Quadratic Equations
Target IIT JEE
MATHEMATICS
QUADRATIC EQUATIONS
1
Quadratic Equations
Quadratic Equations 1.
POLYNOMIAL
(i)
The expression of the form a0 x n a1 x n 1 ........ an 1 x an is called polynomial where a0 , a1 , .............. an are real number and n is a non negative integer. Quadratic Polynomial The polynomial where n 2 and a0 0 is called a quadratic polynomial i.e. polynomial with degree 2. such as a0 x 2 a1 x a2 or ax 2 bx c .
(ii)
Quadratic Equations The general form of quadratic equation is ax 2 bx c 0, where a, b, c are real numbers and a 0.
For example: x2 5 x 10 0 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: x2 4 x 4 becomes zero if we put x 2
(iv)
2 is a zero of this polynomial.
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. ax2 bx c 0 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, x2 5 x 6 0
Let us put x 2, 3. 22 5 2 0
2. (a)
and
32 5 3 6 0
x 2, 3 are roots of quadratic equation.
FINDING SOLUTION OF QUADRATIC EQUATION Factorisation method: Let us consider the quadratic equation: ax2 bx c 0
To find roots, we factorise it and get: ax 2 bx c (lx m)(nx p ) 0
lx m 0 or nx p 0
x
m l
or
x
p n
Hence, these are the roots of the given quadratic equation. Illustration: Find the roots of the following equation: a 2 x2 3abx 2b 2 0
Sol.
a 2 x2 2abx abx 2b 2 0
2
Quadratic Equations ax(ax 2b) b(ax 2b) 0 (ax 2b)(ax b) 0 x 2b / a or 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 ax2 bx c 0 b c a x2 0 a a 2b c a x2 x 0 2a a 2
b On adding and subtracting , we get: 2a 2 2 2b b c b a x2 x 0 2a 2a a 2a 2 b 4ac b 2 a x 0 2a 4a 2
[Using ( x y )2 x 2 2 xy y 2 ]
2
b b 2 4 ac x 2a 4a 2
b b 2 4ac x 2a 2a
x
b b 2 4ac , 2a 2a
b b 2 4ac 2a
Depending on the sign of b 2 4ac, we make the following cases. Case I:
b 2 4ac 0
Then the two roots of quadratic are: a
b b 2 4 ac b b 2 4 ac and 2a 2a
Case II:
b 2 4ac 0
Then the equation does not have any real roots. Illustration: Solve the equation 6 x2 x 2 0 . Sol.
x
1
2 2 Expressing 6 x x 2 6 x 6 3
2x 1 6 x2 0 12 3
2x 1 1 1 6 x2 0 12 144 144 3
3
Quadratic Equations 2
1 (1 48) 0 x 12 144
1 49 x 12 144
x
2
(c)
1 7 12 12
x
1 7 2 12
2 1 x , 3 2
Direct formula Method From the previous method, we can derive the formula for quadratic equation : ax2 bx c 0
(i) If b 2 4ac 0, then the equation has real and distinct roots i.e.,
b b 2 4ac , 2a
b b 2 4ac 2a
b
(ii) If b 2 4ac 0, then the equation has equal roots 2a (iii) If b 2 4ac 0, then the equation has no real roots.
Note: b 2 4ac is called the Discriminant of the quadratic equation and is represented by or D. Illustration: Find the roots of the equation a 2 x2 3abx 2b 2 0 . B 2 4 AC (3ab) 2 4(2b 2 )( a 2 )
Sol.
(d)
D 9a 2b 2 8b2 a 2 a 2 b2
The equation has two distinct roots i.e.
B D , 2A
B D 2A
3ab a 2 b 2 , 2a 2
3ab a 2 b 2 2a 2
x
b , a
D0
2b a
Sum and product of roots of a Quadratic Equation Let and be the roots of a quadratic equation: ax2 bx c 0 where a, b, c R and a 0 . Using the direct formula method,
b b 2 4 ac , 2a
2b
b
b b 2 4ac 2a
coeff. of x
Sum of roots 2a a coeff. of x 2
Product of roots =
( b ) 2
b 2 4ac
2
4a 2
b 2 b 2 4 ac c constant term 4a 2 a coeff. of x 2
4
Quadratic Equations
Illustration: Find the sum and product of roots of the equation 4 x 2 2 3x 5 0 . Sol.
(e)
2 3 4
3 2
5 4
Formation of a quadratic equation when roots are given Let , be the roots of equation 2 ax2 bx c 0 .
b a
&
c a
Consider ax2 bx c 0 b c c b x 0 x2 x 0 a a a a
x2
x 2 ( ) x 0
x 2 (sum of roots) x product of roots 0
Illustration: Find the quadratic equation whose roots are 2 and 3/2. 3 2
Sum of roots 2
x2
7 2
and
7 x3 0 2
Product of roots 2
3 3 2
2 x2 7 x 6 0
Note: If one root of a quadratic equation is in the form p q and a, b and c are rational numbers, then other root must be p q since x 3.
