Solutions to Maths workbook - 1 | Quadratic Equations

MEQB Level - 1

1.(C) p  13, q  2 15  30 ;    x 2  px  q  0  x 2  13x  30  0  x  3,  10

2 2 2 2
2.(C) x  a  b   x  b  c  0  x  a  b  
 0 x b  c   x a b  0  x b c d
 x  a  b, x  b  c  a  b  b  c  a , b, c are in A.P.
1 1 1
3.(B)    
 x  p x  q  r 2x  p  q     x 2  p  q  2r x  pq  qr  pr  0
 
x p x q r
Sum of the roots  0  p  q  2r  0  p  q  r  3r

4.(A) 2
The roots of x  2mx  m  1  0 are 2
2m  4m 2  4 m 2  1    m 1
2
Given 2  m  1, m  1  4  m  1, m  3   1  m  3
2
5.(A) Discriminant  c  a    4 b  c a  b  c 2  a 2  2ca  4ab  4b 2  4ac  4bc
  
2
 a 2  4b 2  c 2  4ab  4bc  2ac  a  2b  c   0  The roots are real.

x 2
6.(D) y  yx 2  5y  1 x  9y  0 
  5y  1  4y 9y  0   11y 2  10y  1  0
 
2
x  5x  9
1
 11y 2  10y  1  0  11y  1y  1  0    y 1  Maximum value = 1
11
7.(B) If ,  are the roots then     a  2,   a  1
2 2
Sum of the squares of the roots, S  2  2        2  a  2   
 2 a  1
S is least  when a 
  1
 2
2 1
2
8.(A) x  74 3  2  3  2 3

1 1
x 
 2 3    2  3   2  3   4
x 2  3 
9.(C) Let the roots be , 2

Then  2    1 and    2   3  k or 2    1  0

3

 
1  5
  3

 1  5    1  5 5  3 5  15   2  5  k or k  2 5
2 8 8
10.(C)     
P x  x  1 x  2 ax  b 
P  1  4   
2 b  a  4  a b  2

P  2   28  4  2a  b   28  2a  b  7

On solving a  3, b  1

11.(A) Remainder  P 3  32  

Solutions 37 Quadratic Equations

12.(A)  1

p 2  4q  1
p 2  4q  4q 2  1  4q 2
2
p 2  4q 2  1  2q
 
2
13.(D) x  2 a  1 x  9a  5  0

For both roots to be –ve

Desired graph is
D0
b
2a
0  
f 0 0

2
a  1  9a  5  0
a   , 1  6,  

b
0 Given
2a

 a 1  0 
a 1  0
a  1
 
f 0 0

9a  5  0
5 5 
a 
9
  , 1  6, 
9  
 
is the final solution.

14.(C)  
3x 2  2 a 2  1 x  a 2  3a  2  0

For roots to be of opposite signs derived graph is f 0  0  

a 2  3a  2  0  a  1a  2   0

a  1, 2 
15.(A) ( x  m )2  1  x  m  1,m  1
Case 1: Both roots lie in (-2,4)
So, m-1 > -2 and m+1<4
m  ( 1,3)
Case 2:
Exactly one root lie in (-2,4)
if (m  1)  ( 2,4)  m  1  4  m  [3,5)
and if

Solutions 38 Quadratic Equations

 (m  1)  ( 2,4)  m  1  2  m  ( 3, 1]
 m  ( 3, 1]  ( 1,3)  [3,5)
 m  ( 3,5)

16.(C) Equation of curve y  ax 2  bx  c

AP 

 b 2  4ac  3 
2b
3
4a 4a
b
So 3  OP = 3
2a
17.(D) 
x 2  x n a 2  3a  2  a 2  4  0 
For defining logarithm
a 2  3a  2  0
a  1a  2  0
a   , 1   2,   . . . .(i)

Roots are of opposite sign so

f 0  a2  4  0
 
a   2, 2  . . . .(ii)
From (i) and (ii)

a  2, 1 
b c
18.(D) Since,      and  
a a
2 b 2  2ac
Also,  2  2        2 
a2
 b 2  2ac   b
a  b 

Now,





a 2  2  b     
 
 a 2 

 a
 

2
a  b a  b  a 2  ab     b 2
  c 2  b a
  a  ab     b 2
a   a
   
x 2
19.(D) Let y   2yx 2  3y  1 x  6y  2  0
 
2
2x  3x  6
2
But x is real, then 3y  1     
 4 2y 6y  2  0  D  0 

1 1
 13y  13y  1  0  
13
y
3
2
20.(A) Let  and  be the roots of equation x  (a  2) x  a  1  0
Then,     a  2 and   a  1
 3  3  (a  2)3  3(a  1)(a  2)
The value of 3  3 will be least, if a  2  0
2
21.(B) Given that, 32 x  7 x  7  32  2x 2  7 x  7  2  2x 2  7 x  5  0

Now, D  b 2  4ac  7  2  4  2  5  49  40  9  0
Hence, it has two-real roots.

