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ii. Common roots and Signs of the Quadratic Expressions
EXERCISE ii
Very Short Answer Questions
1. If the quadratic equations 2 2 0ax bx c+ + = and 2 2 0ax cx b+ + = ( )b c have a
common root then show that a + 4b + 4c = 0
Sol: Let be the common root of 2 2 0ax bx c+ + = and 2 2 0ax cx b+ + =
2 2 0a b c + + =
2 2 0a c b + + =
Subtracting 2 ( ) 0b c c b + =
2 ( ) ( )b c b c =
1
2 =
Sub1
2 = in 2 2 0a b c + + =
1 12 0
4 2a b c
+ + =
a + 4b + 4c = 0
2. If 2 6 5 0x x + = and2
12 0x x p + = have a common root then find p
Sol: Let be the common root of 2 6 5 0x x + = and 2 12 0x x p + =
26 5 0 (1) + =
212 0 (2)p + =
From (1) ( 1) ( 5) 0 =
1 = pr 5=
Substituting 1= in (2) we have
1 12 0 11p p + = =
Substituting 5 = in (2) we have
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25 (2)(5) 0 11p p + = =
3. If2
6 5 0x x + = and2
3 35 0x ax + = have a common root then find a
Sol: Let be the common root of 2 6 5 0x x+ + = and 2 3 35 0x ax + =
26 5 0 (1) + =
23 35 0 (2)a + =
From (1) we have
( 1)( 5) 0 =
1 = or 5 =
If 1 = then
21 3 (1) 35 0 12a a + = =
If 5 = then
25 3 (5) 35 0 4a a + = =
4. If equations 2 0x ax b+ + = and 2 0x cx d+ + = have a common root and the first
equation has equal roots then prove that 2(b + d) = ac
Sol: Let be the common root of 2 0x ax b+ + = and 2 0x cx d+ + = since 2 0x ax b+ + =
has equal roots
, be the roots of 2 0x ax b+ + =
2
aa + = =
2b b = =
is also a root of 2 0x cx d+ + =
2
0c d + + =
02
ab c d
+ + =
2
acb d+ =
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2 ( b +d ) = ac
5. Discuss the signs of the following quadratic expressions when x is real
i) 2 5 4x x +
here2 2
4 ( 5) 4(1) (4)b ac =
= a > 0
Roots are real
Coefficient of2
1 0x = >
Roots of2
5 4 0x x + =
Are 1, 4
By theorem 2
If x < < then 2ax bx c+ + and a have opposite sign here 1 = , 4 =
Coefficient of 2 0x >
21 4 5 4 0x x x < < + <
If x = 1 or x = 4 then
25 4 0x x + =
x < or x < then x < and a have same sign
1x < or 4x < then
25 4 0x x + >
ii)2
3x x +
Here2 2
4 ( 1) 4(1) (3)b ac =
= 1 12 = - 11 < 0
There fore roots of the equation are imaginary coefficient of 2x is 1 > 0
Hence x R 2ax bx c+ + and a have same sign
23 0x x x R + >
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6. For what values of x the following expressions are positive
i)2
5 6x x + ii)2
3 4 4x x + iii)2
4 5 2x x + iv)2
5 14x x +
Sol:2
5 6 0x x + = Have real roots 2 and 3
When x < 2 or x > 32
5 6 0x x + >
25 6 0x x + > is possible when ( , 2) (3, )x
ii) 23 4 4x x+ +
24 16 (3)(4) 32 0b ac = = = <
Roots of 23 4 4 0x x+ + = are imaginary
Hence 23 4 4 0x x+ + > x R because coefficient of 2 3 0x = >
iii)2
4 5 2x x +
2 24 (4) 4( 5) (2)b ac = =
= 16 + 40 =56
Roots of 24 5 2 0x x + =
