# Is my method of solving equation correct?

The problem in question is $$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$

using $$a+b=2$$ where $$a=\sqrt[5]{16+\sqrt{x}}$$ and $$b=\sqrt[5]{16-\sqrt{x}}$$

$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$ $$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$ $$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$ $$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$

Got $$\frac{32}{(a+b)^2}$$ from the fact that $$(a+b)^2(a+b)^3=32$$ and $$a+b=2$$

$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$ $$a^5+b^5+40ab+20a^2b^2=32$$ From when I defined a and b earlier, I substitute and get $$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$ $$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$ $$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$ $$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$

Then let $$u=\sqrt[5]{256+{x}}$$, $$20(2u+u^2)=0$$ $$u(u+2)=0$$ $$u=0,-2$$

Substituting u to get x from $$u=\sqrt[5]{256+{x}}$$, I get $$x=\cancel{-288},256$$

However, since the original equation has a $$\sqrt{x}$$, which can't be negative, I eliminate $$x=-288$$, leaving just $$x=256$$ as my answer.

So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know. Thank you!

• +1 for your impressive typesetting Dec 7, 2020 at 7:45
• haha thanks :) it took awhile since I'm new to this. Dec 7, 2020 at 7:46

Spoiler alert: this is the solution.

Let's take $$a = \sqrt[5]{16+\sqrt x}$$ and $$b = \sqrt[5]{16-\sqrt x}$$ . Note that because $$\sqrt x \ge 0$$, $$a$$ is always positive (technically greater than or equal to $$\sqrt[5]{16}$$), but $$b$$ can be positive, zero, or negative. We have: $$a + b = 2$$ $$(a + b)^5 = 32$$ We also have: $$a^5 = 16 + \sqrt x$$ $$b^5 = 16 - \sqrt x$$ $$a^5 + b^5 = 32$$ Aha! Let's see where this takes us: $$(a + b)^5 = a^5 + b^5$$ $$a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 = a^5 + b^5$$ $$5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 = 0$$ For this sum to be zero, either all of its terms should be zero, or some of them should be positive and some others negative. The former case happens if $$b$$ is zero (we noted earlier that $$a$$ is always positive, so it can't be zero). The latter case can only happen if $$b$$ is negative, because again as we noted earlier, $$a$$ is always positive and cannot make negative terms. Let's keep that in mind and continue. $$5ab(a^3 + 2a^2b + 2ab^2 + b^3) = 0$$ The expression in the parentheses looks a bit like $$(a+b)^3$$, but not exactly! Let's add $$a^2b + ab^2 - a^2b - ab^2$$ to it: $$5ab[(a+b)^3 - a^2b - ab^2] = 0$$ $$5ab[(a+b)^3 - ab(a+b)] = 0$$ $$5ab(a+b)[(a+b)^2-ab] = 0$$ And recalling that $$a+b = 2$$: $$10ab[4-ab]=0$$ Now, for this equality to hold, we must have either $$ab = 4$$ or $$ab = 0$$.

But $$ab = 4$$ is not possible, because if it were so, considering that $$a$$ is positive, $$b$$ should necessarily be positive as well (otherwise $$ab$$ would not be positive). And if both $$a$$ and $$b$$ were positive, then the sum $$5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4$$ would necessarily be positive, whereas it should be zero.

So we are left with the conclusion that $$ab = 0$$. As we mentioned before, $$a$$ is positive and not zero. So: $$b = 0$$ $$\sqrt[5]{16-\sqrt x} = 0$$ $$16 - \sqrt x = 0$$ $$x = 16^2 = 256$$

• Can you explain why "b" can be positive, zero, or negative. I am having a difficult time why? Thanks! Dec 8, 2020 at 1:06
• Other than that, like the solution! Dec 8, 2020 at 1:06
• @EdTwo Thanks for the compliment. About 'b', in the end it turned out to be 0. The 1st paragraph statement about the sign of 'b' was only an initial argument at the beginning of the solution, and it was based on its formula: it is the fifth root of the difference between two numbers. That difference may be negative, zero, or positive; and so can its fifth root. An initial look at the formula doesn't rule out any of these possibilities. In contrast, 'a' is the fifth root of the sum of a non-negative number and 16. So just by looking at its defining formula we can say 'a' has to be positive. Dec 8, 2020 at 1:29
• got it! thanks for the explanation! Dec 8, 2020 at 1:55
• You're welcome! Dec 8, 2020 at 3:06