Solutions 39 Quadratic Equations

22.(C) Since, 1  p is a root of quadratic equation x 2  px  1  p  0
    …… (i)

 
So, 1  p satisfied the above equation  1  p 2  p 1  p   1  p   0
 1  p 1  p  p  1  0  1  p  2  0  p 1

On putting this value in equation (i), we get : x 2  x  0  x  0,  1

23.(B) Given equation x 2  3 x  2  0

Case I : When x  0 , then x  x  x 2  3x  2  0   x  1 x  2   0  x  1, 2

Case II :When x  0 , then x   x  x 2  3x  2  0   x  1 x  2   0  x   1,  2

Hence, four solutions are possible.
 x 1 x 2  5x  50 
24.(C) 2 1

 
( x  1) x 2  5x  50  0   x  1 or x 2  5x  50  0

Sum of all real values  1  ( 5)   4

25.(C) p(0)  1
p ( x ) leave remainder 4 when divided by ( x  1)
 p( x )  4
Similarly p( 1)  6

Assume p (1)  ax 2  bx  c
p(0)  1  c 1
p(1)  a  b  1  4
( 1)  a  b  1  6
 a  4, b  1

p( x )  4 x 2  x  1
p(2)  4  4  2  1  15
p( 2)  16  2  1  19

26.(B) Given equations are 2x 2  3 x  4  0 …… (i) and ax 2  bc  c  0 …… (ii)

Since, equation (i) has imaginary roots so, equation (ii) will also have both roots same as equation (i)
a b c
Thus,   Hence a : b : c is 2 : 3 : 4
2 3 4

27.(C) x 2  2x  3  4  0

3
Case 1 : x 
2

x 2  2x  7  0
2  32 3
x  x  1  2 2,  1  2 2  (Rejcet)
2 2
3
Case 2: x 
2

x 2  2x  1  0

Solutions 40 Quadratic Equations

 2 44
x 
2
x 1 2
3
x  1  2, 1  2  (Reject)
2
Sum of roots  1  2 2  1  2  2

28.(A) 3x 2  x  5  x  3
Squaring x  3

3x 2  x  5  x 2  6 x  9

x 2  7x  4  0

2x 2  8x  x  4  0
2x ( x  4)  1 ( x  4)  0
1
x  4 or x 
2

Solutions 41 Quadratic Equations

 Solutions to Maths workbook - 1 | Quadratic Equations
MEQB Level - 2

29.(A) Let two consecutive integers n and (n + 1) be the roots of x 2  bx  c  0 . Then, n + (n + 1) = b and


n n 1  c   b 2  4c  2n  1  2  4n n  1  1
30.(A) Let f x  x 2n  1
 
 
At x  1, f x  0 Hence, for no other real value of x, f (x) is zero.

31.(A) ,  are the roots of ax 2  bx  c  0  a 2  b   c  0, a 2  b   c  0

c c
   
 a   b  c,  a   b  c  a   b    
, a  b  

3 3 3 3 3 3
  

  
 
  
 
  
   
    
  0
 a  b   a  b   c    c  
        c3 c3

1 1  a2
32.(B) 2 ,  2 are the roots of a 2 x 2  x  1  a 2  0   2  2  
a2
 
,  2 2 
a2
 a2 1  2
2 2 1
  1  4  4   2  1   2  2  2  1
 
  2  2     2  2
a4
 4 
 a2  a4
 4 22 
a2 
 
a2  a2
 
1  1 1 1 
2
  
2   2  2  2  2   2 
 2  a 2
  1
a 2 

33.(D) From the given equation tan 30  tan15   p and tan 30 tan15  q
tan 30  tan15 p
Now tan 30  15    1  tan 30 tan15
 1 
1q
 1  q  p  q  p  1  2  q  p  3

34.(D) Since roots are real, we have cos p  1  0 and discriminant  0  cos2 p  4 cos p  1 sin p  0  
 cos2 p  4 1  cos p sin p  0  sin p  0
   cos p  1  0,  1  cos p  1  1  cos p  0 
 p  0,    p  0,     cos p  1
35.(D) We have b  c x 2  c  a x  a  b  0
      . . . .(i)

a , b, c are in A.P. Let common difference be k.

 (i)  kx 2  2kx  k  0  x 2  2x  1  0  x  1, 1

 1 is a root of 2 c  a x 2  b  c x  0  2 c  a  b  c  0  2a  b  3c  0
       
a b c a b c
a , b, c are in A.P.  a  2b  c  0         (say)
1 6 32 4  1 7 1 5

 a  7, b  , c  5  a 2  49 2 , b 2  2 , c 2  252  a 2  b 2  2c 2  a 2 , c 2 , b 2 are in A.P.