b b 2 4ac 2a
FINDING THE VALUES OF SYMMETRIC EXPRESSION If and are the roots of ax2 bx c 0 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 ( )( ) ( )
(iii)
2
4 2
2 3 3 ( ) 2 2 3
(iv)
2 2 3 3 2 2 4
(v)
4 4 2 2 2 2 2 2 2 2 2
(vi)
2 2 4 4 2 2 2 4
(vii)
5 5 2 2 3 3 2 3 3 2
2
2
2
5
Quadratic Equations 2 2 2 2 2 2 2 2 2 3 2 2 ( )
(viii)
6 6 2 2 4 4 2 2 2 2 2 2 2 2
(ix)
2
a 2 b c 0
&
a b
c
1 1 ( ) a b a b a c c
&
b
c
1 1 b( ) 2c b c b c b 2 bc( ) c 2 a 2 b c 0
or using b c a 2
&
a 2 b c 0
b c 2
2 2 ( ) 2 2 1 1 1 a 2 a 2 a a 2 2 2 2
Illustration: Let , be the roots of ax 2 bx c 0. Find the values of: (1)
1 2 2 2
3 3 3( ) 2 2 2 4 2 2 2( 2 ) 5
2
(2)
2
a 2b 1 1 2 a b a b a ab ( ) b2
or using s a 2 b c 0
(x)
2
3
3( ) ( ) 2 2 5
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
b 2 2c 2c 2 2 a a a c2 a2
6
Quadratic Equations 2
2
b 2 2c 2c 2 b 2 2ac 2a 2 c 2 2 2 a a a a4 2 c c2 a 2 a2 a4
4.0
b 4 4a 2 c 2 4b 2 ac 2a 2 c 2 c2 a 2
b 4 2a 2 c 2 4b 2 ac c2 a2
FACTORISATION OF A QUADRATIC POLYNOMIAL Let f ( x ) ax 2 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 b
Then the corresponding equation ax2 bx c 0 has equal roots; 2a
The factors are: a x
b b x 2a 2a
Case III: D 0 Then roots of the corresponding quadratic equation are
b D b D , 2a 2a
5. (1)
The factors are a( x )( x ) a x
b D b D x 2a 2a
COMMON ROOTS BETWEEN TWO QUADRATIC EQUATIONS Both roots common Consider two quadratic equations ax 2 bx c 0 and a x 2 b x c 0 in such a case two equatins should be identical. For that, the ratio of coefficients of x2 , x and x 0 must be same, i.e.
(2)
a b c a b c
One root common Consider two quadratic equatins ax2 bx c 0 and ax 2 bx c 0 . Let be the common root of two equations. So should satisfy both the equations.
a 2 b c 0
and
... (1) ... (2)
ax 2 bx c 0
Solving the two equations by cross multiplication method 1 2 bc bc ac ac ab a b
bc bc a c ac a c ac ab a b
ac ac
2
bc bc ab a b
This is the required condition for two equations to have a common root.
7
Quadratic Equations
Illustration: For what value of k , equation: 2 x 2 kx 5 0 & x 2 3 x 4 0 may have one root common. Let be the common root. 2 2 k 5 0
2 3 4 0
6.
2 1 4k 15 8 5 6 k
4k 2 39k 81 0
k 3
4 k 15 3
and
4 k 15 3 3 6k
k
or
3 6k 24k 90 4k 2 15k 9
27 4
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 2
Equation of the type: a f ( x) b f ( x ) c 0 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: at 2 bt c 0
(2)
Find roots of the quadratic equation at 2 bt c 0 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 x2 / 3 2 x1/ 3 15 .
Step 1 Put x1/ 3 t to get
t 2 2t 15
Step 2 t 2 2t 15 0
(t 5)(t 3) 0 t 5 or t 3
Step 3
x1/ 3 5
and
x1/ 3 3
Taking cube on both sides, we get:
x 125 and x 27
Type - 2 b
Equation of the type: a f ( x ) f ( x ) c 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)
b t
Replace f ( x) by t to get at c
8
Quadratic Equations
at 2 ct b 0
(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.
3x 1 x 1
5
Solve . x 1 3x 1 2
Step 1 Put
3x 1 1 5 t to get t x 1 t 2
2t 2 5t 2 0
Step 2 Solve 2t 2 5t 2 0 (t 2)(2t 1) 0 t 2 or t 1/ 2
Step 3
3x 1 2 x 1
3x 1 1 x 1 2
or
3x 1 2 x 2
x 1 or x 1/ 5
6x 2 x 1
or
Theefore roots of the given equation are x 1 & x 1/ 5. Type - 3 Equation of the tpe: 1 1 a x2 2 b x c 0 x x
1 1 a x2 2 b x c 0 x x
or
The following steps are useful to reduce the above equation into quadratic equation and hence find its solutions. 1.
2 1 1 a x 2 b x c 0 and second in the form: Write first equation in the form: x x
2 1 1 a x 2 b x c 0 . x x
2.
Replace x
1 1 t in first equation and x t in second equation, to get a (t 2 2) bt c 0 and x x
a (t 2 2) bt c 0 .
1
1
1
1
Solve the quadratic equations x and x for the first equation or x and x x x x x for the second equation to obtain solution of the given equation. Illustrations: 3.
1.
