Mathematics Quadratic

7/31/2019 Mathematics Quadratic

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Quadrat ic Equations

1

QUADRATI C EQUATI ON S

Target IIT JEE

MATHEMATICS


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( 2 ) ( 2 ) 0ax ax b b ax b

( 2 )( ) 0ax b ax b

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 0ax bx c

20

b ca x

a a

2 20

2

b ca x x

a a

On adding and subtracting

2

,2

b

a

we get:

2 2

2 20

2 2 2

b b c ba x x

a a a a

2 2

2

40

2 4

b ac ba x

a a

[Using 2 2 2( ) 2x y x xy y ]

2 2

2

4

2 4

b b acx

a a

24

2 2

b b acx

a a

2 2

4 4,2 2 2b b ac b b acxa a a

Depending on the sign of 2 4 ,b ac we make the following cases.

Case I: 2 4 0b ac

Then the two roots of quadratic are:

2 24 4and

2 2

b b ac b b aca

a a

Case II: 2 4 0b ac

Then the equation does not have any real roots.

Illustration:

Solve the equation 26 2 0x x .

Sol. Expressing2 2 1

6 2 66 3

xx x x

2 2 16 0

12 3

xx

2 2 1 1 16 0

12 144 144 3

xx


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21 (1 48)

012 144

x

21 49 1 7

12 144 2 12x x

1 7 2 1,

12 12 3 2x x

(c) Direct formula MethodFrom the previous method, we can derive the formula for quadratic equation :

2 0ax bx c

(i) If 2 4 0,b ac then the equation has real and distinct roots i.e.,

2 24 4,

2 2

b b ac b b ac

a a

(ii) If 2 4 0,b ac then the equation has equal roots2

b

a

(iii) If 2 4 0,b ac then the equation has no real roots.

Note: 2 4b ac is called the Discriminant of the quadratic equation and is represented by or .D

Illustration:

Find the roots of the equation 2 2 23 2 0a x abx b .

Sol. 2 2 2 24 ( 3 ) 4(2 )( )B AC ab b a

2 2 2 2 2 29 8 0D 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 bx

a a

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

Let and be the roots of a quadratic equation: 2 0ax bx c where , , and 0a b c R a .

Using the direct formula method,

2 2

4 4,2 2

b b ac b b aca a

Sum of roots 22 coeff. of

2 coeff. of

b b x

a a x

Product of roots =

22 2

2

( ) 4

4

b b ac

a

2 2

2 2

4 constant term

4 coeff. of

b b ac c

aa x


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Illustration:

Find the sum and product of roots of the equation 24 2 3 5 0x x .

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 0ax bx c .

&b c

a a

Consider 2 0ax bx c

2 20 0

b c b cx x x x

a a a a

2 ( ) 0x x

2

(sum of roots) product of roots 0x x Illustration:

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

Sum of roots3 7 3

2 and Product of roots 2 32 2 2

2 27 3 0 2 7 6 02

x x x x

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

root must be p q since

2 4

2

b b acx a

3. FINDING THE VALUES OF SYMMETRIC EXPRESSION

If and are the roots of 2 0ax bx c where , ,a b c R and 0,a 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 22 2 ( )( ) ( ) 4 (iii) 23 3 2 2( ) 3

(iv) 2 23 3 2 2

4

(v) 22 24 4 2 2 2 2 2 22 2 2

(vi) 2 24 4 2 2

2 4

(vii) 5 5 2 2 3 3 2 3 3 2


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2 2 2 2 22

2 2 2 2

2 3 ( )

(viii) 6 6 2 2 4 4 2 2

2

22 2 2 2 22 2 3

(ix)

2 2

21 1

( )

a b

a b a b a ab b

or using s 2 20 & 0a b c a b c

&c c

a b b

1 1 ( )

a b a b a c c

(x) 2 21 1 ( ) 2( )b c

b c b c b bc c

or using 2 20 & 0a b c a b c

2 2b c a b c

2 2 22 2 2 2 2 2

( ) 21 1 1

aa 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

22 2 2

2 2

( ) 2 2

22 2

2

2

2

2 2b c c

a a a

c

a


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2 22 2 2 2 2

2 24

2 2 2

2 4

2 2 2 2b c c b ac a c

aa a a

c c a

a a

4 2 2 2 2 2

2 2

4 4 2b a c b ac a c

c a

4 2 2 2

2 22 4b 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: 0D

Then the polynomial cannot be factorised into linear factors.