36.(C) 2x  3 2x  5  x  1 x  2   30    

  2x  3 x  1   2x  5 x  2   30
    
  2x 2 x 3
2x 2  x  10  30 
 y  3y  10   30 where y  2x 2  x

Solutions 42 Quadratic Equations

  y 2  13y  0  y  0 or 13  2x 2  x  0 or 2x 2  x  13  x  0

1 1  1  104
or  or x   The number of rational roots = 2.
2 4

37.(C) Note that f a  a 2 , f b  b 2 and f c  c 2

     
If g x  f x  x 2 , then g a  g b  g c  0
         
That is, f x  x 2 is quadratic expression which vanishes at three distinct points a, b and c.
 
Thus, f x  x 2  0 or f x  x 2
   
x a
38.(C) Let y  . Then yx 2  3yx  2y  x  a  yx 2  3y  1 x  2y  a  0
   
2
x  3x  2

x is real 
 3y  1 2  4y  2y  a   0  9y 2  6y  1  8y 2  4ay  0  y 2  6  4a y  1  0  
Since y is real, y 2  6  4a y  1  0 
  The roots of y 2  6  4a y  1  0 are real and equal or
 
imaginary

 Discriminant  0  6  4a  2  4  0  36  16a 2  48a  4  0  16a 2  48a  32  0

16a 2  48a  32  0  a 2  3a  2  0  a  1 a  2  0  a  1   
or 2 162  48a  32  0  1  a  2

39.(B) Remainders of f (x) when divided by x  1, x  2, x  3 are f 1 , f 2 , f 3 respectively.     

  
Then f 1  3, f 2  7, f 3  13  
Since  x  1 x  2  x  3  is a third-degree expression, the remainder of f (x) when divided by

 x  1 x  2 x  3 is of second degree  f x  q x  1 x  2 x  3  ax 2  bx  c

     
f 1  3  a  b  c  3 . . . .(i),  
f 2  7  4a  2b  c  7 . . . .(ii)

f  3   13  9a  3b  c  13 . . . .(iii)

(ii) – (i)  3a  b  4 . . . .(iv)

(iii) – (ii)  5a  b  6 . . . .(v)

(v) – (iv)  2a  2  a  1  b  1, c  1  Remainder  x 2  x  1

40.(C) x 2  4y 2  8x  12  0  x 2  8x  4y 2  12  0   . . . .(ii)

2
(ii) is a quadratic in x, x is real  Discriminant  0   8   4 4y 2  12  0  
 16y 2  16  0  y 2  1  0   1  y  1  y must lie between 1 and 1.

x x
 5   12  5
41.(C)     1  cos x   sin x   1 , where cos  
 13   13  13
   
Equality holds for x = 2. If x  2 , both cos  and sin  increase (being positive fractions)

Solutions 43 Quadratic Equations

 So, cos x   sin x   1 if x < 2.
2a x  1 sin 2 
 
42.(D) y  yx 2  y sin 2   2ax sin 2   2a sin 2   yx 2  2ax sin 2   2a  y sin 2   0  
2
x  sin 2 
x  R  4a 2 sin4   4y 2a  y sin 2   0  a 2 sin2   2ay  y 2  0  y 2  2ay  a 2 sin2   0
 
2a  4a 2  4a 2 sin 2 
Let y 2  2ay  a 2 sin2   0  y   y  a  a cos   a 1  cos  
2
 
 y  2a sin2 or 2a cos2
2 2
 
 y  2ay  a sin   0  y cannot lie between 2a sin 2
2 2 2
, 2a cos2
2 2

43.(A) ax  by  1  by  1  ax  y 
1  ax 
b
2
(1  ax )
cx 2  dy 2  1  cx 2  d  1  b 2cx 2  d (1  a 2 x 2  2ax )  b 2
b2
 (b 2c  a 2d )x 2  2adx  d  b 2  0

Both the equations have only one solution  Discriminant = 0  4a 2d 2  4(b 2c  a 2d )(d  b 2 )  0
   
a 2d 2  b 2c  a 2d d  b 2  0  a 2d 2  b 2cd  b 4c  a 2d 2  a 2b 2d  0  b 4c  a 2b 2d  b 2cd

b2 a2
  1
d c

44.(C) 0  sin   sin   sin 
2
   
Now, the given equation is x  sin  x  sin   x  sin  x  sin   x  sin  x  sin   0     
        
Let f x  x  sin  x  sin   x  sin  x  sin   x  sin  x  sin   0  
 f  sin     sin   sin   sin   sin    0  f  sin     sin   sin   sin   sin    0

 f  sin     sin   sin   sin   sin    0

Hence, equation f  x   0 has 1 root between sin  and sin  and other root between sin  and sin  .