1
1
2 Solve 4 x 2 8 x 3 0 x x
Let x
1 t x
Take square of (1) to get:
... (1) x2
1 2 t2 2 x
9
Quadratic Equations x2
1 t2 2 x2
... (2)
Using (1) and (2) in the given equation, we get 4(t 2 2) 8t 3 0
4t 2 8t 5 0 4t 2 10t 2t 5 0 (2t 5)(2t 1) 0
t 5/ 2
and
t 1/ 2
Putting the values of t in (1), we get: x
1 5/ 2 x
1 1/ 2 x
and
x
2( x 2 1) 5 x
and
2( x 2 1) x 0
2( x 2 1) 5 x
and
D 0 (so no solution)
x 2 and x 1/ 2 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 . 1.
The following steps are useful to reduce the above equation into quadratic equation and hence find its solutions. Multiply first two brackets and last two brackets to get
x 2.
2
( a b) x ab)( x 2 (c d ) x cd k 0
As (a b) (c d )
we can replace
x 2 ( 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 x 2 (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 ... (1)
9( x 1)( x 2)( x 2 / 3)( x 1/ 3) 21
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. (3 x 2 5 x 2)(3 x 2 5 x 2) 21
Let
3 x 2 5x t
(t 2)(t 2) 21
t 2 25 0
t 5
On combining (2) and (3) we get: 3 x2 5 x 5 0
x
5 25 4 5 3 6
or
or
3x2 5 x 5 0
no real roots.
10
Quadratic Equations
5 85 6
The solution is :
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:
ax2 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 ax 2 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 ax2 bx c 0 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
2.
Take square on both side to get: ax b e 2 (cx d ) 2e cx d
3.
Take square again to get: (a c) x b d e 2 4e2 (cx d )
ax b e cx d
(a c) x b d e2 2e cx d
2
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:
x2 x
Illustrations: 1.
Solve
2x2 2 x 1 2x 3 0
The equation can be written as
2x2 2 x 1 2 x 3
Take square to get 2 x 2 2 x 1 4 x 2 9 12 x
2 x2 10 x 8 0
x2 5x 4 0
( x 1)( x 4) 0
x 1
For
x 1,
Solving:
LHS 2 1 2 1 2 4 3
or
x4
1 2 3 0
Hence x 1 is not a solution. For Solgin:
x 4,
LHS 2 42 2 4 1 32 8 1 8 3 25 5 0 RHS
The only roots is x 4.
11
Quadratic Equations
SUMMARY 1.
Quadratic Equation The standard form of a quadratic equation is: ax2 bx c 0
2.
where a, b, c are real numbers and a 0 Roots of a quadratic equation roots of a quadratic equation (a 0, a, b, c R )
ax2 bx c 0
are given by
b b 2 4ac b b 2 4ac ; 2a 2a
* sum of the roots = * Product of roots =
b a
c a
* factorised form of ax 2 bx c a ( x )( x ) * If S be the sum and P be the product of roots, then quadratic equation is: x 2 sx p 0
3.
Nature of roots of a quadratic equation ax2 bx c 0
means whether the roots are real or complex. By analysing the expression D b2 4ac
(D called as discriminant), one can get an idea about the nature of the roots as follows: (i)
(a) If D 0
(b 2 4 ac 0)
then the roots of the quadratic equation are non-real or complex roots (b) If D 0
(b 2 4 ac 0)
then the roots are real and equal. Equal roots (c) If D 0
b 2a
(b 2 4ac 0)
then the roots are real and unequal. (ii)
If D i.e., (b 2 4 ac ) is a perfect square and a, b and c are rational, then the roots are rational.
(iii)
If D i.e., (b 2 4 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 D 0 i.e., (b 2 4 ac 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 .
4.
Condition for common root(s)
12
Quadratic Equations
consider two quadratic equations: a1 x 2 b1 x c1 0 and a2 x 2 b2 x c2 0
(a) For two common roots: In such a case, two equations should be identical. For that, the ratio of coefficients of x 2 , x and x0 must be same, a
b
c
1 1 1 i.e., a b c 2 2 2
(b) For one common root: Let be the common root of two equations. So should satisfy the two equations.
a1 2 b1 c1 0
and
a2 2 b2 c2 0
Solving the two equations by using cross multiplication method
1 2 b1c2 b2 c1 a1c2 a2 c1 a1b2 a2b1
a2 c1 a1c2 a1b2 a2b1
2
b1c2 b2 c1 a1b2 a2 b1
(b1c2 b2 c1 )(a1b2 a2 b1 ) (a2 c1 a1c2 ) 2 . This is the condition for one root of two quadratic equations to be
common.
5.
Note: To find the common root between the two equations, make the coefficient of 2 common and then subtract the two equations. 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., (ii)
b c 0, 0, b 2 4 ac 0 a a
Both roots of f ( x ) 0 are positive, if sum of the roots > 0, product of the roots > 0 and D 0 b a
i.e., 0, (iii)
c 0, b 2 4ac 0 a
Roots of f ( x ) 0 are opposite in sign, if product of the roots < 0, c 0. a
i.e.,
EXERCISE 1.
2.
