How do I turn this answer into a radical?

Question

Starting from the top of my work, I did 2(7)$^{5\over2} - 10(7)^{3\over2}$ + $6\sqrt7$ and then did $(259.28 - 185.20 + 6\sqrt7)$ and finally got to ($74.08+6\sqrt7$). The solution for this problem in exact form is $34\sqrt7$ but how do I get to that point with the work I've done? I've checked if what I was doing equals that solution, and it's close, but how do I write that answer from the work I did? Where does the $34$ come from and how do I get it from my last step?

Answer 1

$7^{5/2}=7^27^{1/2} $ and $7^{3/2}=7\times7^{1/2}$,

so it's $2\times7^2\sqrt7-10\times7\sqrt7+6\sqrt7=(98-70+6)\sqrt7=34\sqrt7$

Answer 2

One can easily see that the multiplication of exponents is resolved by adding the powers, if the powers are whole numbers [e.g. just count the number of times that the base is being multiplied]. $$\text{e.g. }7^2\cdot 7^3 = (7\cdot 7)(7\cdot 7 \cdot 7) = 7^{2+3}$$

It's not so intuitive for fractional powers, so we just assume the $A^b\cdot A^c = A^{b+c}$ to be true, and then try to relate it to radicles.

Note that $\sqrt{7}\sqrt{7} = 7^1$ and that $7^{1/2}7^{1/2} = 7^1$ as well. This implies that $\sqrt{7} = 7^{1/2}$.

So what does one do with something like $7^{5/2}$? Again, go back to basics, and model the power as the product of two whole numbers. For example, let's take a look at $7^{12}$

$$\begin{align} 7^{3\cdot 4} = 7^{12} &= 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\cdot 7\\ &=(7\cdot 7\cdot 7)\cdot (7\cdot 7\cdot 7)\cdot (7\cdot 7\cdot 7)\\ &= 7^3\cdot 7^3\cdot 7^3\cdot 7^3\\ &= (7^3)^4 \end{align}$$

Just to summarize, we found that $7^{3\cdot 4} = (7^3)^4$. But of course multiplication is commutative, so we could just as easily have shown that $7^{3\cdot 4} = 7^{4\cdot 3} = (7^4)^3$. Additionally, we could also have picked a different factorization of the power, e.g. $7^{3\cdot 4} = 7^{12} = 7^{2\cdot 6}=(7^2)^6$. Now we're to the point where we've convinced ourselves that $A^{bc} =(A^b)^c= (A^c)^b = A^{cb}$, so we apply the principle to exponents with fractional powers. $$7^{5/2} = 7^{\frac{1}{2}\cdot 5} = (7^{1/2})^5 = \sqrt{7}^5 = \sqrt{7}^4\sqrt{7}=\sqrt{7}^{2\cdot 2}\sqrt{7}=(\sqrt{7}^2)^2\sqrt{7}=7^2\sqrt{7} = 49\sqrt{7}$$ Note that we could also have done it like this. $$7^{5/2} = 7^{5\cdot \frac{1}{2}} = (7^5)^\frac{1}{2} = \sqrt{7^5}=\sqrt{7^4}\sqrt{7}=7^2\sqrt{7} = 49\sqrt{7}$$

So applying what we've deduced, let's apply it to the problem originally presented.

$$\begin{align} 2\cdot 7^{5/2} - 10\cdot 7^{3/2}+6 \sqrt{7} &= 2\cdot (7^{1/2})^5 - 10\cdot (7^{1/2})^3 + 6\sqrt{7}\\ &= 2\cdot (7^{1/2})^4\cdot 7^{1/2} - 10\cdot (7^{1/2})^2\cdot 7^{1/2} + 6\sqrt{7}\\ &= 2\cdot (7^{1/2})^4\cdot 7^{1/2} - 10\cdot (7^{1/2})^2\cdot 7^{1/2} + 6\sqrt{7}\\ &= 2\cdot (7^{4/2})\sqrt{7} - 10\cdot (7^{2/2})\cdot \sqrt{7} + 6\sqrt{7}\\ &= (98-70+6)\sqrt{7}\\ &= 34\sqrt{7}\\ \end{align}$$