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Title: Algebra _ Quadratic Equations
Description: Algebra _ Quadratic Equations To solve a quadratic equation by factoring, Put all terms on one side of the equal sign, leaving zero on the other side. Factor. Set each factor equal to zero. Solve each of these equations. Check by inserting your answer in the original equation.

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Algebra
Quadratic Equations
completion of square method

Solution of quadratic equation by formula method
I
...
x2 + 10x + 9 = 0
Solution:
x2 + 10x + 25 – 25 + 9 = 0
( x + 5)2 – 16 = 0
( x + 5)2 = 16

I
...
5x2 + 14x = 55
Solution:

5x2 + 14x = 55

 5
x+

14
x2 + 5 x 11 = 0
7
5

2



49
11 = 0
25

x+ 7
5

x+

7
5

2

=

2

=

7
x+ 5
x

18
5

+ 11 =

324
25

2

=

18
5

=

18
7

5
5

=

11
5

 The roots are 11 and – 5
5

49
25

(or)

x +

7
5

= 

x

=

x =

18
5

18
7

5
5
 25
5 = 5

I
...
x2 – 4x – 45 = 0
Solution:
(x – 2)2 – 4 – 45 =

0

( x  2)2 = 49

x–2 =
x–2
x
I
...
15 = 17x + 4x2
Solution:

4

4x2 + 17x – 15 = 0
15
x2 + 17 x –
= 0
4
4

17 2 289
15
 x + 8  64  4
x+

17
8

17
x+
8
x+

= 0

2

=

289
15
+
64
4

2

17
8
x

x

=

23
8

= +

23
8

=

529
64

2

=

23
17

8
8

=

6
3
=
8
4

=

3
,5
4

(or) x +

17
23
= 
8
8

(or) x = 

23

8

 40
8

17
8

= 5

I
...


5x + 7
= 3x + 2
x–1

Solution:
5x + 7

3

3x2 – 6x – 9

= 0

x2 – 2x – 3

= 0

(x – 1)2 – 1 – 3

= 0

2

( x – 1)

= 4

( x – 1)

= 2

x–1
x

II
...
x2 + 2x – 2 = 0
Solution:

a = 1, b = 2, c =  2
x

=

 b  –b2 – 4ac
2a

x

=

 2 – 4 + 8
2

x

=  1  –3

=

 2  –12
2

=

2  2–3
2

II
...
x2 – 6x – 3 = 0
Solution:

II
...
2x2 3x – 5 = 0
Solution:

a = 2,

x

b =  3, c =  5
=
=

x =
II
...
4x2 + 7x + 2 = 0
Solution:
a = 4,









b = , c = 2

 –49 – 32
x =
8

=

7  17
8


6  4 –3
2

II
...
(x – 3)2 = 2 (x + 4)
Solution:
x2 – 6x + 9 – 2x – 8

= 0

x2 – 8x + 1

= 0

a = 1, b = 8, c = 1

x

II
...
3x2 + 2 5 x – 5 = 0
a = 3, b = 2 5 , c = 5

Solution:

x

=

2 5  20 + 60 2 
5 80
=
6
6

=

2 5  45
6

=

2 5
6

or

65
6

x = 5 or 
5
3
II
...
– x + 5 =

2x + 3

Solution:
Squaring

= (2x + 3)2

x+5

x + 5 = 4x2 + 12x + 9
4x2 + 11x + 4 = 0
a = 4, b = 11, c = 4

x

=

11  –121 – 64
8

=

11  – 57
8

II
...
a (x2 + 1) = x (a2 + 1)
Solution:

ax2 – x (a2 + 1) + a = 0

x =

II
...
3 a2 x2 – a b x – 2 b2 = 0
Solution:

x =

=
=

=


ab  – a2 b2 + 24 a2 b2
6a2
ab  – 25a2 b2
6a2
6ab or
6a2
b
a

or

4ab
6a2
2b
3a

=

ab  5ab
6a2

II
...
4x2 – 2 (a2 + b2) x + a2 b2 = 0
Solution:

x

2(a2 + b2)  4(a2 + b2)2 –16a2b2
=
8
=

2(a2 + b2)  4a4 + 4b4 – 8a2b2
8

=

2(a2 + b2)  2  (a2 – b2)2
8

=

2 (a2 + b2)  2(a2 – b2)
8

=

(a2+ b2)  (a2 – b2)
4

2
= 2a or
4

=

II
...
4 x2 – 4 a2 x + (a4 – b4) = 0
Solution:

x =

4a2  16a2 – 16(a4– b4)
8

=

4a2  16b4 =
8

=

a2  b2
2



4a2  4b2
8

a2 + b2
2

2
2
, a –b

2

II
...
p2 x2 + (p2 – q2) x – q2 = 0
Solution:

(p2 – q2)  ( p2– q2)2 + 4 p2 q2

= 

x

2p2
(p2– q2)   p4 + a4  2 p 2q2 + 4 p2 q2

=

=

2p2
(p2– q2)   p4 + q4 + 2p2 q2

= 

2p2
(p2 + q2)  (p2– q2)
2p2

2p2
=   2 or
2p
=  1 or

II
...
36 x2 – 12 a x + (a2 – b2) = 0
Solution:

x =

12a  144a2 – 144(a2 – b2)
72

12a  144b2
12a  12b
=
72
72
a + b or
a–b
= a b 
6
6
6

=


Title: Algebra _ Quadratic Equations
Description: Algebra _ Quadratic Equations To solve a quadratic equation by factoring, Put all terms on one side of the equal sign, leaving zero on the other side. Factor. Set each factor equal to zero. Solve each of these equations. Check by inserting your answer in the original equation.