2 2 2 2 b c c a a a c a               





2 2 2 2 2 2 2 2 2 ₄ 2 2 2 2 4 2 2 2 2 b c c b ac a c a a a _a c c a a a             4 2 2 2 2 2 2 2 4 4 2 b a c b ac a c c a     4 2 2 2 2 2 2 4 b a c b ac c a   

4.0 FACTORISATION OF A QUADRATIC POLYNOMIAL

Let 2

( )

f x ax bxc be a quadratic polynomial. The factor of f x( ) depends on the discriminant.

Case I: D 0

Then the polynomial cannot be factorised into linear factors.

Case II: D 0

Then the corresponding equation 2

0

ax bx c  has equal roots; ₂

b a

  

 The factors are:

2 2 b b a x x a a               Case III: D 0

Then roots of the corresponding quadratic equation are

,

2 2

b D b D

a a

     

 The factors are a x( )(x ) a x b_2a D x b_2aD

       

   _  _{ }  _

   

5. COMMON ROOTS BETWEEN TWO QUADRATIC EQUATIONS

(1) Both roots common

Consider two quadratic equations _ax2_{bx c 0 anda x} 2_{b x c0 in such a case two equatins should be} identical. For that, the ratio of coefficients of _x2_{, xand x}0 _{must be same, i.e.} a b c

abc

(2) One root common

Consider two quadratic equatins 2

0

ax bx c and a x 2b x c  0. Let  be the common root of two equations. So  should satisfy both the equations.

2 0 a b c     and ... (1) 2 0 a x b x c   ... (2)

Solving the two equations by cross multiplication method 2 1 bc b c ac a c ab a b           bc b c a c ac a c ac ab a b              



a c ac



2 



bcb c



aba b



Illustration:

For what value of k, equation: _2x2_{kx 5 0 & x}2_{3x40 may have one root common.} Let  be the common root.

2 2 k 5 0 2 3 4 0     2 1 4k 15 8 5 6 k            4 15 3 and 3 6 k k          2 4 15 3 24 90 4 15 9 3 6 k k k k k            2 4k 39k 81 0     27 3 or 4 k k     

6. EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS

In the previous sections we studied quadratic equation and how to find its solution. There are other equations which are not quadratic equations but can be reduced to quadratic equations after making a substitution in the equations. These equations are known as equations reducible to quadratic equations. In this section we will explore the various categories in which we can divide these equations and also methods to reduce them to quadratic form and hence solve them.

Type - 1

Equation of the type:





2





( ) ( ) 0

a f x b f x c

The following steps are useful to reduce this equation into a quadratic equation. (1) Replace f x( )t and reduce the equation into quadratic equation: 2

0

at bt c

(2) Find roots of the quadratic equation 2

0

at bt c using methods discussed in previous section. Let roots be

and

 .

(3) Solve f x( ) and f x( ) to get all roots of the given equation.

Illustrations: 1.. Solve 2 / 3 1/ 3 2 15 x  x  . Step 1 Put 1/ 3 x t to get t22t15 Step 2 2 2 15 0 ( 5)( 3) 0 5 or 3 t  t   t t   t t  Step 3 _x1/ 3 _{5 and x}1/ 3 _{ 3}

Taking cube on both sides, we get: x125 and x 27

Type - 2

Equation of the type:



( )



( )

b a f x c

f x

 

Where f x( ) is an expression in x and a, b, c are real coefficients. The following steps are useful to reduce the above equation into quadratic equation and hence find the solution.

(1) Replace f x( ) by t to get at b c t

2

0

at ct b

   

(2) Solve the quadratic equation obtained in step (1) to find roots say  and. (3) Solve f x( ) and f x( ) to obtain roots of the given equation.

Illustration: 1. Solve 3 1 1 5 1 3 1 2 x x x x                   .

Step 1 Put 3 1 to get 1 5

1 2 x t t x t      2 2t 5t 2 0     Step 2 Solve 2 2t 5t 2 0 (t2)(2t1)0 2 or 1/ 2 t t Step 3 3 1 2 or 3 1 1 1 1 2 x x x x       3x 1 2x 2 or 6x 2 x 1        1 or 1/ 5 x x    

Theefore roots of the given equation are x1 &x 1/ 5.

