Question

Find the indicated real nth root(s) of a. n=9, a=−512 The root is .

Answer 1

The real ninth root of -512 is -2, which is found by simplifying (-512)^(1/9) considering that 512 is 2 raised to the power of 9.

Step-by-step explanation:

To find the indicated real nth root of a number, we are looking for a value that, when raised to the nth power, gives us the original number. For n=9 and a=−512, we are looking for the ninth root of -512. We can use the rule that a negative number raised to an odd power is negative and a positive number raised to any power remains positive to understand that only odd roots of negative numbers are real numbers.

Therefore, we can express the ninth root of -512 as (-512)^(1/9). Since 512 is 2 raised to the power of 9 (2^9=512), we can rewrite the expression as (-2)^9^(1/9). Applying the property that (x^n)^(1/n) equals x, the ninth root of -512 simplifies to -2. Thus, the real ninth root of -512 is -2.

Answer 2

-2

Step-by-step explanation:

"Find the indicated real nth root(s) of a. n=9, a=−512" becomes
`\sqrt[9]{-512}`. The nth root is odd so its root will be negative. THis is true since -512 is the number. The root of this is a number when multiplied by itself 9 times will give -512. 512 is even so it is likely an even number.

For example , 2*2*2*2*2*2*2*2*2 = 512.

Therefore the number is -2.