If and are the roots of equation ax2 bx c 0 , find the value of the following expressions: (i) 2 2
(ii) 3 3
(iv) ( ) 2
(v) 4 4
If and are the roots of equation ax2 bx c 0 , form an equation whose roots are: (i)
3.
(iii) 4 4
1 1 1 ,
(ii)
1 1 1 ,
Form an equation whose roots are squares of the sum and the difference of the roots of the equation. 2 x 2 2( m n ) x m 2 n 2 0
13
Quadratic Equations
4.
Comment upon the nature of the following equation: (i) x 2 (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) x 2 2(3a 5) x 2(9a 2 25) 0 (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 x 2 kx 5 0 and x2 3x 4 0 many have one root in common.
7.
If ax2 bx c 0 and bx 2 cx a 0 have a root in common, find the relation between a, b and c.
8.
If the equations x2 ax b 0 and x2 cx d 0 have one root in common and second equation has equal roots, prove that ac 2(b d ) .
9.
If , are the roots of x 2 px q 0 and , are the roots of x2 rx s 0 , 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 x 2 px q 0 be equal to the ratio of roots of the equation x2 bx c 0 , then prove that p 2 c b 2 q .
11.
Find the roots of the equation 2log x a log ax a 3log a2 x a 0 if a 0, a 1 .
12.
Find the condition for the equation
1 1 1 1 has real roots that are equal in magnitude but opposite x x b m mb
in sign. 13.
For what value of a does the equation log ( x 2 2ax ) log(8 x 6 a 3) have only one solution?
14.
Find the real roots of the equation x 3 4 x 1 x 8 6 x 1 1
15.
Solve the equation:
16.
Solve the equation
x2 4 x 6 x
log x 2 6 x 8 log 2 x2 2 x 3 ( x 2 2 x) 0
17.
Solve the following equation for x : log 2 x 3 (6 x 2 23 x 21) log 3 x 7 (4 x 2 12 x 9) 4
x a x b a b xb x a b a
18.
Solve
19.
Find the roots of the equation x3 4 x 2 4 x 1 0
20.
Solve the equation x4 x3 x 2 x 1 0 Solve the equation
21.
x4 x 2 6 0
22.
Solve: x4 10 x3 26 x 2 10 x 1 0
23.
Solve the equation 4 x 3.2 x 2 0 Solve the equation:
24.
3log3 ( x
2
4 x 3)
x 3
14
Quadratic Equations
25.
Solve the equation ( a x ) a x (b x ) x b a x xb
26.
a b
Solve the equation 4a b 5 x 4b a 5 x 3 a b 2 x 0
27.
Prove that the roots of the equation ( x a)( x c) ( x b)( x d ) 0
28.
are real for any if a b c d . 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 x 2 px q 0 x 2 p1 x q1 0
has real roots if p1 p 2( q1 q) 30.
Prove that at least one of the roots of equation a( x b)( x c) b( x a)( x c ) c ( x a)( x b) 0 are always real.
31.
Find the values of p and q for which the roots of the equation x 2 px q 0 are equal to p and q . 2
32.
1 3 1 Solve x x 4 , when x 0 x 2 x
33.
Solve x4 2 x3 x2 2 x 1 0
34.
Solve ( x 1)( x 3)( x 5)( x 7) 9
35.
Find ’k’ if one root of kx2 14 x 8 0 may be six times other..
36.
If , are the roots of the equation x2 x 1 0 then equation whose roots are 2, 2 .
37.
If 2 3 is one root of x 2 px q 0 then find ' p ' , ' q '
38.
Solve
39.
If the equations (a 2 4a 3) x 2 (a 1) x a 2 1 0 has infinite roots then find ' a ' .
40.
Equations px2 qx r 0 and qx 2 2 pr x q 0 have real roots then show that ' p ', ' q ',' r ' are in G.P..
41.
If ' a ' and ' b ' are the roots of 11x 2 4 x 2 0 then compute the product (1 a a 2 ...... ) (1 b b 2 ..... )
42.
If the quadratic ax 2 bx c 0(a 0) a, b, c are integers, has natural numbers as it roots, then S.T. a divides
2x 6 x 4 5
'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 (1 a 2 ) x 2 2( x a )(1 ax) 1 0 has no real roots.
15
Quadratic Equations
44.
Let , are the roots of the quadratic equation x2 ax b 0 and , be the roots of the equation x2 ax b 2 0 .
Given that 1/ 1/ 1/ 1/ 5 /12 and 24 . Find the value of the coefficient ' a ' . 45.
Solve 3 x2 4 3x 2 4 x 1 4 x 4
Solution To Exercise 1.
(i)
b c and a a 2
b 2c 2 2 ( )2 2 a a
b2 2ac a2
(ii) 3 3 ( )3 3 ( ) 3
3 b c b b 3abc 3 a3 a a a
(iii) 4 4 ( 2 2 ) 2 2 2 2
2
b2 2ac (b2 2ac )2 2c 2 a 2 c 2 2 a4 a a 2 2 (iv) ( ) ( ) 4
b2 4c b2 4ac a2 a a2
(v) 4 4 ( 2 2 )( )( ) b 2 2ac b b 2 4ac 2 a2 a a
2.
b 2 (b 2 ac) b 2 4ac a4
1
1
(i) Sum(s)
( )
b ( a c ) ac
1
1
1
Product (p) = 2
(c a ) 2 ca
2 The equation is = x sx p 0
x
2
(c a ) b ( a c ) x2 0 x ac ac
16
Quadratic Equations ac x 2 b (c a ) x (c a ) 2 0
1
1
1
1
( )
(ii) Sum(s) =
(ac b) 2 bc
1
1
1
1
a
Product (p) = c The equation is: x 2 sx p 0
3.