Type - 3

Equation of the tpe: 2 2 1 1 0 a x b x c x x                  or 2 2 1 1 0 a x b x c x x                 

The following steps are useful to reduce the above equation into quadratic equation and hence find its solutions.

1. Write first equation in the form:

2 1 1 2 0 a x b x c x x _{ }  _{ }                 

  and second in the form:

2 1 1 2 0 a x b x c x x _{ }  _{ }                    . 2. Replace x 1 t x

  in first equation and x 1 t x

  in second equation, to get 2

( 2) 0

a t  bt c and 2

( 2) 0

a t  bt c .

3. Solve the quadratic equations x 1 andx 1

x  x 

 

   

 

  for the first equation or

1 1 and x x x  x           

for the second equation to obtain solution of the given equation.

Illustrations: 1. Solve 2 2 1 1 4 x 8 x 3 0 x x                  Let x 1 t x   _{... (1)}

Take square of (1) to get: 2 2 2

1 2

x t

x

2 2 2 1 2 x t x     _{... (2)}

Using (1) and (2) in the given equation, we get 2 4(t 2) 8 t 3 0 2 4t 8t 5 0 2 4t 10t2t 5 0 (2t5)(2t1)0 5 / 2 and 1/ 2 t t    

Putting the values of t in (1), we get:

1 1 5 / 2 and 1/ 2 x x x x      2 2 2(x 1)5x and 2(x 1)x0 2

2(x 1)5x and D0 (so no solution)

2 and 1/ 2

x x are two solutions of the given equation.

Type - 4

Equation of the type: (xa x b x c x)(  )(  )( d)k0 where a b c d k, , , , R such that ab c d.

The following steps are useful to reduce the above equation into quadratic equation and hence find its solutions. 1. Multiply first two brackets and last two brackets to get



_x2_{(ab x) ab x)(} 2_{(cd x) cd}



_k0

2. As (a b )(c d ) we can replace 2 _{( )}

x  ab xt to get (tab t)( cd)k0

3. Solve the above quadratic equation in t to get roots  , .

4. Solve _x2_{(ab x)  and x}2_{(ab x)  to get solution of the required equation.}

Illustrations:

(1) Solve (x1)(x2)(3x2)(3x1)21

The given equation is : (x1)(x2)(3x2)(3x1)21

Take 3 as common factor from last two factors to get

9(x1)(x2)(x2 / 3)(x1/ 3)21 ... (1)

The sum of constant terms of first factor and third factor is same as the sum of the constant terms of second factor and fourth factor.

 In equation (1), multiply first factor with third and second factor with fourth to get.

2 2 (3x 5x2)(3x 5x2)21 Let 2 3x 5xt (t 2)(t 2) 21     2 25 0 5 t    t 

On combining (2) and (3) we get:

2 2 3x 5x 5 0 or 3x 5x 5 0 5 25 4 5 3 or no real roots. 6 x    

5 85 The solution is : 6   Type - 5(a)

Equation of the type: ax b cxd; a b c d, , , R

The following steps are useful to reduce the above equation into quadratic equation and hence find the solution of it.

1. Take square on both sides to get: _{axb(cxb)}2 2. Solve quadratic equation to get roots  , .

3.  and are roots of the given equation if they satisfy axb0 and cxd 0.

Type - 5(b)

Equation of tye type: 2

ax bx c dxe

The following steps are useful to reduce the above equation into quadratic equation and hence find the solution of it.

1. Take square on both sides to get _ax2_{bx c (dxe)}2_. 2. Solve the above quadratic equation in x to get roots  & . 3.  and are roots of the given equation if they satisfy 2

0

ax bx c  and dx e 0.

Type - 5(c)

Equation of the type: _{ax b  cxd e}.

The following steps are useful to reduce the above equation into quadratic equation and hence find the solution of it.

1. The given equation can be written as _{ax b  e cxd} 2. Take square on both side to get: 2

( ) 2

ax b e  cxd  e cxd 2

(a c x b d) e 2e cx d

       

3. Take square again to get: 2 2 2

(a c x) b d e 4e cx( d)

       

 

Solve this quadratic equation in x to get roots  & .