Case II: 0D

Then the corresponding equation 2 0ax bx c has equal roots;2

b

a

The factors are:2 2

b ba x x

a a

Case III: 0D

Then roots of the corresponding quadratic equation are

,2 2

b D b D

a a

The factors are ( )( )2 2

b D b Da x x a x x

a a

5. COMMON ROOTS BETWEEN TWO QUADRATIC EQUATIONS(1) Both roots common

Consider two quadratic equations 2 20 and 0ax bx c a x b x c in such a case two equatins should be

identical. For that, the ratio of coefficients of 2 0, andx x x must be same, i.e.a b c

a b c

(2) One root common

Consider two quadratic equatins 2 0ax bx c and 2 0a x b x c . Let be the common root of two

equations. So should satisfy both the equations.

2 0a b c and ... (1)

2 0a x b x c ... (2)

Solving the two equations by cross multiplication method

21

bc b c ac a c ab a b

bc b c a c ac

a c ac ab a b

2

a c ac bc b c ab a b

This is the required condition for two equations to have a common root.


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Illustration:

For what value of ,k equation: 2 22 5 0 & 3 4 0x kx x x may have one root common.

Let be the common root.

22 5 0k

2 3 4 0

21

4 15 8 5 6k k

4 15 3and

3 6

k

k

24 15 324 90 4 15 9

3 6

kk k k

k

24 39 81 0k k

273 or

4k k

6. EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS

In the previous sections we studied quadratic equation and how to find its solution. There are other equationswhich 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

( ) ( ) 0a 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 0at bt c

(2) Find roots of the quadratic equation 2 0at bt c using methods discussed in previous section. Let roots beand .

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

Illustrations:

1.. Solve 2 / 3 1/ 32 15x x .

Step 1 Put 1/ 3x t to get 2 2 15t t

Step 2 2 2 15 0 ( 5)( 3) 0 5 or 3t t t t t t

Step 3 1/ 3 1/ 35 and 3x x

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

Type - 2

Equation of the type: ( )( )

ba 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 getb

at ct


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2 0at ct b

(2) Solve the quadratic equation obtained in step (1) to find roots say and .

(3) Solve ( ) and ( )f x f x to obtain roots of the given equation.

Illustration:

1. Solve3 1 1 5

1 3 1 2

x x

x x

.

Step 1 Put 3 1 1 5to get1 2

x t tx t

22 5 2 0t t

Step 2 Solve 22 5 2 0t t

( 2)(2 1) 0t t

2 or 1/ 2t t

Step 33 1 3 1 1

2 or1 1 2

x x

x x

3 1 2 2 or 6 2 1x x x x 1 or 1/5x x

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

Type - 3

Equation of the tpe:

2

2

1 10a x b x c

xx

or

2

2

1 10a x b x c

xx

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 12 0a x b x cx x

and second in the form:

21 1

2 0a x b x cx x

.

2. Replace1

x tx

in first equation and1

x tx

in second equation, to get 2( 2) 0a t bt c and

2( 2) 0a t bt c .

3. Solve the quadratic equations

1 1andx x

x x

for the first equation or

1 1andx x

x x

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

Illustrations:

1. Solve2

2

1 14 8 3 0x x

xx

Let1

x tx

... (1)

Take square of (1) to get:2 2

2

12x t

x


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2 2

2

12x t

x ... (2)

Using (1) and (2) in the given equation, we get

24( 2) 8 3 0t t

24 8 5 0t t

24 10 2 5 0t t t

(2 5)(2 1) 0t t

5/ 2 and 1/ 2t t

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

1 15 / 2 and 1/ 2x x

x x

2 22( 1) 5 and 2( 1) 0x x x x

22( 1) 5 and 0x x D (so no solution)

2 and 1/ 2x x are two solutions of the given equation.

