1

6
=
3 10
6310
6
2
16 3 28
=
108 100
16 3 28
=
8
7
= 2 3
2
310
32
3
(b)
5 3
32
3 5 3
=
5 3 5 3
2
3 3 5 3
=
5 2 15 15 3 2
2
=
3
15 2 3
5 3
2
5 3
6 2 15 3 5 3 3
=
2
3
= 3
15 5 3
2
Example 5: Simplify
1 1 1
(a)
1 x 1
x 1
x
(b)
x
Solution:
(a)
1
x
11
x
1
x 1
x
1
=
1
x
x
1
1
x 1
1
x
3
3
11
x 1
1
x 1
x 1
x
x
x
1
1
x
1
x 1
x 1
=
=
x 1
3 3
x 1
1 x
(b)
x
x
1
x
x 1
x
x
2
2
x
x
x
x 1
x 1
x
x x
x
x x x 1
x
1 1
x x
x x
=
x
x
x
x
1
x
1
x
x
1
1.25 Equations with Surds
They are solved by isolating surds on one side
of the equation, then squaring throughout until
surd has disappeared.
However this method introduces extra solutions
because a negative sign when squared is equal
to a positive when squared i.e.
2 2
x
x x.
Hence solution of the equation with surds
should be checked by substituting the solutions
got in the original equation before giving final
solution.
NB: This part should be learnt after the
quadratic equation is taught.
Example 1Simplify 14 4 6
Solution
Suppose 14 4 6 a 2 b
Squaring both sides of the equation we have
2
14 4 6 a 2 b 2
4 6 a 4b
4 ab
14
By comparing R.H.S and L.H.S.
14 a 4b……………………..(i)
ab 6……………………………(ii)
a 14 4b Substituting for a
14
4b
b
6
7b
2b 2 3
2b
2 7b 3 0