4.  and  are roots of the given equation if they satisfy ax b 0 and cxd0. Note: 2

x  x

Illustrations:

1. Solve 2

2x 2x 1 2x 3 0

The equation can be written as 2

2x 2x 1 2x3

Take square to get 2 2

2x 2x 1 4x  9 12x 2 2 2x 10x 8 0 x 5x 4 0         (x 1)(x 4) 0 x 1 or x 4        For x 1, Solving: _{LHS  2 1 2 1 2 4 3     }  1 2 3  0

Hence x 1 is not a solution.

For x 4,

Solgin: 2

2 4 2 4 1

LHS       32 8 1 8 3     25 5 0RHS

SUMMARY

1. Quadratic Equation

The standard form of a quadratic equation is: 2

0

ax bx c 

where a b c, , are real numbers and a 0

2. Roots of a quadratic equation roots of a quadratic equation

2 0 ax bx c  (a0, , ,a b cR) are given by 2 ₄ 2 ₄ ; 2 2 b b ac b b ac a a        

* sum of the roots = b

a    * Product of roots = c a  * factorised form of _ax2 _{bx c a x( )(x )}       

* If S be the sum and P be the product of roots, then quadratic equation is:

2 ₀

x sxp

3. Nature of roots of a quadratic equation 2

0

ax bx c 

means whether the roots are real or complex. By analysing the expression

2

4

Db  ac

(D called as discriminant), one can get an idea about the nature of the roots as follows: (i) (a) If _{D0 (b}2_4ac0)

then the roots of the quadratic equation are non-real or complex roots (b) If _{D0 (b}2_4ac0)

then the roots are real and equal. Equal roots 2 b a     (c) If _{D0 (b}2_4ac0) then the roots are real and unequal.

(ii) If D i.e., (b24ac) is a perfect square and a b and c, are rational, then the roots are rational. (iii) If D i.e., (b24ac) is not a perfect square and a b and c, are rational, then the roots are of the form

m n and m n.

(iv) If D 0 i.e., _(b2_{4ac0), and the coefficients}

,

a b and c are real then the roots are complex conjugate of each other i.e., the roots are of the form

p iq and p iq (p q, R and _{i  1}).

(v) If a quadratic equation in x has more than two roots, then it is an identity in x (i.e. true for all real values of x) and abc0.

consider two quadratic equations:

2 2

1 1 1 0 2 2 2 0

a x b xc  and a x b xc 

(a) For two common roots:

In such a case, two equations should be identical. For that, the ratio of coefficients of _{x x}2_{, and} 0

x must be same, i.e., 1 1 1 2 2 2 a b c a b c

(b) For one common root:

Let  be the common root of two equations. So  should satisfy the two equations. 2 1 1 1 0 a b c     and 2 2 2 2 0 a bc 

Solving the two equations by using cross multiplication method 2 1 2 2 1 1 2 2 1 1 2 2 1 1 b c b c a c a c a b a b         2 1 1 2 1 2 2 1 a c a c a b a b      2 1 2 2 1 1 2 2 1 b c b c a b a b     2 1 2 2 1 1 2 2 1 2 1 1 2 (b c b c)(a b a b) (a c a c )

     . This is the condition for one root of two quadratic equations to be common.

Note: To find the common root between the two equations, make the coefficient of _2 _{common and then} subtract the two equations.

5. Some more result on roots of quadratic equation.

(i) Both roots of f x ( ) 0 are negative, if sum of the roots < 0, product of the roots > 0 and D 0.

i.e., b 0, c 0,b2 4ac 0

a a

    

(ii) Both roots of f x ( ) 0 are positive, if sum of the roots > 0, product of the roots > 0 and D 0

i.e., b 0, c 0,b2 4ac 0

a a

    

(iii) Roots of f x ( ) 0 are opposite in sign, if product of the roots < 0,

i.e., c 0

a  .