Type - 4

Equation of the type: ( )( )( )( ) 0x a x b x c x d k 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

2 2( ) )( ( ) 0x a b x ab x c d x cd k

2. As ( ) ( )a b c d we can replace

2 ( )x a b x t to get ( )( ) 0t ab t cd k

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

4. Solve 2 ( )x a b x and 2 ( )x a b x to get solution of the required equation.

Illustrations:

(1) Solve ( 1)( 2)(3 2)(3 1) 21x x x x

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

Take 3 as common factor from last two factors to get

9( 1)( 2)( 2 / 3)( 1/ 3) 21x x x x ... (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(3 5 2)(3 5 2) 21x x x x

Let 23 5x x t

( 2)( 2) 21t t

2 25 0 5t t

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

2 23 5 5 0 or 3 5 5 0x x x x

5 25 4 5 3or no real roots.

6x


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5 85The 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: 2( )ax b cx b

2. Solve quadratic equation to get roots , .

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

Type - 5(b)

Equation of tye type: 2ax bx c dx e


solution of it.

1. Take square on both sides to get 2 2( )ax bx c dx e .

2. Solve the above quadratic equation in x to get roots & .

3. and are roots of the given equation if they satisfy 2 0ax bx c and 0.dx e

Type - 5(c)

Equation of the type: ax b cx d e .


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

3. Take square again to get:22 2( ) 4 ( )a c x b d e e cx d

Solve this quadratic equation in x to get roots & .

4. and are roots of the given equation if they satisfy 0 and 0ax b cx d .

Note: 2x x

Illustrations:

1. Solve 22 2 1 2 3 0x x x

The equation can be written as 22 2 1 2 3x x x

Take square to get2 2

2 2 1 4 9 12x x x x 2 22 10 8 0 5 4 0x x x x

( 1)( 4) 0 1 or 4x x x x

For 1,x

Solving: 2 1 2 1 2 4 3LHS 1 2 3 0

Hence 1x is not a solution.

For 4,x

Solgin: 22 4 2 4 1LHS 32 8 1 8 3 25 5 0 RHS

The only roots is 4.x


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SUMMARY

1. Quadratic Equation

The standard form of a quadratic equation is:

2 0ax bx c

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

2. Roots of a quadratic equation

roots of a quadratic equation

2 0ax bx c ( 0, , , )a a b c R are given by

2 24 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

2

( )( )ax bx c a x x

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

2 0x sx p

3. Nature of roots of a quadratic equation

2 0ax bx c

means whether the roots are real or complex.

By analysing the expression

24D b ac

(D called as discriminant), one can get an idea about the nature of the roots as follows:

(i) (a) If 20 ( 4 0)D b ac

then the roots of the quadratic equation are non-real or complex roots

(b) If 20 ( 4 0)D b ac

then the roots are real and equal.

Equal roots2

b

a

(c) If 20 ( 4 0)D b ac

then the roots are real and unequal.

(ii) If D i.e., 2( 4 )b ac is a perfect square and ,a b and c are rational, then the roots are rational.

(iii) If D i.e.,2( 4 )b ac 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 0D i.e.,2( 4 0)b ac , 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 1i ).

(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 ofx )

and 0a b c .

4. Condition for common root(s)


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consider two quadratic equations:

2 2

1 1 1 2 2 20 0a 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 2 ,x x and 0x 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 10a b c and 22 2 2 0a 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 ( ) 0f x are negative, if sum of the roots < 0, product of the roots > 0 and 0D .

i.e.,20, 0, 4 0b c b ac

a a

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

i.e.,2

0, 0, 4 0b c

b aca a

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

i.e., 0c

a .

EXERCISE

1. If and are the roots of equation 2 0ax 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 0ax 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

2 2( ) 0x m n x m n


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4. Comment upon the nature of the following equation:

(i) 2 2( ) 0x a b x c

(ii) 2( ) 2( ) ( ) 0a b c x a b x a b c

(iii) 2( ) ( ) ( ) 0b c x c a x a b

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

(i) 2 22(3 5) 2(9 25) 0x a x a

(ii) ( )( ) ( )( ) ( )( ) 0x a x b x b x c x c x a

6. Find the values of k, so that the equations 22 5 0x kx and 2 3 4 0x x many have one root in common.

7. If 2 0ax bx c and 2 0bx cx a have a root in common, find the relation between , .a b and c

8. If the equations 2 0x ax b and 2 0x cx d have one root in common and second equation has equal

roots, prove that 2( )ac b d .

