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To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:

\( \frac{-6a^8}{a^4} \)
\( \frac{-6}{1} \) a^{(8 - 4)}
-6a⁴

First, solve for \( \left|a - 8\right| \):

\( \left|a - 8\right| \) + 9 = -6
\( \left|a - 8\right| \) = -6 - 9
\( \left|a - 8\right| \) = -15

The value inside the absolute value brackets can be either positive or negative so (a - 8) must equal - 15 or --15 for \( \left|a - 8\right| \) to equal -15:

So, a = 23 or a = -7.

To subtract these radicals together their radicands must be the same:

7\( \sqrt{28} \) - 3\( \sqrt{7} \)
7\( \sqrt{4 \times 7} \) - 3\( \sqrt{7} \)
7\( \sqrt{2^2 \times 7} \) - 3\( \sqrt{7} \)
(7)(2)\( \sqrt{7} \) - 3\( \sqrt{7} \)
14\( \sqrt{7} \) - 3\( \sqrt{7} \)

Now that the radicands are identical, you can subtract them:

The equation for this sequence is:

a_n = a_n-1 + 3(n - 1)

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 3(6 - 1)
a₆ = 31 + 3(5)
a₆ = 46