25 4 2 0x x =
4 16 4(5)( 2)
10x
=
4 56
10x
=
If4 56 4 56
10 10x
+<
Then2
4 5 2 0x x + >
2x ax bx c < < + + and a have opposite sign
iv) 2 5 14x x +
Here2
4b ac =
= 25 4(1) (14) = -31 < 0
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Roots of 2 5 14 0x x= + =
Are imaginary
25 14 0x x x R + >
7. For what values of x the following expressions are negative
i) 2 7 10x x + ii) 215 4 3x x+ + iii) 22 5 3x x+ iv) 2 5 6x x
i) 2 7 10x x +
Here 2 4 49 40 9b ac = = =
Roots are real coefficient of2
1 0x = >
Roots of the equation2
7 10 0x x + = are 2, 5
If x < < then 2ax bx c+ + and a have opposite sign
22 5 7 10 0x x x < < + <
ii) 215 4 3x x+
a = -3, b = 4, c = 15
24 16 4( 3)(15)b ac = = =
= 196 > 0
Roots are real roots of2
15 4 3 0x x+ =
23 4 15 0x x =
23 9 5 15 0x x x + =
x = -5/3, x = 3
Coefficient of2
3 0x = <
215 4 3 0x x + < When
5
3x < and x > 3
x < or 2x ax bx c> + + and a have same sign
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iii) 22 5 3x x+
a = 2, b = 5, c = - 3
24 25 24 49b ac = = + =
Roots are real
2 22 5 3 0 2 6 3 0x x x x x+ = + =
(2x-1)(x + 3) = 0
Roots of2
2 5 3 0x x+ =
Are1
3,2
Coefficient of 2 2 0x = >
22 5 3 0x x + < When
13
2x < <
2x ax bx c < < + + and a have opposite sign
iv) 2 5 6x x
a = 1, b = - 5, c = -6
24 25 24 49b ac = = + =
Roots are real
25 6 0 ( 6)( 1) 0x x x x = + =
x = 6; x = -1
Coefficient of 2 1x =
25 6 0 6 1x x x < < <
2x ax bx c < < + + and a have opposite sign
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8. Find the changes in sign of the following expressions and find their extreme values
i)2
5 6x x +
Here a = 1, b = - 5, c = 6
24 25 24 1b ac = = =
Hence roots are real
25 6 0 2,3x x x + = =
22 3 5 6 0x x x < < + <
22 3 5 6 0x or x x x= = + =
22 3 5 6 0x or x x x< > + >
Here a = 1 > 0
25 6x x + has absolute minimum value
absolute minimum value2
4 24 25 1
4 4 4
ac b
a
= = =
ii) 215 4 3x x+
a = -3, b =4, c = 15
24 16 4( 3)(15)b ac = =
= 196 > 0
Roots are real roots of2
15 4 3 0x x+ = are5
3 and 3
{Refer to prob (ii) of 7}
25 3 15 4 3 0
3
x x x < < + >
25 3 15 4 3 03
x or x x x< > + <
253 15 4 3 0
3x or x x x< > + =
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Here a = - 3 < 0
Absolute maximum value
24 180 16
4 4 3
ac b
a x
= =
196 4912 3
= =
9. Find the maximum or minimum values of following expressions as x varies over R
i)2
7x x +
Here a = 1, b = -1, c = 7
a > 0 absolute minimum2
4
4
ac b
a
=
28 1 27
4 4
= =
ii) 212 32x x
a = -1, b = 12, c = -32
a < 0 hence absolute maximum value2
4 49 1)( 32) 144
4 4
ac b
a
= =
= 4
iii) 22 5 3x x+
Here a = -3, b = 2, c = 5
a < 0 hence maximum value2
4
4
ac b
a
=
60 4 16
12 3
= = +
iv)
2
ax bx a+ +
( , )a b R
a > 0 minimum value2 2
4
4
a b
a
=
a < 0 maximum value2 2
4
4
a b
a
=