EXERCISE

1. If  and  are the roots of equation 2

0

ax bx c  , find the value of the following expressions:

(i) _2_2 _{(ii) }3_3 _{(iii) }4_4 (iv) _()2 _(v) 4 4

 

2. If  and  are the roots of equation 2

0

ax bx c  , form an equation whose roots are:

(i)  1    1, 1     _(ii) 1 , 1 1  

3. Form an equation whose roots are squares of the sum and the difference of the roots of the equation.

2 2 2

4. Comment upon the nature of the following equation: (i) _x2_{(ab x) c}2 _0

(ii) _{(a b c x)} 2_{2(ab x) (a b c)0} (iii) _{(bc x)} 2_{(ca x) (ab)0}

5. What can you say about the roots of the following equations?

(i) 2 2

2(3 5) 2(9 25) 0

x  a x a  

(ii) (x a x b )(  ) ( x b x c )(  ) ( x c x )( a)0

6. Find the values of k, so that the equations 2

2x kx 5 0 and x23x 4 0 many have one root in common.

7. If 2

0

ax bx c  and 2

0

bx cx a  have a root in common, find the relation between a b and c, .

8. If the equations 2

0

x ax b  and 2

0

x cx d  have one root in common and second equation has equal

roots, prove that ac2(bd).

9. If  , are the roots of _x2_{pxq0 and  , are the roots of} 2

0

x rx s  , evaluate the value of () (  ) () () in terms of p q r s, , , . Hence deduce the condition that the equations have a common root.

10. If the ratio of roots of the equation 2

0

x pxq be equal to the ratio of roots of the equation 2

0

x bx c  ,

then prove that _{p c}2 _{b q}2 _. 11. Find the roots of the equation

2

2 logxalogaxa3 loga xa0 if a0,a1.

12. Find the condition for the equation 1 1 1 1

xxb mm b has real roots that are equal in magnitude but opposite

in sign.

13. For what value of a does the equation log _(x2_{2ax)log(8x6a3) have only one solution?} 14. Find the real roots of the equation

3 4 1 8 6 1 1

x  x  x  x 

15. Solve the equation: _{x 2 4x 6x} 16. Solve the equation

2 2

2 6 8 2 2 3

log_{x  x} log _{x x} (x 2 )x 0

17. Solve the following equation for x:

2 2 2 3 3 7 log _x (6x 23x21)log _x (4x 12x9)4 18. Solve x a x b a b x b x a b a       

19. Find the roots of the equation 3 2

4 4 1 0

x  x  x 

20. Solve the equation 4 3 2

1 0

x x x   x

21. Solve the equation 4 2 6 0 x x   22. Solve: 4 3 2 10 26 10 1 0 x  x  x  x 

23. Solve the equation ₄x _3.2x _{2 0}

  

24. Solve the equation:

2 3

log ( 4 3)

25. Solve the equation (a x) a x (b x) x b a b a x x b          

26. Solve the equation

4a b 5x 4b a 5x3 a b 2x0

27. Prove that the roots of the equation

(x a x c )(  )(x b x )( d)0

are real for any  if abcd.

28. Show that the roots of the equation

(x a x b )(  ) ( xa x c)(  ) ( x b x c )(  )0 are always real. 29. Prove that at least one of the equations

2 ₀

x pxq

2

1 1 0

x p xq 

has real roots if p p1 2(q1q)

30. Prove that at least one of the roots of equation

( )( ) ( )( ) ( )( ) 0

a x b x c  b xa x c c x a x b   are always real. 31. Find the values of p and q for which the roots of the equation

2 ₀

x pxq are equal to p and q.

32. Solve 2 1 3 1 4 2 x x x x                 , when x 0 33. Solve 4 3 2 2 2 1 0 x  x x  x  34. Solve (x1)(x3)(x5)(x7)9

35. Find ’k’ if one root of 2

14 8 0

kx  x  may be six times other..

36. If  , are the roots of the equation 2

1 0

x   x then equation whose roots are 2,2.

37. If _{2 3} is one root of _x2_{pxq0 then find ' 'p , ' 'q} 38. Solve _{2x 6 x45}

39. If the equations _(a2_4a3)x2_{(a1)xa}2_{ 1 0 has infinite roots then find}

' 'a .