When you went from : $$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$ to $$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$ You incorrectly assumed that $$a^3 + b^3 = (a+b)^3 = \biggl(\frac{32}{(a+b)^2}\biggr)$$ Therefore you cannot say that $$a^3 + b^3 = \frac{32}{2^2}$$

• you're right. Man, I thought I did good. lol. looks like its back to square one. Dec 7, 2020 at 7:58

You can correct the error that Ryan points out. Instead, note that $$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 8 - 6ab.$$ So, $$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$ becomes $$a^5 + b^5 + 5ab(8 - 6ab) + 20a^2b^2 = 32.$$ Using $$a^5 + b^5 = 32$$, then expanding and collecting like terms, $$40a^2b^2 - 10ab = 0 \implies ab = 0 \text{ or } ab = 4.$$ So, now we know that $$(r - a)(r - b) = r^2 - (a + b)r + ab$$ equals either $$r^2 - 2r + 4$$, or $$r^2 - 2r$$, depending on the value of $$ab$$. Note that the former has no real roots (though it does have complex solutions, but that's whole other kettle of fish), but the latter does. The two numbers would have to be the roots of the former polynomial: $$0$$ and $$2$$. We must therefore have, putting the numbers in ascending order, $$\sqrt[5]{16 - \sqrt{x}} = 0 \quad \sqrt[5]{16 + \sqrt{x}} = 2.$$ It's not hard to see that $$\sqrt{x} = 16 \implies x = 256$$.

Note: this is the same answer as your method. The reason is, because as it turned out $$b = 0$$, we actually had $$(a + b)^3 = a^3 + b^3$$ for that particular $$a$$ and $$b$$.

When you know $$a+b$$, you can express $$a^n+b^n$$ in terms of powers of $$c=ab$$. A few cases

• $$a^2+b^2=(a+b)^2-2ab$$
• $$a^3+b^3=(a+b)^3-3ab(a+b)$$

What about $$a^5+b^5$$? The relations are obtained from the expansion of $$(a+b)^5$$: \begin{align} (a+b)^5&=a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)\\ &=a^5+b^5+5ab((a+b)^3-3ab(a+b))+10a^2b^2(a+b)\\ &=a^5+b^5+5ab(a+b)^3-5a^2b^2(a+b) \end{align} In your case you get $$a^5+b^5=32-40c+10c^2$$ which yields $$32=32-40c+10c^2$$ hence $$c=0$$ or $$c=4$$.

The first case yields $$16-\sqrt{x}=0$$; the second case yields $$256-x=4^5$$

Alternative approach. Since $$a=\sqrt[5]{16+\sqrt{x}}\ne0$$, we can set $$t=b/a$$ and from $$(a+b)^5=a^5+b^5$$, we get $$f(t)=(t+1)^5-t^5-1=0$$ We have $$f'(t)=5(t+1)^4-5t^4$$, which vanishes only at $$t=-1/2$$, and $$f(-1/2)<0$$. Thus $$f$$ only vanishes twice, but one root is negative. Since $$0$$ is a root, it is the only nonnegative one. Thus $$b=0$$.

There is another solution.

Let $$\sqrt[5]{16+\sqrt{x}}=1+t$$, where $$t>0$$; therefore, $$\sqrt[5]{16-\sqrt{x}}=1-t$$. Hence $$(1+t)^5+(1-t)^5=32,$$ or $$10\,t^4+20\,t^2+2=32.$$ Immediately we have $$t^4+2\,t^2-3=0,$$ and $$t^2=1$$. Since $$t>0$$, $$t=1$$ is the single solution. So we have $$\sqrt[5]{16-\sqrt{x}}=0\Rightarrow \sqrt{x}=16\Rightarrow x=256.$$