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SHORT ANSWER QUESTIONS:
Determine the range of the following expressions
1.2
2
1
1
x x
x x
+ +
+
Sol: let2
2
1
1
x xy
x x
+ +=
+
2 21x x x y xy y + + = +
2( 1) ( 1) 1 0x y x y y + + =
x is real
24 0b ac
2( 1) 4( 1) ( 1) 0y y y+
23 10 3 0y y +
23 10 3 0y y +
1(3 1)( 3) 0 3
3y y y
ii) 22
2 3 6
x
x x
+
+ +
Let2
2
2 3 6
xy
x x
+=
+ +
22 2 3 6x x y xy y+ = + +
22 (3 1) 6 2 0x y x y y+ + =
Since x is real 2 4 0b ac
2(3 1) 4(2 )(6 2) 0y y y
2 2(9 6 1) (48 16 ) 0y y y y +
239 10 1 0y y + +
239 10 1 0y y
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13 (3 1) 1(3 1) 0y y y +
(13 1)(3 1) 0y y+
1 1
13 3y
iii) ( 1)( 2)3
x xx
+
+
Let( 1)( 2)
3
x xy
x
+=
+
22 3x x xy y + = +
2(1 ) (2 3 ) 0x x y y+ + =
Since x is real
24 0b ac
2(1 ) 4(2 3 ) 0y y + +
22 1 8 12 0y y y + + +
210 7 0y y +
( 1)( 9) 0y y
1y or 9y
( , 1] [9, )y
iv)2
2
2 6 5
3 2
x x
x x
+
+
Let2
2
2 6 5
3 2
x xy
x x
+=
+
2 22 6 5 3 2x x x y xy y + = +
2(2 ) 6 3 5 2 0x y x xy y + + =
2(2 ) 3 ( 2) 5 2 0x y x y y + + =
Since x is real2
4 0b ac
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29( 2) 4(2 )(5 2 ) 0y y y
2 29 36 36 4{10 9 2 } 0y y y y + +
24 0 ( 2)( 2) 0y y y +
( , 2] [2, )y
2. Prove that1 1 1
3 1 1 (3 1)( 1)x x x x+
+ + + +does not lie between 1 and 4 if x is real
Sol: Let1 1 1
3 1 1 (3 1)( 1)y
x x x x+ =
+ + + +
1 3 1 1
(3 1)( 1)
x xy
x x
+ + + =
+ +
24 1
3 4 1x y
x x+ =
+ +
24 1 3 4x x y xy y+ = + +
23 4 ( 1) 1 0x y x y y+ + =
Since x is real
24 0b ac
2{4( 1)} 4(3)( )( 1) 0y y y
2 216 32 16 12 12 0y y y y + +
24 20 16 0y y +
25 4 0y y +
( 1)( 4) 0y y
( ,1] [4, )y
Hence y does not lie between 1 and 4
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3. If x is real then prove that2 5 9
x
x x +lies between
1
11 and 1
Sol: Let2 5 9
xy
x x=
+
25 9x x y xy y= +
2(5 1) 9 0x y y x y + + =
Since x is real
24 0b ac
2(5 1) 4 (9 ) 0y y y+
2 225 10 1 36 0y y y+ +
211 10 1 0y y + +
211 10 1 0y y
211 11 1 0y y y +
(11 1)( 1) 0y y+
11
11
y
Hence y lies between1
11 and 1
4. If the expression2
3 2
x p
x x
+takes all real values x R then find the bounds of p
Sol: Let2
3 2
x py
x x
=
+
2
3 2x p x y xy y = +
2(3 1) 2 0x y x y y p + + + =
Since x is real
24 0b ac
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2(3 1) 4 (2 ) 0y y y p+ +
26 4 1 0y y py+ +
2(6 4 ) 1 0y y p+ +
Since coefficient of2
y and expression have same sign y R
2(6 4 ) 4 0p
236 16 48 4 0p p+
216 48 32 0p p +
23 2 0p p +
( 1)( 2) 0p p
(1, 2)p
{ for p 1, 2, given expression is not defined}
5. If 2c ab and the roots of2 2 2 2
( ) 2( ) ( ) 0c ab x a bc x b ac + = are equal then show
that3 3 3
3a b c abc+ + = (or) a = 0
Sol: Given equation is
2 2 2 2
( ) 2( ) ( ) 0c ab x a bc x b ac + =
Roots are equal
2 2 2 24( ) 4( )( ) 0a bc c ab b ac =
