5 Chapter 1
Exercise 1.1
Q3i.Sovle x 2- 8x + 15 = 0 by quadratic formula
Q3iv. Solve 3x ( x - 2 ) + 1 = 0 by quadratic formula Sol: Since x 2- 8x + 15 = 0 Comparing we the
Sol: Since 3x ( x - 2 ) + 1 = 0 general quadratic eq i.e.,
ax 2 + bx + c = 0 we get a = 1, b = - 8, c =
15 3x 2 - 6x + 1 =
0 Comparing we the general quadratic equation
Using quadratic formula
x =
- b ± b 2 - 4ac 2a
i.e.,
ax 2 + bx + c = 0 we get a = 3,b = - 6,c = 1
Substituting the values of a,b and c we get
Using quadratic formula
x ( 8 ) ( 8 ) 2 4 ( 1 )( 15 )
2 ( 1 )x 8 64 60 2 x 8 2 x =
- b ± b 2 - 4ac 2a
= - - ± - - =
± - Substituting the values of a,b and c we get
x =
- ( - 6 ) ± ( - 6 ) 2 - 4 ( 3 )( 1 ) 2 ( 3
)
= ± 4 =
8 2
± 2 x = 6 ± 36 6 - 12 =
6 ± 6 24 x = 8 2 ± 2 2
= 4 ±
1 x = 6 ± 6 4 × 6 =
6 ± 6 4 6 Either or x = 4 + 1 = 5 x = 4 - 1 =
3 Solution set = { 5,3 } Q3ii.Solve x 2- 2x - 4 = 0 by quadratic formula Sol: Since x 2- 2x - 4 = 0 Comparing with ax 2 + bx + c = 0 we get a = 1, b = - 2, c = -
4 x = 6 ± 6 2 6 = 6 6 ±
2 6 6 x = 1 ±
3 6 Solution set = ⎧ │ ⎨ │ ⎩ 1 + 3 6 ,1 - 3
6 ⎫ │ ⎬ │ ⎭
Using quadratic formula
x =
- b ± b 2 - 4ac 2a Substituting the values of a,b and c we get
Q3v.Solve 6x 2 - 17x + 12 = 0 by quadratic formula Sol: Since 6x 2 - 17x + 12 = 0 Comparing we the general quadratic equation
x =
- ( - 2 ) ± ( - 2 ) 2 - 4 ( 1 )( - 4 ) 2 ( 1
)i.e.,
ax 2 + bx + c = 0 we get a = 6, b = - 17,c = 12
Using quadratic formula
x = 2 ± 2 4 + 16 =
2 ± 2
x =
- b ± b 2 - 4ac 2a 20 Substituting the values of a,b and c we get
x = 2 ± 2 4 × 5 =
2 ± 2 4 5 x = 2 ± 22 5 = 1 ±
5 Solution Set = { 1 + 5,1 -
5 } x =
- ( - 17 ) ± ( - 17 ) 2 - 4 ( 6 )( 12 ) 2 ( 6 )x = 17 ± 289 12 - 288 = 17 12 ± 1 =
17 12 ± 1 Either or x =17 12 - Q3iii. Solve 4x 2 + 3x = 0 by quadratic formula Sol: Since 4x 2 + 3x + 0 = 0 Comparing with
= =
1 x = 17 12 + 1 =
18 12 x 16 12 4 3
x =
2 3 ax 2 + bx + c = 0 we get a = 4, b = 3, c = 0 Using quadratic formula
x =
- b ± b 2 - 4ac 2a Substituting the values of a,b and c we get
Solution set = ⎧ ⎨ ⎩ Q3vi.Solve
4 3 ,32
⎫ ⎬ ⎭
( ) ( ) ( )( ) ( )
x 23 - 12 x = 24 1
by quadratic formula
x =
- 3 ± 3 2 - 4 4 0 2 4
Sol: Since
x 23 - 12 x =
24 1 Multiply each term by 24 then x = - 3 ± 8 ( 3 )2 =
- 3 8
± 3 24. x 3 2- 24. 12 x = 24. 24 1 2Either or
2
= - - = - =
- 8x - 2x = 1
x 3 83 x =- 3 8 + 3 x 8 6 3 4
x = 0 8
=0 8x - 2x - 1 =
0 Comparing we general quadratic equation i.e., ax 2 + bx + c = 0 we get a = 8,b = - 2,c = - 1
Using quadratic formula Solution set = ⎧ ⎨ ⎩ - ⎫ ⎬ ⎭
