Posted by **em** on Friday, February 23, 2007 at 7:02am.

Hi, could someone please tell me how I would go about working these out: (the answer must be in the form a(sqroot)b where a and b are integers and b is as small as possible.)

1) Simplify (sqroot)500

2) Simplify (sqroot)50 x (sqroot)60

Also..

Simplify the following fractions so that there is no root in the denominator:

1/(sqroot)5

12/(sqroot)3

Thankyou so much!

1) sqrt500 =sqrt(25 x4 x 5)

=(sqrt 25)x(sqrt4)x (Sqrt5)

=5 x 2 x (sqrt5)

=10sqrt5

therefore a=10 and b=5

2) sqrt50 xsqrt60 =Sqrt300

=sqrt25 x sqrt4 x sqrt3

=5 x 2 x sqrt3

=10sqrt3

therefore a=10 and b=3

## Answer This Question

## Related Questions

- maths - Hi, could someone please tell me how I would go about working these out...
- Math - Two positive integers are in ratio 1:3. If their sum is added to their ...
- maths - a few questions - Simplify 3(x-2)^2/x-2 Is the answer 3x^2-10x+10? If ...
- Maths - I got the last question so please don't worry.. These are the ones I get...
- maths - The rectangular coordinates of a point are given (-5,-5root3). Find ...
- Maths - The sum of the squares of three consecutive positive integers is 365 ...
- please i need help accounting - i have a promblem it gives me the information ...
- algebra - Can someone please help me with this? Solve x^3-3x^2-10x=0 by ...
- Maths- very very simple - can someone tell me what the title of this proof is? a...
- MATHS - Critical help please. Level 3 maths algebraic methods is really hard. Q...

More Related Questions