Solving Math Problem: sqrt (a^2 - a [delta] d) = a - sqrt(a [delta] d)?

[IB]

Suppose I have sqrt (a^2 - a [delta] d). What do I do? Do I do like this: sqrt (a^2 - a [delta] d) = a - sqrt(a [delta] d)? Thanks.

PS: One more thing. How to write mathematics with latex?

 

[Learning Curve]

What is sqrt (a^2 - a [delta] d) equal to? or are you asking to simplify it?

 

[IB]

Yeah. I was just wondering whether it can still be simplified.

 

[Hurkyl]

Depending on what you want to do, a differential approximation, or maybe a Taylor sum, might be useful.

But as for algebraic manipulation, what you did is wrong. roots and exponents (usually) distribute over multiplication (and division), not addition (and subtraction).

 

[IB]

But as for algebraic manipulation, what you did is wrong. roots and exponents (usually) distribute over multiplication (and division), not addition (and subtraction).

Could you give an example to demonstrate that? And how can I correct my wrong algebraic manipulation? Thanks.

 

[rocketboy]

Hey,

Here is the thread on how to use the LaTeX thingy:

https://www.physicsforums.com/showthread.php?t=8997

Have fun!
-Jon

 

[VietDao29]

Say a = 5, and delta d = 9 / 5.
So:
$$\sqrt{a ^ 2 - a \Delta d} = \sqrt{5 ^ 2 - 5 \times \frac{9}{5}} = \sqrt{25 - 9} = \sqrt{16} = 4$$
And:
$$a - \sqrt{a \Delta d} = 5 - \sqrt{5 \times \frac{9}{5}} = 5 - \sqrt{9} = 5 - 3 = 2$$
And 4 is not 2.
Viet Dao,

 

[IB]

Thanks, VietDao29.