square root

[alltimeacheiver]

what is the value

[clock60]

17^1/2>16^1/2. and 16^1/2=4. thus 17^1/2 >4, i hope that it is slightly more than 4
47^1/2<49^1/2, and 49^1/2=7, thus 47^1/2<7, slightly less then 7
7+4=11 rounded value,
i hope that the answr is B

[alltimeacheiver]

why cannt we simply add them as i did 17+47=64 square root = 8

[Whitney Garner]

17^1/2>16^1/2. and 16^1/2=4. thus 17^1/2 >4, i hope that it is slightly more than 4
47^1/2<49^1/2, and 49^1/2=7, thus 47^1/2<7, slightly less then 7
7+4=11 rounded value,
i hope that the answr is B

Perfect estimation work clock60!! The answer is, in fact, B.

why cannt we simply add them as i did 17+47=64 square root = 8

**note, I will abbreviate square root with "sqrt" from here out**

The reason we cannot add the values under the square roots is because powers and roots cannot be expanded over addition or subtraction (only over multiplication or division).

For a real world example to help you see what I mean:

sqrt(25) + sqrt(16)

Because these are perfect squares, it would be just as easy to take the roots first and then add:

sqrt(25) + sqrt(16) = 5 + 4 = 9

Because I have only used very basic rules of algebra I can feel fairly confident in my math (I know definitively that the square root of 25 is 5, the square root of 16 is 4, and that the sum of 5 and 4 is 9.

But let's apply the shortcut you used in your solution to see if it works:

sqrt(25) + sqrt(16) = sqrt(25+16) = sqrt(41) = approx. 6.4

Notice that this is not equal to 9! (in fact, it is actually pretty far off from 9.

So, the shortcut of just adding the numbers under the roots will not work. This is very similar to the ideas that:

(x + y)^2 does not = x^2 + y^2.... we have to FOIL (x+y)(x+y)
sqrt(x+y) does not = sqrt(x) + sqrt(y)

...and so on.

I hope this clears that up a bit for you!!

Whit[/spoiler]