(ac b 2 ) a x2 x 0 bc c
bcx 2 ( ac b 2 ) x ab 0 is the required equation.
Let , are the roots of given equation.
(m n) and
(m2 n 2 ) 2
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 p (m n) 2 ( m n) 2 2( m 2 n 2 ) ( m 2 n 2 ) 2
The equation is: x 2 Sx p 0 The required equation is
x 2 4mnx ( m 2 n 2 ) 2 0
4.
(i) Discriminant (D) D (a b) 2 4(1)(c 2 ) (a b) 2 4c 2
D0
The roots are real.
(ii) D 4(a b )2 4(a b c)(a b c ) 4 ( a b ) 2 ( a b ) 2 c 2 4 ( a b ) 2 ( a b) 2 c 2 4c 2 (2c) 2
D 0 and also a perfect square hence the roots are rational.
(iii) D (c a )2 4(b c )( a b ) c 2 a 2 (2b ) 2 4 ab 4bc 2ac c 2 a 2 (2b )2 4 ab 4bc 2ac ( c a 2b ) 2
D 0 and also a perfect square.
17
Quadratic Equations
The roots are rational.
5.
(i) D 4(3a 5)2 8(9a 2 25) 4(3a 5)2 5 D 0 , so the roots are non real if a 5 / 3 and real and equal if a . 3
(ii) Simplifying the given equation : 3 x 2 2( a b c) x ( ab bc ca ) 0 D 4( a b c) 2 12( ab bc ca) 4( a 2 b 2 c 2 ab bc ca )
(a 2 b 2 c 2 ab bc ca ) 1 ( a b ) 2 (b c ) 2 ( c a ) 2 2
2 ( a b) 2 (b c ) 2 ( c a ) 2
D 0 , so the roots are real.
Note: If D 0 , then (a b )2 (b c) 2 (c a ) 2 0
abc
If a b c , then the roots are equal. Let be the common root of two equations.
6.
2 2 k 5 0
2 3 4 0
Solving the two equations 2 1 4k 15 8 5 6 k
7.
(3) 2 (4 k 15)(6 k )
4k 2 39k 81 0
k 3
or
k
27 4
Using the condition for common root, we have ( a 2 bc ) 2 (ba c 2 )( ac b 2 )
a 4 b 2 c 2 2a 2bc a 2 bc b3 a ac3 b 2 c2
a ( a 3 b 3 c 3 3abc) 0
a 0 or a 3 b 3 c 3 3abc 0
This is the relation between a, b and c . From second relation, we also have the relation a b c 0 8.
The equation x2 cx d 0 has equal root.
D 0 D c2 4d 0
... (i)
and the equal roots are x
x
( c ) c 2 4 d c 0 c 2(1) 2 2
c is the equal root of this equation. 2
18
Quadratic Equations
This is the common root of the both the equations.
x
c2 c a b 0 4 2
c 2 4b 2ac
4d 4b 2 ac
2(d b) ac
9.
c will satisfy the first equation. 2
c
2
ac 2(b d )
Roots of x 2 px q 0 are
,
Roots of x2 rx s 0
,
4 d from (i )
are
p
and
q
r
and
s
( ) ( ) ( ) ( )
2 ( ) 2 ( )
( 2 r s ) ( 2 r s ) ( 2 p q 0 and 2 p q 0)
( p q r s) ( p q r s )
(r p) ( s q) (r p) ( s q) ( r p ) 2 ( s v ) 2 ( s v )( r p )( ) ( r p ) 2 q ( s q ) 2 ( s q )( r p )( p ) ( q p )[ rq pq ps pq ] ( s v ) 2 ( r p )( rq ps ) ( s v) 2
If the equation have a common root then either r
i.e.
10.
or
s
or
r
or
ps
( r )( s)( p r )( s ) 0
( s q ) 2 ( r p )(rq ps) 0
( s q) 2 ( r p)( ps qr )
( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2
( ) 2 ( ) 2 4 4
p 2 b2 4 q 4c
( ) 2 ( ) 2 ( ) 2 ( ) 2
19
Quadratic Equations
11.
p 2 c b2 q
The given equation can be written as 2
log a log a 3log a 0 log x log ax log a 2 x
( a 0 and a 1, log a 0)
2 1 3 0 y b y 2b y
2(b y )(2b y ) y (2b y ) 3 y (b y ) 0
4b 2 11by 6 y 2 0
(where b log a and y log x )
which is quadratic in y. y
12.
11b 121b 2 96 b 2 12
4b b , 3 2
( y log x and b log a)
y
4 log x log a or 3
xa
4 3
or
xa
log x
log a 2
1 2
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
13.