9. If , are the roots of 2 0x px q and , are the roots of 2 0x 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 0x px q be equal to the ratio of roots of the equation2 0x bx c ,

then prove that 2 2p c b q .

11. Find the roots of the equation

22log log 3log 0x ax a xa a a if 0, 1a a .

12. Find the condition for the equation1 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 2( 2 ) log(8 6 3)x ax x a have only one solution?

14. Find the real roots of the equation

3 4 1 8 6 1 1x x x x

15. Solve the equation: 2 4 6x x x

16. Solve the equation

2 2

2

6 8 2 2 3log log ( 2 ) 0

x x x xx x

17. Solve the following equation for x :

2 2

2 3 3 7log (6 23 21) log (4 12 9) 4x xx x x x

18. Solve

x a x b a b

x b x a b a

19. Find the roots of the equation 3 24 4 1 0x x x

20. Solve the equation 4 3 2 1 0x x x x


4 2 6 0x x

22. Solve: 4 3 210 26 10 1 0x x x x

23. Solve the equation 4 3.2 2 0x x

24. Solve the equation:

23log ( 4 3)3 3

x xx


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( ) ( )a x a x b x x ba b

a x x b


4 5 4 5 3 2 0a b x b a x a b x

27. Prove that the roots of the equation

( )( ) ( )( ) 0x a x c x b x d

are real for any if a b c d .

28. Show that the roots of the equation

( )( ) ( )( ) ( )( ) 0x a x b x a x c x b x c are always real.

29. Prove that at least one of the equations

2 0x px q

2

1 10x p x q

has real roots if1 12( )p p q q

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

( )( ) ( )( ) ( )( ) 0a x b x c b x a x c c x a x b are always real.

31. Find the values ofp and q for which the roots of the equation

2 0x px q are equal to p and q .

32. Solve

21 3 1

42

x xx x

, when 0x

33. Solve 4 3 22 2 1 0x x x x

34. Solve ( 1)( 3)( 5)( 7) 9x x x x

35. Find k if one root of2

14 8 0kx x may be six times other..36. If , are the roots of the equation 2 1 0x x then equation whose roots are 2, 2 .

37. If 2 3 is one root of2 0x px q then find ' 'p , ' 'q

38. Solve 2 6 4 5x x

39. If the equations 2 2 2( 4 3) ( 1) 1 0a a x a x a has infinite roots then find ' 'a .

40. Equations 2 20 2 0px qx r and qx prx q have real roots then show that ' ', ' ', ' 'p q r are in G.P..

41. If ' 'a and ' 'b are the roots of 211 4 2 0x x then compute the product2 2(1 ...... ) (1 ..... )a a b b

42. If the quadratic 2 0( 0) , ,ax bx c a 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

(1 ) 2( )(1 ) 1 0a x x a ax

has no real roots.


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44. Let , are the roots of the quadratic equation 2 0x ax b and , be the roots of the equation

2 2 0x ax b .

Given that 1/ 1/ 1/ 1/ 5 /12 and 24 . Find the value of the coefficient ' 'a .

45. Solve 2 23 4 3 4 1 4 4x x x x

Solution To Exercise

1. (i) ba

and ca

2

2 2 2 2( ) 2b c

a a

2

2

2b ac

a

(ii) 3 3 3( ) 3 ( )

3 3

3

33

b c b b abc

a a a a

(iii) 4 4 2 2 2 2( ) 2

2 22 2 2 2 2

2 4

2 ( 2 ) 22

b ac c b ac c a

aa a

(iv)2 2

2 2

2 2

4 4( ) ( ) 4

b c b ac

aa a

(v) 4 4 2 2( )( )( )

2 2

2 2

2 4b ac b b acaa a

2 2

4( 2 ) 4

bb ac b ac

a

2. (i)1 1

Sum(s)

( )( )

b a c

ac

Product (p) =1 1 1

2

2( )c a

ca

The equation is =2 0

xx sx p

22 ( ) ( ) 0

b a c c ax x

ac ac


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2 2( ) ( ) 0ac 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: 2 0x sx p

22 ( ) 0

ac b ax x

bc c

2 2( ) 0bcx ac b x ab is the required equation.

