Why does the square root of (64a^2+81b^2) not equal 8a + 9b ?

Question

This was on my homework, and I initially thought that it DID equal that, but it does not equal that and I'm not really sure why it doesn't. I feel like I'm just missing something simple. Can anyone possible explain this to me?

Answer 1

Work backwards. Square 8a+9b.

(8a+9b)(8a+9b)=64a²+72ab+72ab+81b²⁼

64a²+144ab+81b²

 

to go one step further...

(64a²+81b²)=(8a+9bi)(8a-9bi) where i=√-1 and i²=-1

multiply the binomials together to get...

(64a²+72abi-72abi-81b²i²)=64a²+81b²

 

Answer 2

Here's another way to look at it.

Squares and square roots are related to multiplication/division, but not to addition/subtraction.

A square root can undo multiplication, but not addition.

 

This is a legal operation:

√(a² • b²) = √(a²) • √(b²) =  a • b

This is not legal:

√(a² + b²) ≠ √(a²) + √(b²)

 

Plug in some numbers and look at what happens:

√(4 • 9) = √(36) = 6

or

√(2² • 3²) = 2 • 3 = 6

or

√(36) = √(2•2•3•3) = 2 • 3 = 6   There are two factors of 36 that are squares, 2² and 3².

 

You just can't do that with added quantities.

√(4 + 9) = √(13)