4 2 2 2 2 2 3 3 22 0a b c a bc b c ac ab a bc+ + + =
4 3 3 23 0a ab ac a bc+ + =
3 3 3( 3 ) 0a a b c abc+ + =
0a = or3 2 3
3 0a b c abc+ + =
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III) QUADRATIC INEQUATIONS
EXERCISE -- 1(c)
1. Solve the in equations by algebraic method
i) 215 4 4 0x x+
215 10 6 4 0x x x+
5 (3 2) 2(3 2) 0x x x+ + <
(3 2)(5 1) 0x x+
2 / 3 2 / 5x
ii) 2 2 1 0x x + <
2( 1) 0x <
Square of any real number cannot be negative
Hence x
iii)2
2 3 2 0x x
22 3 2 0x x+
22 4 2 0x x x+
2 ( 2) ( 2) 0x x x+ +
( 2)(2 1) 0x x+
12
2x
iv)2
4 21 0x x
2
7 3 21 0x x x +
( 7) 3( 7) 0x x x +
( 3)( 7) 0x x+
3x or 7x
( , 3] [7, )x
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2. Solve the following in equations by graphical method
i) 2 7 6 0x x + >
let2
7 6y x x= +
The values of y for some selected values of x are given below
x -5 -4 -3 -2 -1 0 1
y 66 50 36 24 14 6 0
x 2 3 4 5 6
y -8 -6 -6 -4 0
Hence solution set ( , 1] [6, )=
ii) 24 0x
Let2
4y x=
x -4 -3 -2 -1 0 1 2 3 4
y -12 -5 0 3 4 3 0 -5 -12
27 6 0y x x= + 1 6when x or x
5 4 3 2 1 1 2 3 4 5
24 0y x= 2 2when x
5 4 3 2 1 1 2 3 4 5
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Hence solution set [ 2, 2]x
iii) 215 4 4 0x x+
let2
15 4 4y x x= +
x -3 -2 -1 2
3
0 2
5
1
y 89 48 7 0 -4 0 15
215 4 4 0y x x= +
2 2
3 5x
Hence2 2
,3 5
x
iv) 2 4 21 0x x
24 21y x x=
x -3 -2 -1 0 1 2 7
y 0 -9 -16 -21 -24 -25 0
2
52
5
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24 21 0y x x=
When 3x or 7x
2. Solve the following in equations
3 8 2x <
83 8 0 (1)
3x x
And 3 8 2x <
Here 3 8 0x
But given 3 8 2x <
No solution
ii) 2 6 5 8 2x x x + >
2 6 5 0x x +
And i) 8 2x < 0 or2 2
6 5 (8 2 )x x x + > and 8 2 0x
We have 2 6 5 0x x +
( 1)( 5) 0 1 5x x x
And 2 8 0 4x x > >
73
0
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Or 2 26 5 64 4 32x x x x + > +
20 5 38 69x x> +
25 38 69 0x x + <
And x < 4
25 23 15 69 0x x x + <
(5 23) 3(5 23) 0x x x <
( 3)(5 23) 0x x <
233
5x< <
1 5x and x > 4
Or23
35
x< < and x < 4
3 5x <
iii) Solve the in equation2
2 8x x+ >
When a, b R and 0a , 0b then 0a b a b> >
22 8x x + >
22 (8 ) 0x x + >
2, 2 2x x> <
2 6 0x x+ >
( , 3) (2, )x
Solution
(2, 2 2)x
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iv) Solve the in equation
2( 3)(2 ) 4 12 11x x x x < + +
( 3)(2 ) 0x x and2
( 3)(2 ) 4 12 11x x x x < + +
2 3x and2 2
5 6 4 12 11x x x x < + +
2 3x and 20 5 7 17x x< + +
2 3x and 25 7 17 0x x+ +
25 7 17x x o+ + > x R
27 4(5)(17) 0 <
Solution set [2, 3]
v) Solve the in equation
2 26 6
2 5 4
x x x x
x x
+ +
+ +
26 0x x+ = , 2 5 0x + , and 4 0x +
(or)2
6 0x x+ > and1 1
2 5 4x x
+ +
26 0 2x x x+ = or x = 3
26 0 2 3x x x+ > < <
1 12 5 4
2 5 4x x
x x + +
+ +
1x
{ }2,3 [ 2, 1]
[ 2, 1] {3}
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vi) Solve the in equation