3. Let , are the roots of given equation.

( )m n and2 2

( )

2

m n

We have to get the equation whose roots are 2( ) and 2( )

Sum (s) = 2 2( ) ( )

2 2 22( ) 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:2

0x Sx p The required equation is

2 2 2 24 ( ) 0x mnx m n

4. (i) Discriminant (D)

2 2 2 2( ) 4(1)( ) ( ) 4D a b c a b c

0D

The roots are real.

(ii) 24( ) 4( )( )D a b a b c a b c

2 2 2

4 ( ) ( )a b a b c

2 2 2 2 24 ( ) ( ) 4 (2 )a b a b c c c

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

(iii) 2( ) 4( )( )D c a b c a b

2 2 2(2 ) 4 4 2c a b ab bc ac

2 2 2(2 ) 4 4 2c a b ab bc ac

2( 2 )c a b

0D and also a perfect square.


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The roots are rational.

5. (i) 2 2 24(3 5) 8(9 25) 4(3 5)D a a a

0D , so the roots are non real if 5 / 3a and real and equal if5

3a .

(ii) Simplifying the given equation :

23 2( ) ( ) 0x a b c x ab bc ca

24( ) 12( )D a b c ab bc ca

2 2 24( )a b c ab bc ca

2 2 22 ( ) ( ) ( )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

0D , so the roots are real.

Note: If 0D , then2 2 2( ) ( ) ( ) 0a b b c c a

a b c

If a b c , then the roots are equal.

6. Let be the common root of two equations.

22 5 0k

2 3 4 0

Solving the two equations

21

4 15 8 5 6k k

2( 3) (4 15)(6 )k k

24 39 81 0k k

273

4k 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 22a b c a bc a bc b a ac b c

3 3 3( 3 ) 0a a b c abc

3 3 30 3 0a or a b c abc

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

8. The equation 2 0x cx d has equal root.

20 4 0D D c d ... (i)

and the equal roots are

2( ) 4 0

2(1) 2 2

c c d c cx

2

cx is the equal root of this equation.


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19

This is the common root of the both the equations.

2

cx will satisfy the first equation.

2

04 2

c ca b

2 4 2c b ac

4 4 2d b ac

2 4 ( )c d from i

2( )d b ac

2( )ac b d

9. Roots of 2 0x px q are ,

Roots of 2 0x rx s are ,

p and q

r and s

( ) ( ) ( ) ( )

2 2( ) ( )

2 2( ) ( )r s r s

2 2( 0 0)p q and p q

( ) ( )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. ( )( )( )( ) 0r s p r s

2( ) ( )( ) 0s q r p rq ps

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


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20

2 2p c b q

11. The given equation can be written as

2

log log 3log2 0

log log log

a a a

x ax a x ( 0 1, log 0)a and a a

2 1 30

2y b y b y

(where log logb a and y x )

2( )(2 ) (2 ) 3 ( ) 0b y b y y b y y b y 2 24 11 6 0b by y

which is quadratic in y.

2 211 121 96

12

b b by

4,

3 2

b by

( log log )y x and b a

4 loglog log log

3 2

ax a or x

4 1

3 2x 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 22b m b mmb m

2 2 22 2 2m b m

2 22m b

13. 2log( 2 ) log(8 6 3)x ax x a

2 2 8 6 3x ax x a

2 (2 8) 3(2 1) 0x a x a

2(2 8) 4 3(2 1)D a a

For one solution to exist 0D 2( 4) 3(2 1) 0a a

2 14 13 0a a

( 1)( 13) 0a a

1, 13a

14. Let 21x t

3 4 1 8 6 1 1x x x x


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21

2 24 4 9 6 1t t t t

2 2( 2) ( 3) 1t t

2 3 1t t

Case (i) 2t

2 3 1t t

5 2 1t

2t

1 4x

5x ... (i)

Case (ii) 2 3t

2 3 1t t

1 1 (2,3)true t

24 9t

5 10x ... (ii)