45.(B) ,  are roots of x 2  px  q  0       p and   q

,  are also roots of x 2n  pn x n  qn  0  2n  p n n  qn  0 . . . .(i)

2n  pn n  qn  0 . . . .(ii)
(i) – (ii)  2n  2n  pn n  n  0     n  n
 
n  n  p n  0  n  n   pn . . . .(iii)

n
  n n   n
and are roots of x n  1   x  1  0   1    1  0  n  n      0  
  n   
  

  pn   p  n  0  n
 pn  1  1   0 This is possible only when n is even.
 
 
46.(C) If  is the common root then 1  2a 2  6a   1  0   . . . .(i)

a 2    1  0 . . . .(ii)

Solutions 44 Quadratic Equations

 2  1 6a  1 a 1
   2  ,
6a  1 a  1  2a 2 2 2
1  2a  6a 6a  2a  1 6a  2a  1
2
a  1 6a  1

2

2
 a  12   6a  1  6a 2  2a  1
6a  2a  1
6a 2
 2a  1 
 a 2  2a  1  36a 3  12a 2  6a  6a 2  2a  1  36a 3  19a 2  6a  0
2 3
  
a 36a 2  19a  6  0  a 4a  3 9a  2  0  a     ,
9 4
47.(A) p, q, r are in A.P.  p  r  2q
2
p r 
The roots of px 2  qx  r  0 are real  q 2  4 pr  0     4 pr  0  p 2  r 2  14 pr  0
 2 
 
2 2 2
r  r  r  r  r
 1     14    0    7   49  1  0    7   48  7  4 3
p p p  p  p
       
48.(AC) ,  are the roots of x 2  px  q  0       p,   q

 4 , 4 are the roots of x 2  rx  s  0   4  4  r ,  44  s

2
Discriminant of x 2  4qx  2q 2  r  0 is  4q   4 2q 2  r  
2

 16q 2  8q 2  4r  8q 2  4r  822  4 4  4  4  4  4  222  4 2  2      0

49.(A) Since, sin A, sin B, cos A are in GP.

 sin 2 B  sin A cos A . . . .(i)

Also, x 2  2x cot B  1  0 [Given]

Now, b 2  4ac  4 cot 2 B  4 

4 cos2 B  4 sin 2 B


4 1  2 sin 2 B 
sin 2 B sin 2 B
2


4 1  2 sin A cos A   sin A  cos A 
 4  0 [Using (i)]
 
sin 2 B  sin B 
 Roots of given equation are always real.

50.(A) Since, a, b and c are the sides of a ABC , then : a b  c  a 2  b 2  2ab  c 2

Similarly, b 2  c 2  2bc  a 2 , c 2  a 2  2ca  b 2

a 2  b2  c2
On adding, we get :  
a 2  b 2  c 2  2 ab  bc  ca
  
ab  bc  ca
2

a 2  b2  c2

Also, D  0, a  b  c 2  3 ab  bc  ca   0 
ab  bc  ca
 3  2 . . . .(ii)

Solutions 45 Quadratic Equations

 4
From equations (i) and (ii), 3  2  2   
3
2
x  2x  11
51.(C) Let m
x 3
 x 2  2x  11  mx  3m  x 2  (2  m )x  (3m  11)  0 ............(i)
Since x is real, therefore, discriminant of equation (i) must be non-negative.
i.e. (2  m )2  4  1  (3m  11)  0
 4  4m  m 2  12m  44  0  m 2  16m  48  0
 (m  4)(m  12)  0  m  4 or m  12

52.(C) a  b  c  0 ; a 2  b2  c2  1
2

a 4  b 4  c 4  a 2  b2  c 2   2 a b 2 2
 
 b 2c 2  c 2a 2  1  2 a 2b 2  b 2c 2  c 2a 2  . . . . .(i)

a  b  c 2  a 2  b 2  c 2  2 ab  bc  ca 
1

0  1  2 ab  bc  ca   ab  bc  ca 
2
2
ab  bc  ca  
 a 2b 2  b 2c 2  c 2a 2  2 ab 2c  abc 2  bca 2 
1
4
 
 a 2b 2  b 2c 2  c 2a 2  2abc a  b  c  
1
a 2b 2  b 2c 2  c 2a 2  Put in (i), we get :
4
1 1 1
a 4  b4  c4  1  2  1 
4 2 2
1
a 4  b4  c4 
2
53.(D)  
2 x  p x q  p q p x q x      p  q 2   x  p 2   x  q 2
     2   x  p 2   x  q 2
2 x  p x q  p q

 2x 2  2x  p  q   2 pq  p 2  q 2  2 pq  2x 2  p 2  q 2  2x  p  q 
It is an identity so infinite values of x are possible.