1 1 2 bm bm m
b mb m 2 b2 m2 m
2m2 2b2 2m2
2m2 b2
log( x 2 2 ax) log(8 x 6 a 3)
x2 2ax 8 x 6a 3
x 2 (2a 8) x 3(2a 1) 0
D (2 a 8) 2 4 3(2 a 1)
For one solution to exist D 0 ( a 4) 2 3(2 a 1) 0
14.
a 2 14a 13 0
(a 1)(a 13) 0
a 1, 13
Let x 1 t 2 x 3 4 x 1 x 8 6 x 1 1
20
Quadratic Equations
t 2 4 4t t 2 9 6t 1
(t 2) 2 (t 3) 2 1
t 2 t 3 1
Case (i)
t2
2t 3t 1
5 2t 1
t 2
x 1 4
x5
... (i)
Case (ii)
2t 3
t 2 3t 1
1 1 true t (2,3)
4 t2 9
5 x 10
Case (iii)
... (ii) t 3
t 2t 3 1
2t 6
t2 9
x 1 9
t 3
x 10
... (iii)
Combining (i), (ii) and (iii); x [5, 10] 15.
x2 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 4( x 2)(4 x) (4 x) 2
(4 x) 4 x 8 4 x 0
(4 x )(5 x 12) 0
x 4, x
12 5
Substitute x 4 , in (i) L.H.S =
42 44 2
R.H.S =
6 4 2
x 4 is a solution.
Substitute x
12 in (i) 5
21
Quadratic Equations
L.H.S =
12 12 2 4 5 5 2 8 5 5
R.H.S = 6 x
2 2 2 2 3 5 5 5
12 18 2 3 5 5 5
12 is also a solution. 5
Note: Whenever we square a equation and find the roots, verify whether the roots satisfy initial equation or not. 16.
log x 2 6 x 8 log 2 x2 2 x 3 ( x 2 2 x ) 0
log 2 x 2 2 x 3 ( x 2 2 x ) ( x 2 6 x 8) 0
log 2 x2 2 x 3 ( x 2 2 x ) 1
x 2 2 x (2 x 2 2 x 3)1
x2 4x 3 0
( x 1)( x 3) 0
x 1
x 3
or
x 1 and x 3 satisfy the condition x2 2 x 0
But at x 3, x 2 6 x 8 9 6(3) 8 1 0 which is not possible
x 3 is not the solution.
x 1 satisfies x2 6 x 8 0 and 2 x 2 2 x 3 0
Also at x 1 ,
x2 6 x 8 3 1
and
2x2 2 x 3 1
Hence x 1 is the only solution. 17.
log 2 x 3 (6 x 2 23 x 21) log 3 x 7 (4 x 2 12 x 9) 4
log (2 x 3) (2 x 3)(3 x 7) log(3 x 7) (2 x 3)2 4 log (2 x 3) (2 x 3) log (2 x 3) (3 x 7) 2 log (3 x 7) (2 x 3) 4 1 log (2 x 3) (3 x 7) 2 log (3 x 7) (2 x 3) 4
Let log (2 x 3) (3x 7) a then log (3 x 7) (2 x 3)
1 a
2 3 a
a
a 2 3a 2 0
(a 1)(a 2) 0
a 1
or
a2
Consider: a 1 log (2 x 3) (3 x 7) 1
22
Quadratic Equations
3x 7 2 x 3
x 4
But x 4 does not satisfy 2 x 3 0 and 3x 7 0 . Hence x 4 is not a solution. Case (ii) a 2 log (2 x 3) (3 x 7) 2
(3 x 7) (2 x 3) 2
4 x2 9 x 2 0
(4 x 1)( x 2) 0
x
1 4
x 2
or
x 2 does not satisfy 2 x 3 0
Hence x 2 is not a solution. x
1 satisfies 3x 7 0 and 2 x 3 0 4 1 4
Also at x , 2 x 3 1
18.
Put
x
1 is the only solution. 4
xa t xb
1 a b t t b a
a b t2 t 1 0 b a 2
a b a b 4 b a b a t 2
a b a b b a b a 2
t
a b
or
b a
a Case (i) t b xa a xb b
bx ab ax ab x (a b) 0 x 0 b a
Case (iii) t ;
23
Quadratic Equations xa b xb a
ax a 2 bx b 2
x ( a b) a 2 b 2
x ab
x 0, a b
19.
Given equation can be written as ( x 1)( x 2 3 x 1) 0
x 1
or
x 2 3x 1 0 x
x 1,
20.
3 9 4 2 3 5 3 5 , 2 2
The given equation can be written as 2 1 13 1 13 x 1 x 2 x 1 0 x 2 2
x2
1 13 x 1 0 2
... (i)
x2
1 13 x 1 0 2
... (ii)
The first equation has two roots x
13 1 2 13 2 13 1 2 13 2 , 4 4
Second equation has no real roots. 21.
Let x2 t
t2 t 6 0
(t 2)(t 3) 0
t 2
Case (i): t 2,
22.
