Question

(v)If √(9+√48 - √32 - √24 )= √a-√b+2, where a,b € N, then find the value of a +b.​

Answer 1

If √(9+√48 - √32 - √24 )= √a-√b+2, where a,b € N, then find the value of a+b

Let's first simplify a little bit. Start from the last sq root: 13+sqrt(48))))

13+√48=12+4√3+1= (2√3)2+2 *1*2*√3+12=(2*√3+1)2

Now

√5-(13+√48)=√5-(2*√3+1)=√4-2*√3)=√12-2*1*√3+(√3)2=

=√(1-√3)2=(1-√3),

Now the last (basically the first)

2(√3+1-√3) = 2(√(4-√3))= √16-4√3= √12-4√3+4 =

√(2√3)2-2*2*√3+22)=√(2√3-2)2=2√3-2.

So far what we have is

2(sqrt(3+sqrt(5-sqrt(13+sqrt(48))))=2(2√3-2)=2(√3-1)

Therefore

2(√3-1)=√a+√b (*)

Square both sides of (*)

4(4-2√3)=a+b-2√(ab)

Since a and b are natural numbers, I think you can figure our what should be a+b.

Answer 2

Step-by-step explanation:

√32 + √48 /√8 + √12

√4*√8 + √4*√12/√8+√12  

 (√4*√8=√32,√4 * √12 = √48)

2√8 + 2√12/√8+√12      ( √4=2 )

2(√8+√12)/1(√8+√12)   ( 2 taken common and 1 taken common)

ans = 2/1= 2                 (√8+√12 cancelled out)