Solution to 54 to 56
54.(A) 55.(B) 56.(C)

p
 3,

 p 2  8q  1
4 8

p  12, 8q  p 2  8
8q  144  8
152
q
8
q  19

f x  2x 2  12x  19
 
D  144  152  0

Solutions 46 Quadratic Equations

57.(D)  m  0 and D  0
16  4m 3m  1  0 

4  m 3m  1  0 
3m 2  m  4  0
3m  4 m  1  0
 4 
m   ,

  1, 
3 
 

But m 0
 m  1,   
Least integral value = 2

58.(B) a  6  x 2  2  2ax  1  a  6  x 2  2ax  3  0

D0

4a 2  12 a  6  0
 
a 2  3a  18  0
a  6 a  3  0
a   6, 3 

Least integer  5

59.(C) i.e. x 2  6 x  8  0 . . . .(i)

and 8x 2  6 x    0 . . . .(ii)

Have a common root.
Solving (i) and (ii) for common root
(i) – (ii)

   8 x 2  8     x  1

So, from (i)   2, 14


If both roots are common
8
 1    8 So 
  2, 14,  8

60.(C) 3

2 x 2 x 6   2.3 x  x 6   1  0
2

 3
x  x 6   1  x 2  x  6  0
2
  x  3 x  2  0  x  2, 3

61.(A)  
Given, log10 x 3  y 3  log10 x 2  y 2  xy  2  
 log10
x 3
 y3  2   
log10 x  y  2  x  y  100
2 2
x  y  xy
Using AM  GM
x y x y 100
  xy  xy    xy  2500
2 2 2

Solutions 47 Quadratic Equations

62.(A) Let  be the common root of x 2  px  qr  0 and x 2  qx  rp  0 .

Then 2  p   qr  0 . . . .(i) and 2  q   rp  0 . . . .(ii)

Subtracting (ii) from (i), we get :
 p  q    r q  p   0 or  p  q    r   0   r (Since p  q).

Similarly, p and q are other two common roots.

 Product of common roots = pqr

63.(D)   
Let, f x  x  a1 2   x  a2 2  . . . .   x  a n 2

 nx 2  2 a1  a 2  . . . .  an x  a12  a 22  . . . .  an 2
  
    1 2 
  2 a1  a 2  . . . .  an  a  a  . . . .  a
n
f (x) assumes least value when x 
2n n

64.(AB) Let  be a root of x 2  x  m  0 . Then 2 is a root of x 2  3x  2m  0

 2    m  0 and 4 2  6  2m  0 or 22  3  m  0 . . . .(ii)

or 22  2  2m  0 . . . .(i)
Subtracting (ii) from (i), we get :   m  0 or  = –m

Putting value of  in (i), we get : 2m 2  4m  0 or m  0,  2

65.(C) ,  are roots of the equation x  a   x  b   c  0 or x 2  a  b x  ab  c  0

 
    a b . . . .(i)
and   ab  c . . . .(ii)
Let p, q be roots of the equation x  c   x  c    c   
or x 2      2c x  c 2  c     1    0
   
 p  q      2c . . . .(iii) and pq  c 2  c     1  
  . . . .(iv)

Putting value of  +  from (i) into (iii), (iv) and value of  from (ii) into (iv), we get :


p  q  a  b  2c  a  c  b  c    and pq  c 2  c a  b  1  ab  c  a  c . b  c
     
 p  a  c and q  b  c

66.(D) 2e 2 log k  1  31 or 2e 2 log k  32 or e 2 log k  16

or k 2  16 or k =  4.

x 2  2x  a
67.(B) y gives y  1 x 2  4y  2 x  3ay  a  0
     
x 2  4 x  3a
Since x is real D  0 ,

 4y  22  4 y  13ay  a   0 or  4y 
2  4y  1  a y  1 3y  1  0
  
or  4  3a  y 2  4a  4  y  1  a   0
Since y  R 4  3a  0 and  4a  4 2  4  4  3a 1  a   0
4
or a
3
and   
16 a 2  2a  1  4 3a 2  7a  4  0 
Solutions 48 Quadratic Equations
 4 4
 a and 4a 2  4a  0  a and  
a a 1  0
3 3
4
 a and a  0, 1  a  0, 1
3
68.(C) Determinant of f x  D1  b 2  4ac and Determinant of g x  D 2  b 2  4ac
   
D1  D 2  2b 2  0  Atleast one of D1 and D2

    
f x . g x  0 has at least two-real roots.

69.(D)          2         
Since ,  are roots of x 2  px  r  0,      p and   r           2  p  r
Since  is a root of x 2  px  q  0, 2  p   q  0 or 2  p   q          q  r
2x  1 2x  1
70.(AD)  0 gives 0
2x 3
 3x 2
x  
x x  1 2x  1 
+ – + – +

–1 –1/2 0 1/2
 1  1 
Solution : x   ,  1    , 0    ,  
 2  2 
   

71.(B) x 4  4 x  1  0 gives x 4  4 x  1

From the figure it is clear that the given

equation has 2 positive real roots.