4 3,0 x =
- b ± b 2 - 4ac 2a Substituting the values of a,b and c we get
6 Chapter 1
Exercise 1.1
x =
- ( - 2 ) ± ( - 2 ) 2 - 4 ( 8 )( - 1 ) 2 ( 8
)
x = 2 ± 16 4 + 32 = 2 ± 16 36 =
2 16 ± 6 Either Or x = 2 16 - 6 =
- 16
4 x = 2 16 + 6 =
16 8 x =
- 4
1 x =
2 1 Solution Set = ⎧ ⎨ ⎩ - 4 1 , 2 1 ⎫ ⎬ ⎭ Q4i. Find all solution of t 2 - 8 t + 7 = 0 Solution: we have t 2 - 8 t + 7 =
0 2 - 7 - + 7 = 0 ( - 7 ) - 1 ( - 7 )
=
0 ( - 1 )( - 7 )
=
0 1 0 1
t t t t t t
t t t- = t
= or t- 7 =
0 t
=
7 Solution set = { 1,7 } Q4ii. Find all solution of 72 + 6x = x 2 Solution; We have 72 + 6x = x 2 Or x 2 - 6 x - 72 =
0 2 - 12 + 6 - 72 =
0 ( - 12 ) + 6 ( - 12 )
=
0 ( + 6 )( - 12 )
=
0 6 0 6
x x x x x x
x x ∴ x
+ = x= - or x- 12 =
0 x=
12 Solution set = { - 6,12 } Q4iii. Find all solution of r 2 + 4 r + 1 = 0 Solution; We have r 2 + 4 r + 1 = 0 Comparing we the general quadratic equation i.e. ax 2 + bx + c = 0 we get a = 1, b = 4, c = 1 using
x = - b ± b 2 2a
- 4ac putting the values
x=
- 4 ± 4 2 - 41 ( )( 1
) 2 ( 1
)
x=
- 4 ± 2
16 - 4
x= - 4 ± 2 12 =
- 4 ( ± 4 × 3 2
x= - 4 ± 2 2 3
=
2 - 2 ± 2
3 )
x
= - 2 ± 3
Solution set = { - 2 - 3, - 2 + 3 } Q4vi Find all solution of x ( x + 10 ) = 10 ( - 10 - x ) Solution: we have x ( x + 10 ) = 10 ( - 10 - x ) x 2 + 10 x = - 100 -
10 x
2x + 10 x + 10 x
+ 100 =
0
x 2
+ 20 x
+ 100 =
0 ( x ) 2 + 2 ( x
)( 10 ) + ( 10 )
2
=
0 ( x
+ 10 )
2
=
0 ⇒ x
+ 10 =
0 x
= -
10
Solution set = { - 10 } Q5. The equation ( y + 13 )( y + a ) has no linear term. Find the value of a Solution; we have ( y + 13 )( y + a ) Or y 2 + ay + 13 y + 13 a y 2 + ( a + 13 ) y + 13 a According to condition { no linear term } Means a+ 13 = 0 or a = - 13 Q6. The equation ax 2
+ 5 x = 3 has x = 1 as a solution. What is the other solution. Sol: we have ax 2 + 5 x = 3 has solution x =
1 a( 1 ) 2 5 ( 1 ) 3 a5 3 a3 5 a
2
+ = + = = - = - Therefore given equation becomes
- 2 x 2+ 5 x
=
3 - 2 x 2
+ 5 x
- 3 =
0 Or 2 x 2 - 5 x + 3 =
0 2 x 2 - 3 x - 2 x
- 3 =
0 x ( 2 x - 3 ) - 1 ( 2 x
- 3 )
=
0 ( 2 x - 3 )( x
- 1 )
=
0 Either 2 x- 3 = 0 or x- 1 =
0 2 x=3x
= 32
x = 1 Solution set = ⎧ ⎨ ⎩ 3 2,1 ⎫ ⎬ ⎭ Q7. What is the positive difference of the roots of x 2 - 7 x - 9 = 0 Solution we have x 2 - 7 x - 9 = 0 Comparing we the general quadratic equation i.e. ax 2 + bx + c = 0 we get a = 1, b = - 7, c = 9 using
x = - b ± b 2 2a
- 4ac putting the values
x=
- ( - 7 ) ± ( - 7 ) 2 - 4 ( 1 )( - 9
) 2 ( 1 )x= 7 ± 49 + 36 2
x
=
7 ± 2
85
7 Chapter 1
Exercise 1.1
