# Representing $\frac{3 + \sqrt{5}}{2}$ as a square of a quadratic surd

How can I represent $\frac{3 + \sqrt{5}}{2}$ as square of a quadratic surd? Actually, I was solving a question where $\frac{3 + \sqrt{5}}{2}$ was converted to $(\frac{1+\sqrt{5}}{2})^2$. How did the solution writer think of it?

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$\sqrt{\frac{7 + 3\sqrt 5}{2}}$ – SomeStrangeUser Feb 2 '14 at 10:06
@SomeStrangeUser: I think you missed the point. – TonyK Feb 2 '14 at 10:11
Sorry, Didn't read it right the first time. we can set $(a+b\sqrt{5})^2=\frac{3}{2} + \frac{1}{2}\sqrt{5}$. Then, solve for $a$ and $b$. This gets us: $$a^2 + 5b^2 = \frac{3}{2},\ 2ab = \frac{1}{2}$$ – SomeStrangeUser Feb 2 '14 at 10:17

We can set $(a+b\sqrt{5})^2=\frac{3}{2} + \frac{1}{2}\sqrt{5}$. Then, solve for rational $a$ and $b$.

Comparing the terms we obtain: $$a^2 + 5b^2 = \frac{3}{2},\ \ \ 2ab = \frac{1}{2}$$ Solving for $a$ and $b$, we get $a=b=\frac{1}{2}$ or $a=b=-\frac{1}{2}$, thus the two possible squares are: $$\left(\frac{1+\sqrt{5}}{2}\right)^2,\ \ \ \left(\frac{-1-\sqrt{5}}{2}\right)^2$$

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Hint: $1+\frac{1}{\phi}=\phi$, where $\phi$ is the golden ratio. Mathematicians are very familiar with the basic properties of this number.

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So you mean had it not been related to the golden ratio, it would have been difficult to convert like that? – shaurya gupta Feb 2 '14 at 10:15
@shaurya: I think David's answer is completely irrelevant! Why did you accept it? – TonyK Feb 2 '14 at 10:22
I didn't mean to...Coz he hasn't answered my comment... – shaurya gupta Feb 2 '14 at 11:48
This answer is not irrelevant at all...from that identity you find that $\phi^2 = \phi + 1$ (although granted it is easier to start with this equation as defining $\phi$) and the RHS is exactly the number we are wanting the square root of. – fretty Feb 2 '14 at 13:44
Having it in the form $\phi^2 = \phi + 1$ certainly makes more sense for this problem than $1 + \frac{1}{\phi} = \phi$... – Soke Feb 2 '14 at 14:41

Here is a way to 'complete the square'

$\frac{3 + \sqrt{5}}{2} \cdot \frac{2}{2} =$

$\frac{6 + 2 \sqrt{5}}{4} =$

$\frac{1 + 2 \sqrt{5} + 5}{4} =$

$\frac{(1 + \sqrt{5})^2}{4} =$

$(\frac{1 + \sqrt{5}}{2} )^2$

So

$\sqrt{ \frac{3 + \sqrt{5}}{2} } = \frac{1 + \sqrt{5}}{2}$

The other root is obvious by inspection , $- \frac{1 + \sqrt{5}}{2}$

Now , as an exercise , can you do the same procedure to derive the other root?

Hint: You need to place a minus sign in an appropriate position.

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$\frac{1+\sqrt 5}{2}$ is the golden ratio.

It is the solution to the quadratic equation $x^2=x+1$

The solution writer was probably familiar with this number.

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How does this help? – TonyK Feb 2 '14 at 10:22
$\frac{3+\sqrt 5}{2} = \frac{1+\sqrt 5}{2}+1$ – Flowers Feb 2 '14 at 10:24
But how is your answer supposed to explain how to find the square root of $\frac{3+\sqrt 5}{2}$? – TonyK Feb 2 '14 at 10:27
@TonyK Because it's equal to the square of the golden ratio, and so it's square root is the golden ratio. See my answer. – David H Feb 2 '14 at 10:31
The OP wanted to understand the thought process of the "solution writer." That's what I had to offer, sorry. – Flowers Feb 2 '14 at 10:35