72.(C) x 4  4 x  1  0 gives x 4  4 x  1
From the graph it is clear that the given
equation has one negative real root.

Solutions 49 Quadratic Equations

73.(B) From the above graph we see that there are 2-real roots for the equation x 4  4 x  1  0 . The equation
has 4 roots. Hence remaining 2 roots are complex roots.

74.(A) For x 2  x  1  0, D  0 . So, it has non-real roots.

As x2  x 1 is a factor of ax 3  bx 2  cx  d , roots of x2  x 1  0 are also roots of

3 2
ax  bx  cx  d  0
3
 ax  bx 2  cx  d  0 has 2 non-real roots.
d
Let x = p be the remaining real root. Product of roots  p [Product of roots of x 2  x  1  0 ]
a

75.(C) x 3  px 2  qx  r  0

, ,  are roots of equation  2  2  2  2  2  2  2
    
 2   2     2  2      2   2    3 
       
                      3             3  qp  3r 
76.(B) , ,  are roots of x 3  px 2  qx  r  0 ;       p,    q,   r
 1 1 1  22  2  2  2 2       2  2 2  2  2  q 2  2rp
     
 2 2 2 
  22 2 22  2 r2
2
 x 
77.(A) x2    8
 x 1
 

x 2 x 1 2  x 2  8  x  12
x 4  2x 3  2x 2  8x 2  16 x  8

x 4  2x 3  6 x 2  16x  8  0

 x  2 x 3  6x  4   0
2
 x  2 x 
2  2x  2  0

Number of roots = 3

Solutions 50 Quadratic Equations

78.(B) mx 2  9mx  5m  1  0
   x R

It is possible only if m  0 and D  81m 2  4m 5m  1  0  

 61m 2  4m  0

m 61m  4  0 
 4 
m   0, 
 61 
 
 4 
If m = 0 then expression is also positive so m  0, 
 61 

79.(B) Since, x 4  20  x 4  20  22  20

Let x 4  20  y

 y 2  y  42  0  y 2  y  42  0

 y  6 y  7   0  y6  y  7 
 x 4  20  6  x 4  20  36

 x 4  16  x  2
Hence, the number of real roots of the equation is 2.
80.(C) Given equation is x 4  x 3  4 x 2  x  1  0.
We should try to factorize the given equation by combining terms.
( x 4  2x 2  1)  ( x 3  2x 2  x )  0

 ( x 2  1)2  x ( x 2  2x  1)  0  ( x  1)2 ( x  1)2  x ( x  1)2  0

 ( x  1)2[( x  1)2  x ]  0  ( x  1)2 ( x 2  3x  1)  0

3  5
 Roots are x  1, 1, Positive roots are 1, 1
2
 3  5 3  5 
 Number of positive roots = 2 Note : , are negative values 
 2 2 

81.(B) Let a be common root of x 3  ax  1  0 and x 4  ax 2  1  0.

a will satisfy both x 3  ax  1  0 and x 4  ax 2  1  0.

 a 3  aa  1  0 … (i)
4 2
a  aa  1  0 … (ii)
Multiply (i) with a and subtract (ii) from (i) to get,
a 1  0  a  1.
Replace a  1 in (i) to get, 1  a  1  0  a  2
Alternate Method :
By inspection, x = 1 satisfies both equations.  x  1 is a common root.
Replace x  1 to get a  2 .

Solutions 51 Quadratic Equations

82.(C) If A  first term, D = common difference of an AP, then
1 1
 t p  A  ( p  1)D  … (i) and t q  A  (q  1)D  … (ii)
q p
Solving equations (i) and (ii), we get :
1
AD  t pq  A  ( pq  1)D  1
pq
Also, sum of the coefficients of the given equation is
( p  2q  3r )  (q  2r  3 p )  (r  2 p  3q )  0
 x  1 satisfies the given equation, [Using (i) and (ii)]
 x  1 is a root of the equation. Hence, x  1  t pq

83.(B) As , ,  and  are roots of x 4  x 2  1  0, we get :

4  2  1  0  (2 )2  (  )2  1  0   2 satisfies x 2  x  1  0 … (i)

4  2  1  0  (2 )2  ()2  1  0  2 satisfies x 2  x  1  0 … (ii)

 4  2  1  0  (  2 )2  (  )2  1  0   2 satisfies x 2  x  1  0 … (iii)
4 2 2 2 2 2 2
   1  0  ( )  ( )  1  0   satisfies x  x  1  0 … (iv)
On combining (i), (ii), (iii) and (iv), we get :
 2, 2 ,  2 , 2 are roots of x 2  x  1  0

But x 2  x  1  0 is a degree two equation. Therefore it cannot have than two roots. It means we have
to use two equations of the same nature to accommodate four roots.
  2, 2 ,  2 , 2 are roots of ( x 2  x  1)( x 2  x  1)  0  2, 2 ,  2 , 2 are roots of