Therefore roots are 7 - 2 85 , 7 +
2 85 Difference of roots are 7 + 2 Either 2 x- 1 = 0 or x- 4 = 0 85 -
7 - 2
85
=7 + 85 - 7 + 85
2
=
2 285
x = 12 x =
4 Solution set = ⎧ ⎨ ⎩ 1 2,4 ⎫ ⎬ ⎭ Exp 8: Solve x x + - 1 3 + x x
+ -
1 3 = 13 6
Difference of roots are = 85
Solution We have x x + - 1 3 + x x
+ -
1 3 = 13 6
Solution of equations reducible to quadratic equation in one variable:
Let y = x x + - 1 3 ⇒ 1 y = x x
+ 3 -
1
Type 1:Equation of the form, ax 2 n + bx n + c = 0 1.Put x n= y 2.Find the value of y from the new equation 3Put the values of y in supposition to get roots. Exp 6 Solve 12 x 4 - 11 x 2 + 2 =
0 Therefore y + 1 y =
136 Multiply each term by 6y
6 y 2 + 6 = 13 y Or 6 y 2 - 13 y + 6 = 0 Solution we have 12 x 4 - 11 x 2 + 2 =
0 6 y 2 - 9 y - 4 y
+ 6 =
0 Let y = x 2 ⇒ y 2 = x 4 given equation becomes
3 y ( 2 y - 3 ) - 2 ( 2 y
- 3 )
=
0 2- + =
( 3 y - 2 )( 2 y
- 3 )
=
0 2
- - + = ( - ) - ( - )
=
Either ( - )( - )
= Either 4 y- 1 = 0 or 3 y - 2 =
0 y = 14 y = 23 Putting back value of y = x 2 therefore,
2 1412
12 y 11 y
2 0 12 y 8 y 3 y
2 0 4 y 3 y 2 1 3 y
2 0
4 y 1 3 y
2 0
3 y- 2 =
0
2 y- 3 =
0
y
= 23
=
Putting back the value of y
1 2 3 3
or
y
32
- + = By cross multiplication
3 ( x 1 ) 2 ( x
3 ) 3 x 3 2 x
6 3 x 2 x
6 3 x
9x- x +
= x1 3
x
3 2
- = + ( - = + - = + =
2 x - 1 ) = 3 ( x
+
3 ) x=x2 =23
2 x - 2 = 3 x
+
9 2 x - 3 x
= 9 +
2 x
= ±
x
= ±
23 Solution set = ⎧ │ ⎨ │ ⎩ - x=11± 1 2 ,± 2 3
⎫ │ ⎬ │ ⎭ Type 2: The Equation of the form, ax + bx =
c x
= -
11 Solution set = { - 11,9 } Type 3 a x 2
1 x 2b x 1 x c 0 (one term is reciprocal to the other)
1. Suppose one term = y 2. Take reciprocal of the equation then put
in question and solve to get value of y. 3. Put y values in supposition to get roots Exp7: Solve 2 x + 4 x = 9 Solution we have 2 x + 4 x = 9 Multiply each term by x
22
⎛ │ ⎝ + ⎞ │ ⎠ + ⎛ │ ⎝ + ⎞ │ ⎠ + = 1 Let y = x + 1 x 2 taking square on bs to get the value of x 2
+ x 1 23 putting these substitution in given equation Exmple 9i). 2 ⎛ │ ⎝ x 2
+ x 1 2⎞ │ ⎠ - 9 ⎛ │ ⎝ x + 1 x
⎞ │ ⎠ + 14 = 0 Sol: we have 2 ⎛ │ ⎝ x 2
+ x 1 2⎞ │ ⎠ - 9 ⎛ │ ⎝ x + 1 x
⎞ │ ⎠ + 14 = 0 x ⎛ │ ⎝ 2 x + 4 x
⎞ │ ⎠
= 9 x 2 x + 4 =
9
x
Let y = x + 1 x ⇒ y 2 = x 2
+ x 1 2+ 2 y 2 - 2 = x 2
+ x 1 22 x - 9 x
+ 4 =
0 2 2 8 1 4 0
Thus ( 2 x - x - x
+ =
2 y given - 2 ) - equation 9 y + 14 =
becomes 0 2 x ( x - 4 ) - 1 ( x
- 4 )
=
0
2 y 2- 4 - 9 y
+ 14 =
0 ( 2 x - 1 )( x
- 4 )
=
0
2 y 2
- 9 y
+ 10 =
0
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