To show this let us use the first example that we solved earlier
i.e.:
2
3
10
0
x
x
+

=
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26
Exercise 2.2.2
Solve the following quadratic equations
:
2
(
) 4
5
1
0
a
x
x
+
+
=
;
2
(
) 2
2
4
0
b
y
y


=
;
2
(c)
12
35
0
y
y
+
+
=
;
2
(
) 9
12
5
0
d
a
a


=
2
2
2
3
10
0
3
3
4
1
(
10)
4
3
9
40
3
49
3
7
2
2
1
2
2
2
3
7
4
2
2
2
3
7
10
5
2
2
+

=
↓
↓
↓
 ±
 × × 
 ±

 ±
+
 ±
 ±
=
=
=
=
=
×
 +
=
=
=
⇒
 

=
=
= 
x
x
a
b
c
b
b
ac
x
a
x
or
x
We have 2 solutions:
2
5
x
or x
=
= 
. If you check the solution we got earlier
using the other method you will see that the results are exactly the same.
Extra Exercises for Practice
Solve these equations
2
2
2
2
2
( )
4
10
6
0; (
) 3
6
0; (
) 12
16
16
0;
(
) 3
6
0; (
) 6
15
18

+
=

=
+

=

=
+
=
a
x
x
b
y
y
c
p
p
d
x
x
e
a
a