( x 2  x  1)2  0

84.(C) For roots to be real, Discriminant  q 2  4 p  0 … (i)

Case I p  1

 q2  4 {p = 1 is replaced in (i)}  q 2  q  2,3,4

 order pairs (1, 2), (1, 3), (1, 4) will make roots real … (ii)
Case II p2

 q2  8  q2 2  q  3, 4
 order pairs (2, 3), (2, 4) will make roots real … (iii)
Case III p3

 q 2  12  q2 3  q4  q4

 order pairs (3, 4) will make roots real … (iv)
Case IV p4

 q 2  16  q4  q4
 order pairs (4, 4) will make roots real
Combining (ii), (iii), (iv), (v) total number of order pairs that will make roots real = 7.

Solutions 52 Quadratic Equations

85.(BD) x  1 is root
a b a b
Let other root    Product of the roots  (1)( )   roots are 1,
b c b c

86.(A) x  2x 2  6 x  9
D  36  4( 2)( 9)  36  72  0 and a  0

So quadratic expression 2x 2  6 x  9 is always negative whereas x is always +ve

 Equation will not hold for any x.  x 

So x  2x 2  6 x  9 has no solution.

87.(A) In the given expression, coefficient of x 2  0.

Hence, the given expression will assume positive values only if
2
 (3a  1)  4  1  (2a 2  2a  11)  0 [Using b 2  4ac  0]

 9a 2  6a  1  8a 2  8a  44  0  a 2  14a  45  0 
(a  5)(a  9)  0
 5  a  9.

88.(B) ax  2bx  c  0, a1x 2  2b1x  c1  0 have a common root  (ac1  a1c )2  4(ab1  a1b )(bc1  b1c )
2

[From condition of one common root]

Let the discriminant of the given equation be  :

  (2bb1  ac1  a1c )2  4(b 2  ac )(b12  a1c1 )

 4b 2b12  (ac1  a1c )2  4bb1(ac1  a1c )  4b 2b12  4acb12  4a1c1b 2  4aca1c1

 (ac1  a1c )2  4aca1c1  4bb1ac1  4bb1a1c  4acb12  4a1c1b 2

 (ac1  a1c )2  4(ab1  a1b )(bc1  b1c )  0  The roots are equal.

q r
89.(D) Since  and  are roots of px 2  qx  r  0, p  0    ,  
p p
Since, p, q and r are in AP  2q  p  r
1 1   q 4r
Also,  4  4      4  
   p p
 2q  p  r   
2  4r  p  r  p   9r
q 4r 4r 4 r r 1
      and    
p p 9r 9 p 9r 9
16 4 16  36 52 2
    2     2  4  81  9 
81
    2  81
  
9
13

90.(A) Given equation are x 2  2x  3  0 …… (i)

and ax 2  bx  c  0 …… (ii)
Since, equation (i) has imaginary roots. So, equation (ii) will also have both roots same as equation (i).
a b c
Thus,   Hence, a : b : c is 1 : 2 : 3.
1 2 3

Solutions 53 Quadratic Equations

91.(B) Let the quadratic equation be ax 2  bx  c  0
Sachin made a mistake in writing down constant terms. So, sum of roots is correct.
i.e.,   7
Rahul made mistake of writing down coefficient of x.
So, product of roots is correct.
i.e.,   6

Correct quadratic equation is x 2  (a  )x    0

x 2  7 x  6  0 having roots 1 and 6.

92.(A) Let  and  be the roots of equation x 2  ax  1  0, then     a and   1

Now, |    | (  )2  4  |    | a 2  4

According to given condition, a 2  4  5  a 2  4  5  a 2  9  |a | 3  a  ( 3, 3)

93.(AD) Use tan15  2  3


 1
tan  22   2  1
 2 


94.(B) Since, both roots of equation x 2  2mx  m 2  1  0 are greater than –2 but less than 4.
b
 D  0,  2    4, f (4)  0 and f ( 2)  0
2a

Now, D  0; 4m 2  4m 2  4  0

 4  0,  m  R … (i)

b  2m 
2   4  2 4
2a  2 1 
 
 2  m  4 … (ii)

f (4)  0  16  8m  m 2  1  0  m 2  8m  15  0  (m  3)(m  5)  0
   m  3 and 5  m   … (iii)

and f ( 2)  0  4  4m  m 2  1  0  m 2  4m  3  0  (m  3)(m  1)  0

   m  3 and 1  m   … (iv)
From equations (i), (ii), (iii) and (iv), we get m lie between –1 and 3
Alternate Solution :
The given equation is x 2  2mx  m 2  1  0

 x 2  2mx  m 2  1  ( x  m )2  1  x  m  1  x  m 1
Since, it is given that the roots of the equation are greater than –2 but less than 4.
 2  x  4   2  m 1  4   2  m  1  4 and 2  m  1  4   3  m  5
and 1  m  5  1  m  3