Case (iii) 3t2 3 1t t

2 6 3t t

2 9t

1 9 10x x ... (iii)

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

15. 2 4 6x x x ... (i)

On squaring both sides

( 2) (4 ) 2 ( 2)(4 ) 6x x x x x

2 2 ( 2)(4 ) 6x x x

2 ( 2(4 ) 6 2x x x

2 ( 2)(4 ) 4x x x

Squaring again on both sides

24( 2)(4 ) (4 )x x x

(4 ) 4 8 4 0x x x

(4 )(5 12) 0x x

124,

5x x

Substitute 4x , in (i)

L.H.S = 4 2 4 4 2

R.H.S = 6 4 2

4x is a solution.

Substitute12

5

x in (i)


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22

L.H.S =12 12

2 45 5

2 8 2 2 22 3

5 5 5 5 5

R.H.S =12 18 2

6 35 5 5

125

x 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 22

6 8 2 2 3log log ( 2 ) 0

x x x xx x

2

2 2 0

2 2 3log ( 2 ) ( 6 8)

x xx x x x

2

2

2 2 3log ( 2 ) 1

x xx x

2 2 12 (2 2 3)x x x x

2 4 3 0x x

( 1)( 3) 0x x

1 3x or x

1 3x and x satisfy the condition 2 2 0x x

But at 3,x 2 6 8 9 6( 3) 8 1 0x x which is not possible

3x is not the solution.

1x satisfies 2 6 8 0x x and 22 2 3 0x x

Also at 1x , 2 6 8 3 1x x and

22 2 3 1x x Hence 1x is the only solution.

17. 2 22 3 3 7log (6 23 21) log (4 12 9) 4x xx x x x

2

(2 3) (3 7)log (2 3)(3 7) log (2 3) 4x xx x x

(2 3) (2 3) (3 7)log (2 3) log (3 7) 2 log (2 3) 4x x xx x x

(2 3) (3 7)1 log (3 7) 2log (2 3) 4x xx x

Let (2 3)log (3 7)x x a

then (3 7) 1log (2 3)x xa

23a

a

2 3 2 0a a

( 1)( 2) 0a a

1 2a or a

Consider: 1a

(2 3)log (3 7) 1

x x


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23

3 7 2 3x x

4x

But 4x does not satisfy 2 3 0x and 3 7 0x .


Case (ii) 2a

(2 3)log (3 7) 2x x

2(3 7) (2 3)x x

24 9 2 0x x

(4 1)( 2) 0x x

12

4x or x

2x does not satisfy 2 3 0x


1

4x satisfies 3 7 0x and 2 3 0x

Also at1

, 2 3 14

x x

1

4x is the only solution.

18. Putx a

tx b

1 a bt

t b a

2 1 0a bt tb a

2

4

2

a b a b

b a b at

2

a b a b

b a b a

a b

t orb a

Case (i)a

tb

x a a

x b b

( ) 0 0bx ab ax ab x a b x

Case (iii) ;b

ta


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24

x a b

x b a

2 2ax a bx b

2 2( )x a b a b

x a b

0,x a b

19. Given equation can be written as2( 1)( 3 1) 0x x x

1x

2 3 1 0or x x

3 9 4

2x

3 5 3 5

1, ,2 2

x

20. The given equation can be written as

2 21 13 1 131 1 02 2

x x x x

2 1 131 0

2x x

... (i)