23 10 8x x x >
Sol: 2 3 10 0x x
and i) 8 x < 0
or (ii)2 2
3 10 (8 )x x x > and (8 ) 0x
( 5) ( 2) 0x x +
[ , 2] [5, ]x
8 0 8x x < >
ii) 2 23 10 64 16x x x x > +
7413 74
13x x> >
And 8 0 8x x
74,
13x
( ,8)x
74(8, ) , ( ,8)13
74 74(8, ) , ,
13 13
=
SOLVE THE FOLLOWING EQUATIONS
1. 22
1 19 27 8 0x x
x x
+ + + =
Let1
x ax
+ =
2 2 2 2
2 2
1 12 2x a x a
x x+ + = + =
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2
2
1 19 27 8 0x x
x x
+ + + =
29( 2) 27 8 0a a + =
29 27 10 0a a =
29 3 30 10 0a a a+ =
3 (3 1) 10(3 1) 0a a a+ + =
(3 1)(3 10) 0a a+ =
1
3a = ,
10
3a =
Case i): suppose1
3
a =
21 1 1 1
3 3
xx
x x
++ = =
23 3 3 3 0x x x x2 + = + + =
1 1 4(3)(3)
2 3x
=
1 35
6
i
x
=
Case ii):10
3a =
21 10 3 3 103
x x xx
+ = + =
23 10 3 0x x + =
2
3 9 3 0x x x + =
3 ( 3) 1( 3) 0x x x =
(x 3) (3x -1) = 0
3x = ,1
3x=
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1
3a = , a = 3
Case i)1
3a =
4 1 1
36 9 4 14 1 3
x
x xx
= = ++
10 532 10
32 16x x= = =
Case ii) a = 3
4 1 4 13 9
4 1 4 1
x x
x x
= =
+ +
4 1 36 9 32 10x x x = + =
5
16x= is only solution
5
16x= Does not satisfy the equation
3. 22
43 1 5
3 1x
x+ + =
+
Let 23 1x a+ =
24 5 5 4 0a a aa
+ = + =
( 4)( 1) 0 4; 1a a a a = = =
Case i) a = 4
2 23 1 4 3 1 16x x+ = + =
2 5 5x x = =
Case ii) a = 1
2 23 1 1 3 1 1 0x x x+ = + = =
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5. 22
1 12 3 1x x
x x
+ + =
This is same model as (1) try your self
6. x (x + 2 ) (x + 3) (x + 5) = 72
X (x +5) (x + 2) (x +3) = 72
2 2( 5 )( 5 6) 72x x x x+ + + =
Let 2 5x x a+ =
2( 6) 72 6 72 0a a a a+ = + =
(a +120 (a 6) = 0
a = - 12, a = 6
case i)a = - 12
2 25 12 5 12 0x x x x+ = + + =
5 25 48 5 23
2 2
ix
= = =
Case ii)a = 6
2 25 6 5 6 0x x x x+ = + =
26 6 0x x x+ =
( 6)( 1) 0x x+ =
6x = , x = 1
7. x (x -1) (x + 2) (x 3) = - 8
2 2( ){ 6} 8x x x x =
Let2
x x a =
2( 6) 8 6 8 0a a a a = + =
(a 4) (a 2) = 0
a 4 or a = 2
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Case i): a = 4
2 24 4 0x x x x = =
1 1 16 1 17
2 2x
+ = =
Case ii) a = 2
2 22 2 0x x x x = =
22 2 0x x x + =
X (x 2) + (x 2) = 0
(x + 1)( x 2) = 0
X = 2; x = -1
8. (x 1) (x + 1) (2x +3) (2x 1) = 3
(x 1) (2x + 3) (x +1) 2x 1) = 3
2 2(2 3 2 3}{2 2 1} 3x x x x x x+ + =
2 2(2 3)(2 1) 3x x x x+ + =
22x x a+ =
2( 3)( 1) 3 4 3 3a a a a = + =
24 0 0,4a a a = =
If 2 22 2 0x x a x x+ = + =
1(2 1) 0 0;
2x x x+ = =
If2 2
2 4 2 4 0x x x x+ = + =
1 1 4(2)( 4) 1 33
4 4x
= =
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9. Find the range of the values of x which satisfy( 1)( 3)
02
x x
x
+
Sol:( 1)( 3)
02
x x
x
+
Case i) ( 1)( 3) 0x x+
And (x 2) > 0
3x
Case ii) ( 1)( 3) 0x x+ and x 2 < 0
1 3x and x < 2
1 2x <
solution set [ 1, 2) [3, )
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