NEW!!! Tough and tricky exponents and roots questions

In question number 1, once I reach: 50 + 2 * (sqr root of (25^2 - 10 * sqr root of 6))^2 could I operate in the following way?
Pass the 25 to the left side and multiply by 2 so: 50 + 50 * (sqr root of -600)?
Then do decompose 600 and take the square out of the square root so that: 100*2*5 * (sqr root of -3)
It does not work as I do not get to the right answer... Can anyone please help me understand what I am doing wrong here?

6. If \(x=\sqrt[5]{-37}\) then which of the following must be true?
A. \(\sqrt{-x}>2\)
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32

MUST KNOW FOR THE GMAT:

• Even roots from a negative number are undefined on the GMAT (as GMAT is dealing only with Real Numbers): \(\sqrt[{even}]{negative}=undefined\), for example, \(\sqrt{-25}=undefined\).

• Odd roots have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

BACK TO THE ORIGINAL QUESTION:

As \(-2^5=-32\), then \(x\) must be a little bit less than -2, hence \(x=\sqrt[5]{-37} \approx -2.1 \lt -2\). Thus \(x^3 \approx (-2.1)^3 \approx -8.something \lt -8\), so option D must be true.

As for the other options:

A. \(\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1} \lt 2\), \(\sqrt{-x} \gt 2\) is not true.

B. \(x \approx -2.1 \lt -2\), thus \(x \gt -2\) is also not true.

C. \(x^2 \approx (-2.1)^2=4.something \gt 4\), thus \(x^2 \lt 4\) is also not true.

Answer: D.

7. If \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\), then which of the following must be true?

A. \(x \gt 12\)
B. \(10 \lt x \lt 12\)
C. \(8 \lt x \lt 10\)
D. \(6 \lt x \lt 8\)
E. \(x \lt 6\)


Here is a useful little trick: any positive integer root of a number greater than 1 will be more than 1. For example: \(\sqrt[1000]{2} \gt 1\).

Now, \(\sqrt{10} \gt 3\) (as \(3^2=9\)) and \(\sqrt[3]{9} \gt 2\) (as \(2^3=8\)).

Therefore:

\(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}=\)

\(=(\text{number more than 3})+(\text{number more than 2})+(\text{7 numbers more than 1})=\)

\(=(\text{number more than 5})+(\text{number more than 7})=\)

\(=(\text{number more than 12})\)

Answer: A
Answer: E.

How about answer choice E? E looks true as well as D

Is there another way to do this exercise? I tried calculating the maximun and the minimum to understand the range of x. We have the sum of 9 terms. If all terms were equal to the largest term (sqrt{10}), we would have: sum = 9* (sqrt{10}) with is aprox 27 (a bit more than 27). So, the actual sum must be less than 27.
Then, we do the same with the smallest: sum = 9* 2^(1/10).... 3^2*2^(1/10). I struggle solving that last expression...