2 1 131 0

2x 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 2x t

2 6 0t t

( 2)( 3) 0t t

2 3t or t

Case (i): 22, 2t x

No real roots

Case (ii): 23, 3t x

3x

22. 2 4 ( 2 )M b a c a

2( 10) 4(1)(26 2) 100 96 4 0

The equivalent equation is

2 ( 10) 100 4(26 2)1

2x x


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25

2 ( 10) 100 4(26 2)1 0

2x x

2 210 2 10 21 1 0

2 2x x x x

2 24 1 0 6 1 0x x or x x

4 16 4 6 36 4

2 2x or x

2 3 3 2 2x or x

23. 2(2 ) 3.2 2 0x x

2(2 ) 3.2 2 0x x

Let 2xt

2 3 2 0t t

( 2)( 1) 0t t

2 1t or t

1 02 2 2 1 2x xor

1 0x or x

24. 2 4 3 3x x x

2 5 6 0x x

( 2)( 3) 0x x

2 3x or x

Either 2x or 3x 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 ba b

a x x b

wherefrom we have,

1 1

2 2( ) ( )a x a x x b x b a b

or ( )( ) 0a x x b

Thus, the required solutions will be

1 2,x a x b

26. We have

4 5 4 5 3 2a b x b a x a b x

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

4 5 . 4 5 2( 2 )a b x b a x a b x

Squaring them once again, we find

2(4 )(4 ) 5 (4 4 ) 25a b b a x a b b a x

2 2 2

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


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26

Hence, 2 0x ax bx ab

and, consequently,1 2,x a x b .

Substituting the found values into the original equation, we get

2 3 0b a b a b a

2 3 0a b a b a b

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 ) ( ) 0x a c b d x ac bd

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 ( ) 0D for any . Since ( )D is a second-degree trinomial in and2(0) ( ) 0D 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 24( 2 2 ) 4( ) ( )ab ad bc dc bd ac a c b d

be less than zero. We have

2 2 24( 2 2 ) 4( ) ( )ab ad bc dc bd ac a c b d

4( 2 2 )ab ad bc dc bd ac ab cb ad cd

( 2 2 )ab ad bc dc bd ac 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

23 2( ) 0x a b c x ab ac bc

Let us the prove that

22( ) 12( ) 0a b c ab ac bc

We have,

2 2 2 24( ) 12( ) 4( )a b c ab ac bc a b c ab ac bc

2 2 22(2 2 2 2 2 2 )a b c ab ac bc

2 2 2 2 2 22 ( 2 ) ( 2 ) ( 2 )a ab b a ac c b bc c

2 2 22 ( ) ( ) ( ) 0a b a c b c

29. Suppose the roots of both equations are imaginary.

Then, 2 21 14 0, 4 0p q p q

Consequently, 2 2 2 2 21 1 1 1 14 4 0, 2 0,( ) 0p p q q p p pp p p

which is impossible.


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27

30. Let us rewrite the given equation as

2( ) 2( ) 3 0a b c x ab ac bc x abc

Prove that its discriminant is greater than or equal to zero.

We have, 24( ) 12 ( )ab ac bc abc a b c

2 2 22 ( ) ( ) ( ) 0ab ac ab bc ac bc

31. By properties of the quadratic equation we have the following system

,p q p pq q From the second equation we get,

( 1) 0q p

Hence, either 0q or 1p . From the first one we find

if 0q , then 0p ; if 1p , then 2q .

Thus, we have two quadratic equations satisfying the set requirements

2 20 2 0x and x x

32.