40. Equations 2 2

0 2 0

px qx r and qx  pr xq have real roots then show that ' ', ' ', ' 'p q r are in G.P.. 41. If ' 'a and ' 'b are the roots of 2

11x 4x 2 0 then compute the product (1 a a2... ) (1  b b2...)

42. If the quadratic _ax2_{bx c 0(a0) , ,a b c are integers, has natural numbers as it roots, then S.T. a divides}

' ' & ' 'b c .

(a) ac can be expressed as the sum of two squares of natural numbers (b) ' 'a divides ' ' & ' 'b c

(c) ' 'b divides ' ' & ' 'c a

(d) ' 'c divides ' ' & ' 'a b

(e) None of these

43. The value of ' 'aR for which the equation 2 2

44. Let  , are the roots of the quadratic equation 2

0

x ax b  and  , be the roots of the equation 2

2 0

x ax b   .

Given that 1/1/1/1/ 5 /12 and  24. Find the value of the coefficient ' 'a .

45. Solve 2 2 3x 4 3x 4x 1 4x4

Solution To Exercise

1. (i) b a   and c a  2 2 2 2 2 ( ) 2 b c a a       _ _    2 2 2 b ac a   (ii) 3 3 _{( )}3 _{3 ( )}        3 ₃ 3 3 3 b c b b abc a a a a          _{ }  _{  } _       (iii) _4_4 _(2_2_{)2 }2 2 2 2 2 2 2 2 2 2 4 2 ( 2 ) 2 2 b ac c b ac c a a a a        _{ }  _{ }      (iv) 2 2 2 2 2 2 4 4 ( ) ( ) 4 b c b ac a a a         (v) 4 4 ₍ 2 2_{)( )( )}        2 2 2 2 2 4 b ac b b ac a a a          _{  } _{ }      _{ } 2 2 4( 2 ) 4 b b ac b ac a     2. (i) Sum(s)  1  1       _  _{ }  _     ( ) ( ) b a c ac             Product (p) = _1  _1_1 2     2 (c a) ca   The equation is = x2 sx p 0x       2 2 ( ) ( ) 0 b a c c a x x ac ac       _{ }    

2 _{( ) ( )}2 ₀ ac x b c a x c a       (ii) Sum(s) = 1 1 1 1 ( )                                 2 (ac b) bc    Product (p) = 1 1 1 1 a_c                           

The equation is: _x2_{sxp0} 2 2 ( ) 0 ac b a x x bc c      _{ }     2 ₍ 2_{) 0} bcx ac b x ab

     is the required equation. 3. Let  , are the roots of given equation.

(m n)        and 2 2 ( ) 2 m n   

We have to get the equation whose roots are _{( )}2

 and _{( )}2  Sum (s) = 2 2 () () 2 2 2 2(  ) 2 ( ) 2 4mn    _   _ product (p) = 2 2 () .() 2 2 ( ) . ( ) 4   _   _ 2 2 2 2 2 2 2 ( ) ( ) 2( ) ( ) p mn _ mn  m n _  m n

The equation is: _x2_{Sxp0}

 The required equation is

2 _{4 (} 2 2 2_{) 0} x  mnx m n  4. (i) Discriminant (D) 2 2 2 2 ( ) 4(1)( ) ( ) 4 D ab  c  ab  c 0 D  

 The roots are real.

(ii) _D4(ab)2_{4(a b c a)(  b c)}





2 2 2 4 (a b) (a b) c   _     _



2 2 2



2 2 4 (a b) (a b) c 4c (2 )c       

 D 0 and also a perfect square hence the roots are rational.

(iii) _D(ca)2_{4(bc a b)(  )} 2 2 _{(2 )}2 _{4 4 2} c a b ab bc ac       2 2 _{(2 )}2 _{4 4 2} c a b ab bc ac       2 (c a 2 )b   

 The roots are rational.

5. (i) _{D4(3a5)}2_8(9a2_{25) 4(3a5)}2

0

D

  , so the roots are non real if a 5 / 3 and real and equal if 5 3

a  _.

