Question

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If √(9+√48 - √32 - √24 )= √a-√b+2, where a,b € N, then find the value of a +b.
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Answer 1

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.

hope this may help you

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Answer 2

Answer:

The value of a+b = 5

Step-by-step explanation:

Left-hand side:

$\sqrt{(9+\sqrt{48}-\sqrt{32}-\sqrt{24} } )$

$= \sqrt{(2+3+4+\sqrt{4*12}-\sqrt{4*8}-\sqrt{4*6} } )$

$= \sqrt{((\sqrt{2})^2+(\sqrt{3})^2 +(\sqrt{4} )^2+2(\sqrt{12}-\sqrt{8}-\sqrt{6} } ))$

$= \sqrt{((-\sqrt{2})^2+(\sqrt{3})^2 +(\sqrt{4} )^2+2(\sqrt{4}*\sqrt{3}- \sqrt{4}*\sqrt{2}-\sqrt{3}*\sqrt{2} ))$

Identity: $(a+b+c)^2= a^{2} +b^{2} +c^{2} + 2(ab+bc+ac)$

$= \sqrt{(\sqrt{3}+\sqrt{4}-\sqrt{2})^{2} } =\sqrt{3} -\sqrt{2} +\sqrt{4}$

$=\sqrt{3} -\sqrt{2} +2 = \sqrt{a} -\sqrt{b} +2$

On comparing with the Right-Hand Side

a=3 and b=2

a+b = 3+2=5

Meaning of  a,b € N

• The symbol represents 'belongs to'
• Here a,b belongs to the natural number.

Natural Number:

• All positive integers from 1 to infinity are considered natural numbers and are therefore a component of the number system.
• Because they don't contain zero or negative numbers, natural numbers are also known as counting numbers.
• They are a subset of real numbers, which only include positive integers and exclude negative, zero, fractional, and decimal numbers.

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