21 3 1

4

2

x x

x x

2 2

1 3 1 1 3 14 4 0

2 2x x x x

x x x x

2 3 102

t t where t xx

30 0 3/ 2

2t t t or t

1 1 30

2

x or x

x x

2 21 2 3 2x or x x

21 2 4 2 0x or x x x

1 ( 2)(2 1) 0x or x x

1 2 1/ 2x or or

1, 2, 1/ 2Solution set

33. 4 3 22 2 1 0x x x x

222 12 1 0x x x x

2

2

1 12 1 0x x

xx

21 1

2 2 1 0x xx x

2 12 3y y where y xx

( 3)( 1) 0y y


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28

3 0 1 0y or y

1 13 0 1 0x or x

x x

2 23 1 0 1 0x x or x x

3 9 4 1 1 4

2 2x or x

3 5 1 3

2 2

ix or

34. ( 1)( 3)( 5)( 7) 9x x x x

( 7)( 1)] ( 3)( 5) 9x x x x

2 2( 8 7)( 8 15) 9x x x x

2( 7)( 15) 9 8t t where t x x

2 22 105 9 0t t

2 2

22 96 0 16 6 96 0t t t t t

( 16) 6( 16) 0 ( 16)( 6) 0t t t t t

2 2( 8 16)( 8 6) 0x x x x

2 28 16 0 8 6 0x x or x x

2 8 64 24( 4) 0

2x or x

4, 4 4 10x or x

4, 4 10Solution set

35. Given equation is 2 14 8 0kx x

Let the roots be , 6

Sum of the roots, 6 14 / 7 14 / 2 / k k k

Product of the roots,

2 2(6 ) 8 / 3 4 / 3(2 / ) 4 / 3k k k k k

36. Let 2( ) 1f x x x

Equation whose roots are 2, 2 is ( 2) 0f x

2( 2) ( 2) ( 2) 1 0f x x x

2 3 1 0x x

37. If one root of the equation is 2 3 then other root is 2 3

Sum of the roots = 2 3 2 3 4 4p p

Product of the roots = (2 3)(2 3) 4 3 1q q

38. Squaring both sides

2 6 4 2 (2 6)( 4) 25x x x x


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29

2 (2 6)( 4) 25 4 6 3 27 3x x x x

Again squaring both sides and simplifying.

We arrive at the equation 2 170 825 0 5, 165x x x

Direct substitution of these values in the original equation shows that 5x is root and 165x is not.

Why 165x is not a root of an equation?

Reason:(Extraneous root appeared due to squaring the equation)

39. Any quadratic equation 2 0 ( 0)ax bx c a has infinite solutions then 0, 0, 0a b c .

20. 0. 0 0 is satisfied for all real ' 'x x x

2 2 24 3 ( 1) 1 0a a x a x a has infinite roots if

2 2( 4 3) 0, ( 1) 0, 1 0a a a a simultaneously..

that is possible only for 1a .

40. 2 0px qx r have real roots

2 24 2 0D q pr q pr ... (1)

2

2 0qx prx q have real roots2 24 4 0 0D qr q pr q

2 2 2pr q q pr q pr ... (2)

From (1) and (2) 2q pr and 2 2q pr q pr

41. Let 2 2(1 ...... ) (1 ..... )S a a b b

1 1.

1 1

aS S

r a a b

1 1,

1 1 ( )S S

a b ab a b ab

a, b are the roots of 211 4 2 0 4 /11, 2 /11x x a b ab

1 1 11

4 2 6 51 1

11 11 11

S S

42. Consider example 2 28 7 0, 0x x ax bx c

7ac which cant be expressed as sum of square of two natural numbers

' 'a is wrong.

Consider same example 2 8 7 0x x where , ,a b c are integers and roots also natural numbers '1' and '7 ' .

1, 8, 7 ' 'a b c b does not divide ' 'c .

Hence c is wrong

' 'c does not divide ' 'b in the same example

Hence ' 'd is wrong.

Let , be the roots of 2 0ax bx c

,b c

a a


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, are integers , are integers

,b c

a a

are integers ' 'a divide ' ' & ' 'b c

' 'b is correct

43. 2 2(1 ) 2( )(1 ) 1a x x a ax

2 2 2 2(1 ) 2( ) 1a x x ax a a x 2 2 2 2(1 ) (2 2 ) 2 1 2a x x a ax a

2 2 2(1 2 ) (2 2 ) (1 2 )a a x a x a 2 2 2(1 ) 2(1 ) (1 2 )a x a x a

2 4D B AC

2 2 24(1 ) 4(1 ) (1 2 )D a a a

2 2(1 4 (1 ) (1 2 )D a a a

2 24(1 ) (1 2 1 2 )D a a a a

2 24 ( 1) 0D a a

For all value of ' 'a roots are real.

44. , are roots of2 0 ,x ax b a b

1 1a a

b b

, are roots of 2 ( 2) 0 , 2x ax b a b 1 1

2 2

a a

b b

24 ( 2) 24b b 2 22 24 0 6 4 24 0b b b b b

( 6) 4( 6) 0 ( 4)( 6) 0b b b b b

4,6b

1 1 1 1 5

2 12

a a

b b

If5

44 6 12

a ab

5 3 2 5 55

4 6 12 12 12 12 12

a a a a aa

45. The equation can be written as

2 23 4 4 4 3 4 1 0x x x x

2 23 4 1 4 3 4 1 3 0x x x x

Put 2 23 4 1x x t

2 4 3 0t t

( 1)( 3) 0t t

1 3t or t

23 4 1 1 3x x or

2 23 4 0 3 4 8 0

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Mathematics Quadratic