(ii) Simplifying the given equation : 2 3x 2(a b c x) (abbcca)0 2 4( ) 12( ) D a b c  abbcca 2 2 2 4(a b c ab bc ca)       2 2 2 2 ( a b) (b c) (c a)   _      _ 2 2 2 2 2 2 ( ) 1 ( ) ( ) ( ) 2 a b c ab bc ca a b b c c a          _{       }        0 D

  , so the roots are real.

Note: If D 0, then (ab)2(bc)2(ca)20

a b c

  

 If abc, then the roots are equal.

6. Let  be the common root of two equations.

 2

2 k 5 0

2

3 4 0

   

Solving the two equations 2 1 4k 15 8 5 6 k           2 ( 3) (4k 15)(6 k)      2 4k 39k 81 0     27 3 4 k or k     

7. Using the condition for common root, we have

2 2 2 2 (a bc) (bac )(ac b ) 4 2 2 2 2 3 3 2 2 2 a b c a bc a bc b a ac b c        3 3 3 ( 3 ) 0 a a b c abc      3 3 3 0 3 0 a or a b c abc      

This is the relation between a b and c, . From second relation, we also have the relation a b c0

8. The equation 2

0

x cxd has equal root.

2

0 4 0

D D c d

      ... (i)

and the equal roots are 2 ( ) 4 0 2(1) 2 2 c c d c c x        2 c x

 This is the common root of the both the equations.

 2

c

x  will satisfy the first equation. 2 0 4 2 c c a  b   _{ }    2 4 2 c b ac    4d 4b 2ac   



c2 4d from i( )



2(d b) ac     ac2(bd) 9. Roots of _x2_{pxq0 are  ,} Roots of 2 0 x rx s  are  , p and q         r and s      (  ) ( ) ( ) ( )      2 _{( )} 2 _{( )}               _    _{ }    _ 2 2 ( r s) ( r s)      2 2 ( pq0 and  pq0) ( p q r s) ( p q r s)         



(r p) (s q)

 

(r p) (s q)



       2 2 (r p)  (s v) (s v r)( p)( )         2 2 (r p q) (s q) (s q r)( p)( p)         2 (q p rq)[ pq ps pq] (s v)        2 (r p rq)( ps) (s v)     

If the equation have a common root then either

r or s or r or p s      i.e. (r)(s p)( r)(s)0 2 (s q) (r p rq)( ps) 0       2 (s q) (r p ps)( qr)      10. 2 2 2 2 ( ) ( ) ( ) ( )                    2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( )                       2 2 ( ) ( ) 4 4           2 2 4 4 p b q c  

2 2

p c b q

 

11. The given equation can be written as

2

log log 3log

2 0

log log log

a a a x ax a x (a0and a1, loga0) 2 1 3 0 2 y b y b y    

  (where bloga and ylogx) 2(b y)(2b y) y b(2 y) 3 (y b y) 0         2 2 4b 11by 6y 0     which is quadratic in y.  2 2 11 121 96 12 b b b y   4 , 3 2 b b y 

   ₍ _{ylogx and blog )a}

4 log

log log log

3 2 a x a or x      4 1 3 2 x a or x a   

12. From the given equation xm is a root.

 The other root must be –m

1 1 1 1 m m b mm b     1 1 2 b m b m m      2 2 2 b m b m m b m       2 2 2 2m 2b 2m    2 2 2m b   13. _log(x2_{2ax)log(8x6a3)} 2 2 8 6 3 x ax x a      2 _{(2 8) 3(2 1) 0} x a x a       2 (2 8) 4 3(2 1) D a   a

For one solution to exist D 0

2 (a4) 3(2a1)0 2 14 13 0 a a     (a 1)(a 13) 0     1, 13 a   14. Let 2 1 x t 3 4 1 8 6 1 1 x  x  x  x 

2 2 4 4 9 6 1 t t t t        2 2 (t 2) (t 3) 1      2 3 1 t t      Case (i) t 2 2   t 3 t 1 5 2t 1    2 t   1 4 x    5 x   ... (i) Case (ii) 2 t 3 2 3 1 t   t 1 1  true t (2,3) 2 4 t 9    5 x 10    ... (ii) Case (iii) t 3 2 3 1 t   t 2t 6 t 3     2 9 t   1 9 10 x x      ... (iii)

