stormbytes
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- Joined
- Jul 19, 2020
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- 10
Hi stormbytes. The two situations are not exactly the same.What's the point of restating the exact same thing …
Thank you. I’ll revisit your answer once I’ve done radical exponents. Will have a better sense of what you are illustrating in your examples.The n-th root of a can be written as [MATH]\sqrt[n]{a}[/MATH]=a1/n. So if you have (-a)n, where n is even, (-a)n=an.
So (an)1/n=|a|. Try and check what happen when n is not even, i.e. ((-3)3)1/3=?
The book gave you an example where (a2)1/2 [MATH]\neq[/MATH] a.
Hi stormbytes. The two situations are not exactly the same.
The point they're making is this: When a negative number has been squared, we loose the original sign. In other words, if I give you the number 16, then you can't know what number was squared to obtain that 16.
Here's an example that shows a common beginner's mistake:
Solve the equation x2 = 16
The student takes the square root of each side and writes
x = 4
They missed the solution x = -4 because they incorrectly assumed that the square root of x2 must be x.
The square root of x2 is not simply x. We need to express it as |x| because there's two possibilities.
Therefore, when we take the square root of each side, we write
|x| = 4
That equation means
x = 4 \(\quad\) OR \(\quad\) x = -4
?
Let a < 0 => sqrt(a^2) > 0 -correct , now if you do sqrt(a^2)=a ,then a < 0 which would be incorrect since you would be squaring it inside the square root(the correct way) or just say sqrt(a^2) = abs(a)Thank you. That makes more sense.