2
x 2
No real roots
Case (ii): t 3,
t 3
or
x2 3
x 3
M b 2 4 a (c 2 a ) ( 10) 2 4(1)(26 2) 100 96 4 0
The equivalent equation is 2 (10) 100 4(26 2) x 1 x 2
24
Quadratic Equations 2 (10) 100 4(26 2) x 1 0 x 2
23.
10 2 2 10 2 x 1 x 2 x 1 0 x 2 2
x2 4 x 1 0
x
x 2 3
x2 6 x 1 0
or
4 16 4 2
x
or
6 36 4 2
x 3 2 2
or
(2 2 ) x 3.2 x 2 0 (2 x ) 2 3.2 x 2 0
Let
t 2x t 2 3t 2 0 (t 2)(t 1) 0
24.
25.
t2
or
t 1
2 x 21
or
2 x 1 20
x 1
x0
or
2
x 4x 3 x 3
x2 5x 6 0
( x 2)( x 3) 0
x2
x3
or
Either x 2 or x 3 is not satisfying the original equation. The original equation has no roots. Rewrite the equation in the form 3
3
(a x ) 2 ( x b) 2 1 2
(a x ) ( x b)
1 2
a b
wherefrom we have, 1
1
a x ( a x) 2 ( x b ) 2 x b a b
or
(a x)( x b) 0
Thus, the required solutions will be x1 a, x2 b
26.
We have 4a b 5 x 4b a 5 x 3 a b 2 x
Squaring both members of the equality and performing all the necessary transformations, we get 4a b 5 x . 4b a 5 x 2(a b 2 x)
Squaring them once again, we find (4 a b)(4b a ) 5 x (4a b 4b a ) 25 x 2 4( a 2 b 2 4 x 2 2ab 4ax 4bx)
25
Quadratic Equations
Hence,
x2 ax bx ab 0
and, consequently, x1 a, x2 b . Substituting the found values into the original equation, we get ba 2 ba 3 b a 0 2 a b a b 3 ab 0
27.
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). Rewrite the given equation as (1 ) x 2 ( a c b d ) x ac bd 0
Set up the discriminant of this equation D( ) . We have D( ) 2 (b d ) 2 2 ( ab ad bc dc 2bd 2 ac ) ( a c) 2
Set up the discriminant of this equation D( ) . We have D ( ) (a c b d ) 2 4(1 l )(ac bd ) .
On transformation we obtain D( ) 2 (b d ) 2 2 ( ab ad bc dc 2bd 2 ac ) ( a c) 2
We have to prove that D( ) 0 for any . Since D( ) is a second-degree trinomial in and D(0) (a c) 2 0 , 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 4( ab ad bc dc 2bd 2ac ) 2 4( a c) 2 (b d ) 2
be less than zero. We have 4(ab ad bc dc 2bd 2 ac) 2 4( a c ) 2 (b d ) 2
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 abcd
28.
The original equation can be rewritten in the following way 3 x 2 2( a b c ) x ab ac bc 0
Let us the prove that 2( a b c) 2 12( ab ac bc) 0
We have, 4( a b c) 2 12( ab ac bc) 4(a 2 b 2 c 2 ab ac bc) 2(2a 2 2b 2 2c 2 2ab 2 ac 2bc) 2 ( a 2 2ab b 2 ) ( a 2 2ac c 2 ) (b 2 2bc c 2 ) 2 ( a b ) 2 ( a c ) 2 (b c ) 2 0
29.
Suppose the roots of both equations are imaginary. Then, p 2 4q 0, p12 4q1 0 Consequently, p 2 p12 4q 4q1 0, p 2 p12 2 pp1 0, ( p p1 )2 0 which is impossible.
26
Quadratic Equations
30.
Let us rewrite the given equation as ( a b c ) x 2 2(ab ac bc ) x 3abc 0
Prove that its discriminant is greater than or equal to zero. We have, 4(ab ac bc) 2 12 abc (a b c) 2 ( ab ac) 2 ( ab bc ) 2 ( ac bc) 2 0
31.
By properties of the quadratic equation we have the following system p q p, pq q
From the second equation we get, q( p 1) 0
Hence, either q 0 or p 1 . From the first one we find if q 0 , then p 0 ; if p 1 , then q 2 . Thus, we have two quadratic equations satisfying the set requirements x 2 0 and x 2 x 2 0 2
32.
1 3 1 x x 4 x 2 x 2
2
1 3 1 1 3 1 x 4 x 4 x x 0 x 2 x x 2 x 3 1 t 2 t 0 where t x 2 x
3 t t 0 t 0 or t 3/ 2 2 x
1 1 3 0 or x x x 2
x 2 1 or 2 x 2 3 x 2 x 1 or 2 x 2 4 x x 2 0
x 1 or ( x 2)(2 x 1) 0 x 1 or 2 or 1/ 2
Solution set 1, 2, 1/ 2
33.
x4 2 x3 x2 2 x 1 0 x2 2x 1
2 1 0 x x2
1 1 x2 2 2 x 1 0 x x 2
1 1 x 2 2 x 1 0 x x
y 2 2 y 3 where y x
1 x
( y 3)( y 1) 0
27
Quadratic Equations
y 3 0 x
y 1 0
or
1 1 3 0 or x 1 0 x x
x 2 3 x 1 0 or x 2 x 1 0
34.