Combining (i), (ii) and (iii); x [5, 10]

15. _x _{2 4}_x ₆_x ... (i)

On squaring both sides

(x 2) (4 x) 2 (x 2)(4 x) 6 x          2 2 (x 2)(4 x) 6 x       2 (x 2(4 x) 6 x 2       2 (x 2)(4 x) 4 x     

Squaring again on both sides 2 4(x2)(4x)(4x)





(4 x) 4x 8 4 x 0       (4 x)(5x 12) 0     12 4, 5 x x    Substitute x 4, in (i) L.H.S = _{4 2}  _{4 4}  ₂ R.H.S = _{6 4  2}  x 4 is a solution. Substitute 12 5 x  _{in (i)}

L.H.S = 12 2 4 12 5    5 2 8 2 2 2 2 3 5 5 5 5 5      R.H.S = 6 12 18 3 2 5 5 5     x 12₅ is also a solution.

Note: Whenever we square a equation and find the roots, verify whether the roots satisfy initial equation or not.

16. 2 2 2 6 8 2 2 3 log_{x x} log _{x x} (x 2 )x 0 2 2 2 0 2 2 3 log _{x x} (x 2 )x (x 6x 8)      2 2 2 2 3 log _{x x} (x 2 )x 1    2 _{2 (2} 2 _{2 3)}1 x x x x      2 4 3 0 x x     (x 1)(x 3) 0     1 3 x or x      1 3

x  and x  satisfy the condition 2

2 0

x  x

But at x  3, x26x  8 9 6( 3) 8    1 0 which is not possible

 x  3 is not the solution. 1 x   satisfies 2 6 8 0 x  x  and 2x22x 3 0 Also at x  1, 2 6 8 3 1 x  x   and 2 2x 2x 3 1

Hence x  1 is the only solution.

17. 2 2 2 3 3 7 log _x (6x 23x21)log _x (4x 12x9)4 2 (2 3) (3 7) log x (2x 3)(3x 7) log x (2x 3) 4       (2 3) (2 3) (3 7)

log _x (2x3)log _x (3x7)2 log _x (2x3)4

(2 3) (3 7) 1 log _x (3x7)2 log _x (2x3)4 Let log(2x3)(3x7)a then (3 7) 1 log _x (2x 3) a    2 3 a a    2 3 2 0 a a     (a 1)(a 2) 0     1 2 a or a    Consider: a 1 (2 3) log _x (3x7)1

3x 7 2x 3

   

4

x

  

But x  4 does not satisfy 2x  3 0 and 3x 70. Hence x  4 is not a solution.

Case (ii) a 2 (2 3) log _x (3x7)2 2 (3x 7) (2x 3)     2 4x 9x 2 0     (4x 1)(x 2) 0     1 2 4 x or x      2

x   does not satisfy 2x  3 0

Hence x  2 is not a solution. 1 4 x   satisfies 3x 70 and 2x  3 0 Also at 1, 2 3 1 4 x  x   1 4

x   is the only solution.

18. Put x a t x b    1 a b t t b a    2 1 0 a b t t b a     _  _     2 4 2 a b a b b a b a t     _  _     2 a b a b b a b a    _  _    a b t or b a    Case (i) t a b  x a a x b b    ( ) 0 0 bx ab ax ab x a b x          Case (iii) t b; a 

x a b x b a    2 2 ax a bx b     2 2 ( ) x a b a b     x a b     x0, a b

19. Given equation can be written as 2 (x1)(x 3x1)0 1 x    2 3 1 0 or x  x  3 9 4 2 x     1, 3 5, 3 5 2 2 x     

20. The given equation can be written as 2 1 13 2 1 13 1 1 0 2 2 x x x x                    2 1 13 1 0 2 x  x     ... (i) 2 1 13 1 0 2 x   x  ... (ii)

The first equation has two roots

13 1 2 13 2 13 1 2 13 2

,

4 4

x       

Second equation has no real roots. 21. Let 2 x t  2 6 0 t   t (t 2)(t 3) 0     2 3 t or t     Case (i): 2 2, 2 t  x    No real roots Case (ii): 2 3, 3 t x   x   3 22. _{M b}2_{4 (a c2 )a} 2 ( 10) 4(1)(26 2) 100 96 4 0        

