help in solving sqrt division

Question

How do I solve:

 

1.   42/√(16/9)^2 + 1    or            ____42___

                                                 √(16)^2 + 1

                                                    (9)

 

2.   Then do the same problem but reverse  16/9 to 9/16

 

 

3.  What are the steps to solve  (4-3√2) (4-3√2)

 

thanks

Answer 1

   42 /√( ( 16/9) ^2 ) +1

 

 

   ( 42  / 16/9 ) +1 =

 

   (42( 9) / 16 ) + 1 =

 

  ( 21 (9) / 8 ) + 8/8 =

 

  ( 189 + 8 ) /8 =

 

   197/ 8 = 24 5/8

  

 2.  You can do the reverse

 

 3 . Use identity  ( a +b ) ^2 = a^2 + 2ab + b^2

 

       Do step by step by distributive law twice.

 

 

  ( 4 - 3 √2 ) ( 4 -3 √2 ) = ( 4 - 3√2 ) ^2 = 16 -24 √2 + 18 = 34 - 24 √2

-

Answer 2

1.

 

42/(√(16/9)^2 + 1)

= 42/(16/9 + 9/9)

= 42/(25/9)

= 42*9/25

= 378/25

 

2.

 

42/(√(9/16)^2 + 1)

= 42/(9/16 + 16/16)

= 42/(25/16)

= 42*16/25

= 672/25

 

3.

 

(4-3√2) (4-3√2)

Distribute:

4(4-3√2)-3√2(4-3√2)

Distribute both terms:

4*4-4*3√2-4*3√2+3*3*√2*√2

16-24√2+9*2

34-24√2