x
3 9 4 1 1 4 or x 2 2
x
3 5 1 3i or 2 2
( x 1)( x 3)( x 5)( x 7) 9 ( x 7)( x 1)] ( x 3)( x 5) 9
( x 2 8 x 7)( x 2 8 x 15) 9 (t 7)(t 15) 9 where t x 2 8 x t 2 22t 105 9 0 t 2 22t 96 0 t 2 16t 6t 96 0
t (t 16) 6(t 16) 0 (t 16)(t 6) 0 ( x 2 8 x 16)( x 2 8 x 6) 0 x 2 8 x 16 0 or x 2 8 x 6 0
( x 4) 2 0 or x
8 64 24 2
x 4, 4 or x 4 10
Solution set 4, 4 10
35.
Given equation is kx2 14 x 8 0 Let the roots be , 6 Sum of the roots, 6 14 / k 7 14 / k 2 / k Product of the roots, (6 ) 8 / k 3 2 4 / k 3(2 / k ) 2 4 / k k 3
36.
Let f ( x ) x 2 x 1 Equation whose roots are 2, 2 is f ( x 2) 0 f ( x 2) ( x 2) 2 ( x 2) 1 0 x 2 3x 1 0
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 p p 4 Product of the roots = (2 3)(2 3) q q 4 3 1
38.
Squaring both sides 2 x 6 x 4 2 (2 x 6)( x 4) 25
28
Quadratic Equations
2 (2 x 6)( x 4) 25 4 6 3 x 27 3x
Again squaring both sides and simplifying. We arrive at the equation x 2 170 x 825 0 x 5, 165 Direct substitution of these values in the original equation shows that x 5 is root and x 165 is not. Why x 165 is not a root of an equation? Reason: (Extraneous root appeared due to squaring the equation) 39.
Any quadratic equation ax 2 bx c 0 (a 0) has infinite solutions then a 0, b 0, c 0 .
0.x 2 0.x 0 0 is satisfied for all real 'x'
a
2
4a 3 x 2 ( a 1) x a 2 1 0 has infinite roots if
( a 2 4a 3) 0, ( a 1) 0, a 2 1 0 simultaneously..
that is possible only for a 1 . 40.
px 2 qx r 0 have real roots D 4q 2 2 pr 0 q 2 pr
... (1)
qx2 2 pr x q 0 have real roots D 4qr 4q 2 0 pr q 2 0 pr q 2 q 2 pr q 2 pr
... (2)
From (1) and (2) q 2 pr and q 2 pr q 2 pr 41.
Let S (1 a a 2 ...... ) (1 b b 2 ..... ) S
S
a 1 1 .S 1 r 1 a a b
1 1 ,S 1 a b ab 1 (a b) ab
a, b are the roots of 11x 2 4 x 2 0 a b 4 /11, ab 2 /11 S
42.
1 1 11 S 4 2 6 5 1 1 11 11 11
Consider example x 2 8 x 7 0, ax 2 bx c 0 ac 7 which can’t be expressed as sum of square of two natural numbers
' a ' is wrong.
Consider same example x2 8 x 7 0 where a, b, c are integers and roots also natural numbers '1' and '7 '. a 1, b 8, c 7 ' 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 ax2 bx c 0
b c , a a
29
Quadratic Equations
, are integers , are integers
b c , are integers ' a ' divide ' b ' & ' c ' a a
' b ' is correct
43.
(1 a 2 ) x 2 2( x a )(1 ax) 1 (1 a 2 ) x 2 2( x ax 2 a a 2 x) 1 (1 a 2 ) x 2 x (2 2a 2 ) 2ax 2 1 2 a (1 a 2 2a ) x 2 (2 2a 2 ) x (1 2 a ) (1 a ) 2 x 2 2(1 a 2 ) x (1 2a )
D B 2 4 AC D 4(1 a 2 ) 2 4(1 a ) 2 (1 2a ) D (1 a 2 4 (1 a ) 2 (1 2a )
D 4(1 a ) 2 (1 a 2 2a 1 2a ) D 4 a 2 ( a 1) 2 0 For all value of ' a ' roots are real.
44.
, are roots of x 2 ax b 0 a, b
a 1 1 a b b
, are roots of x 2 ax (b 2) 0 a , b 2
a 1 1 a b2 b2
24 b(b 2) 24 b2 2b 24 0 b 2 6b 4b 24 0 b(b 6) 4(b 6) 0 (b 4)(b 6) 0
b 4,6
1 1 1 1 a a 5 b b 2 12
If b 4 45.
a a 5 a a 5 3a 2a 5 a 5 a5 4 6 12 4 6 12 12 12 12 12
The equation can be written as 3x 2 4 x 4 4 3x 2 4 x 1 0
Put
3x 2 4 x 1 4 3x 2 4 x 1 3 0
3 x2 4 x 1 t 2
t 2 4t 3 0
(t 1)(t 3) 0
t 1
or
t 3
3 x 2 4 x 1 1 or 3
3x 2 4 x 0
x
or
3x 2 4 x 8 0
4 2 , 0, 1 6 . 3 3
30