 The equivalent equation is

2 ( 10) 100 4(26 2) 1 2 x x              

2 ( 10) 100 4(26 2) 1 0 2 x x  _{   }           2 10 2 2 10 2 1 1 0 2 2 x   x x   x      _   _{ }   _     2 2 4 1 0 6 1 0 x x or x x        4 16 4 6 36 4 2 2 x   or x      2 3 3 2 2 x or x      23. _{(2 )}2 x_3.2x_20 2 (2 )x 3.2x 2 0     Let 2x t  2 3 2 0 t t     (t2)(t1)0 2 1 t or t    1 0 2x 2 or 2x 1 2     1 0 x or x    24. 2 4 3 3 x  x  x 2 5 6 0 x x     (x 2)(x 3) 0     2 3 x or x   

Either x 2 or x 3 is not satisfying the original equation.  The original equation has no roots.

25. Rewrite the equation in the form

3 3 2 2 1 1 2 2 ( ) ( ) ( ) ( ) a x x b a b a x x b         wherefrom we have, 1 1 2 2 ( ) ( ) a x ax x b   x b a b or (ax x b)(  )0

Thus, the required solutions will be 1 , 2

x a x b

26. We have

4a b 5x 4b a 5x3 a b 2x

Squaring both members of the equality and performing all the necessary transformations, we get

4a b 5 . 4x b a 5x2(a b 2 )x

Squaring them once again, we find 2

(4ab)(4ba) 5 (4 x a b 4ba)25x 

2 2 2

4(a b 4x 2ab 4ax 4bx)

Hence, 2

0

x ax bx ab  

and, consequently, x1a x, 2 b.

Substituting the found values into the original equation, we get

2 3 0

b a  b a  ba 2 a b  a b 3 a b 0

Hence, if ab, then the equation has two roots; a and b (strictly speaking, if the operations with complex

numbers are regarded as unknown, then there will be only one root). 27. Rewrite the given equation as

2

(1)x (a c bd x) acbd0

Set up the discriminant of this equation D( ) . We have

2 2 2

( ) ( ) 2 ( 2 2 ) ( )

D  bd   abadbcdc bd ac  ac

Set up the discriminant of this equation D( ) . We have 2 ( ) ( ) 4(1 )( ) D   a c bd  l acbd . On transformation we obtain 2 2 2 ( ) ( ) 2 ( 2 2 ) ( ) D   bd   abadbcdc bd ac  ac

We have to prove that D( ) 0 for any . Since D( ) is a second-degree trinomial in  and 2

(0) ( ) 0

D  ac  ,

it is sufficient to prove that the roots of this trinomial are imaginary. And for the roots of our trinomial to be imaginary, it is necessary and sufficient that the expression

2 2 2

4(abadbcdc2bd2ac) 4(ac) (bd)

be less than zero. We have

2 2 2 4(abadbcdc2bd2ac) 4(ac) (bd)  4(ab ad bc dc 2bd 2ac ab cb ad cd)            (abadbcdc2bd2acab cd adcd) 16(b a d)( c c b d)( )( a)     

The last expression is really less than zero by virtue of the given conditions

abcd

28. The original equation can be rewritten in the following way 2

3x 2(a b c x) abacbc0

Let us the prove that 2 2(a b c) 12(abacbc)0 We have, 2 2 2 2 4(a b c) 12(abacbc)4(a b c abac bc ) 2 2 2 2(2a 2b 2c 2ab 2ac 2bc)      



2 2 2 2 2 2



2 (a 2ab b ) (a 2ac c ) (b 2bc c )         



2 2 2



2 (a b) (a c) (b c) 0       

29. Suppose the roots of both equations are imaginary.

Then, 2 2 1 1 4 0, 4 0 p  q p  q  Consequently, 2 2 2 2 2 1 4 4 1 0, 1 2 1 0, ( 1) 0 p p  q q  p p  pp  pp  which is impossible.