Answer:
The answer above is completely correct, but it may help the student to see how to reach it:
We seek a square root of
3
−
2
√
2
. This must have the form
a
+
b
√
2
.
(
a
+
b
√
2
)
2
=
a
2
+
2
a
b
√
2
+
2
b
2
So
a
2
+
2
b
2
=
3
and
2
a
b
=
−
2
⇒
a
b
=
−
1
These are two simultaneous equations for
a
and
b
.
Rearrange the second:
b
=
−
1
a
Substitute into the first:
a
2
+
2
a
2
=
3
a
4
+
2
=
3
a
2
a
4
−
3
a
2
+
2
=
0
Note that this is a quadratic in
a
2
:
(
a
2
)
2
−
3
(
a
2
)
+
2
=
0
Factorise:
(
a
2
−
2
)
(
a
2
−
1
)
=
0
This gives us two possible solutions for
a
2
:
2
and
1
, and so the four solutions for
a
:
±
√
2
and
±
1
.
We are looking for integer solutions for
a
, and so
±
1
are possible solutions. But the other two are possible too - they can simply be folded in to the
√
2
term. This wouldn't have been possible if we'd had the root of some other number in the solution for
a
, but this solution is a special case.
Now use the second equation to deduce the four equivalent solutions for
b
:
b
=
−
1
a
b
=
¯¯¯¯¯
+
1
√
2
=
¯¯¯¯¯
+
1
2
√
2
and
¯¯¯¯
+
1
.
So we have the four solution pairs
(
a
,
b
)
:
(
√
2
,
−
1
2
√
2
)
(
−
√
2
,
1
2
√
2
)
(
1
,
−
1
)
(
−
1
,
1
)
This is a bit suspicious - we expect only two solutions, positive and negative square roots, so we wonder if some of these are identical to each other: When we substitute them in to our desired expression
a
+
b
√
2
, we get:
√
2
−
1
2
√
2
√
2
=
−
1
+
√
2
−
√
2
+
1
2
√
2
√
2
=
1
−
√
2
1
−
√
2
−
1
+
√
2
So the two solutions with
√
2
are identical to the two simpler solutions, so we can get rid of them. We now have two solutions, positive and negative square roots:
1
−
√
2
−
1
+
√
2
When we take the written square root of a quantity, it is implied that the desired root is the positive root, the "principal value" of the square root function. So we take the single solution that comes from
a
=
+
1
:
1
−
√
2
Double check: Make sure that this produces the desired answer:
(
1
−
√
2
)
2
=
1
−
2
√
2
+
2
=
3
−
2
√
2
I think this will help you
