MHB How can we solve a radical equation without using a calculator?

[mathdad]

Show that the RHS = LHS without using a calculator.

View attachment 7807

 

[MarkFL]

I would observe that:

$$\left(1\pm\sqrt{5}\right)^3=1\pm3\sqrt{5}+3\cdot5\pm5\sqrt{5}=16\pm8\sqrt{5}=8\left(2\pm\sqrt{5}\right)$$

Hence:

$$2\pm\sqrt{5}=\left(\frac{1\pm\sqrt{5}}{2}\right)^3$$

 

[mathdad]

I would observe that:

$$\left(1\pm\sqrt{5}\right)^3=1\pm3\sqrt{5}+3\cdot5\pm5\sqrt{5}=16\pm8\sqrt{5}=8\left(2\pm\sqrt{5}\right)$$

Hence:

$$2\pm\sqrt{5}=\left(\frac{1\pm\sqrt{5}}{2}\right)^3$$

What would the algebra look like if I take the cube root on both sides? Is it tedious?

 

[Greg]

That's not necessary. Put $\left(\frac{1\pm\sqrt5}{2}\right)^3$ instead of $2\pm\sqrt5$ in $\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=1$. What do you get?

If you mean taking the cube root of both sides of $2\pm\sqrt{5}=\left(\frac{1\pm\sqrt{5}}{2}\right)^3$, you'd arrive at the solution.

 

[I like Serena]

What would the algebra look like if I take the cube root on both sides? Is it tedious?

Let's assume that the cube root is in the same Ring of Quadratic Integers.
That is, we can write the cube root $\sqrt[3]{2\pm \sqrt 5}$ as $a+b\sqrt 5$.

Then to take the cube root, we can solve a and b from:
$$(a+b\sqrt 5)^3 = 2 \pm \sqrt 5$$
We have:
$$(a+b\cdot\sqrt 5)^3 = a^3+3a^2b\cdot\sqrt 5 + 3ab^2\cdot 5 + b^2\cdot 5\sqrt 5 = a(a^2+15b^2) + b(3a^2 + 5b^2)\sqrt 5$$
So:
$$\begin{cases}a(a^2+15b^2) = 2 \\ b(3a^2 + 5b^2) = \pm 1
\end{cases} \tag 1$$
Unfortunately this is quite tedious to solve. (Worried)

Luckily we can draw on the theory of Quadratic Integers to make it a bit easier.
That's by using the norm $N$, which is defined as $N(x+y\sqrt 5)=x^2-5y^2$.
The norm $N(a+b\sqrt 5)$, when raised to the power of 3, must then be equal to $N(2\pm \sqrt 5)$.
Thus, we get the additional identity:
$$N(a+b\sqrt 5)^3 = N(2\pm \sqrt 5) \quad\Rightarrow\quad (a^2 - 5b^2)^3 = 2^2 - 5(\pm 1)^2 = -1$$
This implies:
$$a^2-5b^2 = -1 \quad\Rightarrow\quad 5b^2 = a^2+1 \tag 2$$
Substituting in (1) yields:
$$a(a^2+3\cdot 5b^2)= a(a^2+3(a^2+1))=4a^3+3a=2 \quad\Rightarrow\quad 4a^3+3a-2=0$$

Next, from the Rational Root Theorem we know that any rational root must be one of:
$$\pm 2, \pm 1, \pm\frac 12, \pm\frac 14$$
And indeed, $a=\frac 12$ fits.
Substituting back into (2) yields:
$$5b^2 = \Big(\frac 12\Big)^2+1 = \frac 54 \quad\Rightarrow\quad b=\pm\frac 12$$

So we find that:
$$\sqrt[3]{2\pm \sqrt 5} = \frac 12 \pm \frac 12 \sqrt 5 \tag 3$$TL;DR: Yep, it's tedious to take the cube root. (Angel)

 

[mathdad]

Let's assume that the cube root is in the same Ring of Quadratic Integers.
That is, we can write the cube root $\sqrt[3]{2\pm \sqrt 5}$ as $a+b\sqrt 5$.

Then to take the cube root, we can solve a and b from:
$$(a+b\sqrt 5)^3 = 2 \pm \sqrt 5$$
We have:
$$(a+b\cdot\sqrt 5)^3 = a^3+3a^2b\cdot\sqrt 5 + 3ab^2\cdot 5 + b^2\cdot 5\sqrt 5 = a(a^2+15b^2) + b(3a^2 + 5b^2)\sqrt 5$$
So:
$$\begin{cases}a(a^2+15b^2) = 2 \\ b(3a^2 + 5b^2) = \pm 1
\end{cases} \tag 1$$
Unfortunately this is quite tedious to solve. (Worried)

Luckily we can draw on the theory of Quadratic Integers to make it a bit easier.
That's by using the norm $N$, which is defined as $N(x+y\sqrt 5)=x^2-5y^2$.
The norm $N(a+b\sqrt 5)$, when raised to the power of 3, must then be equal to $N(2\pm \sqrt 5)$.
Thus, we get the additional identity:
$$N(a+b\sqrt 5)^3 = N(2\pm \sqrt 5) \quad\Rightarrow\quad (a^2 - 5b^2)^3 = 2^2 - 5(\pm 1)^2 = -1$$
This implies:
$$a^2-5b^2 = -1 \quad\Rightarrow\quad 5b^2 = a^2+1 \tag 2$$
Substituting in (1) yields:
$$a(a^2+3\cdot 5b^2)= a(a^2+3(a^2+1))=4a^3+3a=2 \quad\Rightarrow\quad 4a^3+3a-2=0$$

Next, from the Rational Root Theorem we know that any rational root must be one of:
$$\pm 2, \pm 1, \pm\frac 12, \pm\frac 14$$
And indeed, $a=\frac 12$ fits.
Substituting back into (2) yields:
$$5b^2 = \Big(\frac 12\Big)^2+1 = \frac 54 \quad\Rightarrow\quad b=\pm\frac 12$$

So we find that:
$$\sqrt[3]{2\pm \sqrt 5} = \frac 12 \pm \frac 12 \sqrt 5 \tag 3$$TL;DR: Yep, it's tedious to take the cube root. (Angel)

Wonderfully informative reply. Thanks.