Simplify the radical sqrt(18d ^ 9); 3sqrt(2a ^ 7); 2a ^ 3 * sqrt(3); a ^ 4 * sqrt(18a); 3a ^ 4 * sqrt(2a)

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

(

1

−

√

2

)

2

=

1

−

2

√

2

+

2

=

3

−

2

√

2

I think this will help you