Solutions 54 Quadratic Equations

95.(B) Let f ( x )  x 2  2kx  k 2  k  5
b
Since, both roots are less than 5. Then, D  0,   5 and f (5)  0
2a

Now, D  4k 2  4(k 2  k  5)  4k  20  0

 k 5 … (i)
b
 5  k5 … (ii)
2a

 25  10k  k 2  k  5  0  k 2  9k  20  0  (k  5)(k  4)  0
 k  4 and k  6 … (iii)
From equations (i), (ii) and (iii), we get k  4

96.(A) Since, one of the roots of equation x 2  px  12  0 is 4

 16  4 p  12  0  4 p  28  p  7

So, the order equation is x 2  7 x  q  0 whose roots are equal. Let the roots be  and 
7 7
    
1 2
2
7 49
And product of root     q    q  q
2 4
 
97.(B) Let  and  be two numbers whose arithmetic mean is 9 and geometric mane is 4.
     18 and   16

 Required equation x 2  (  )x  ( )  0  x 2  18x  16  0

98.(C) Given equation ax 2  bx  c  0

b c
Let  and  be the roots of the equation. Then,      and  
a a
2 2
1 1  
Also given,      
2 2
   22
2 2 2
 2  b   b / a  2 b b  2a
 (   )              
     a   c /a  c /a  
a c  c
     
2a b b c  2a b c
       
c 
c c a b c a

c a b a b c
Since, , and are in AP. Hence, , and are in HP.
a b c b a b

99.(A) Since, one root of the quadratic equation (a 2  5a  3)x 2  (3a  1)x  2  0 is twice as large as the other,
then let their roots be  and 2.
(3a  1) (3a  1) 2
   2    3   and   2 
2 2 2
(a  5a  3) (  5a  3) (a  5a  3)

2 (3a  1)2
 22    (3a  1)2  9(a 2  5a  3)
(a 2  5a  3) 9(a 2  5a  3)

Solutions 55 Quadratic Equations

 26 2
 9a 2  6a  1  9a 2  45a  27  45a  6a  27  1  a  
39 3
100.(A) Given that p( x )  f ( x )  g( x ) has only one root –1.

 p( x )  (a  a1 )x 2  (b  b1 )x  (c  c1 ) has one root, –1 only,

 p '( x ) will also has root as –1  p '( x )  0 at x  1

(b  b1 )
 2(a  a1 )x  (b  b1 )  0 at x  1   2(a  a1 )  (b  b1 )  0   2 … (i)
(a  a1 )

Now, p( x )  (a  a1 )x 2  (b  b1 )x  (c  c1 )

p(x ) b  b1 (c  c1 )
  x2  x  p( 1)  0
a  a1 a  b1 a  a1

(b  b1 ) (c  c1 ) (c  c1 )
 0  ( 1)2    0 1 2  [Using equation (i)]
a  a1 a  a1 a  a1

c  c1
 1 … (ii)
a  a1

Also p( 2)  2
 4(a  a1 )  2(b  b1 )  (c  c1 )  2 … (ii)
From equations (i), (ii) and (iii), we have
4(a  a1 )  4(a  a1 )  (a  a1 )  2

(a  a 2 )  2

On substituting a  a1  2 in equations (i) and (ii), we get :

c  c1  2 , b  b1  4

Now, p(2)  4(a  a1 )  2(b  b1 )  (c  c1 )  8  8  2  18

1 1
101.(A) ax 2  bx  1  0 roots ,
 

1 1 b 1 1
  , 
  a  a

a  

b  (    )

x 2   (    )2  3  (    )  3/2  3/2  0
 

( x   3/2 )( x  3/2 )  0

x   3/2, x  3/2

102.(A) 2x  1  2x  1  1 rationalizing
2
 1  2x  1  2x  1  2
2x  1  2x  1
3 1 3
 2x  1  , 2x  1   4x 2  1 
2 2 4

Solutions 56 Quadratic Equations

103.(B) The equation ( x 2  2 x )2  3( x 2  2x )  k  2  0, has four real and distinct roots

if t 2  3t (k  2)  0 … (i) and ‘t’ must lies in ( 1,  ) where t  x 2  2x  1

Let the roots of (i) be  and  .
And for t to be lies in ( 1,  ) ,  both lies greater than –1 (i.e. both roots ,  greater than –1)
b
So, a . f ( 1)  0, D  0,  1
a
Now, 1. f ( 1)  0  1 3 k  2  0  k  6 … (ii)
1
And D 0  D  9  4(k  2)  0  k … (iii)
4
b
Also,  1, Which is always true. … (iv)
a
 1
Now from (ii), (iii) and (iv) we get : k   6, 
 4 


Solutions 57 Quadratic Equations