Probably a rather easy Algebra problem that I am not getting

[The Divine Zephyr]

If \sqrt{9+4\sqrt{5}}=\sqrt{a}+\sqrt{b}, find all ordered pairs (a,b)

The equation looks hauntingly like a elliptic curve, so I tried that but it didnt really help. I tried just simplifing it, but it got really messy really fast. Can someone please help?

 

[dextercioby]

Do you know the formula of DOUBLE radicals??If so,use it.

Daniel.

 

[wais]

Hi there,

I can understand why this problem may seem difficult at first glance. However, there is a simple approach that can help you solve it.

First, let's start by squaring both sides of the equation:

(\sqrt{9+4\sqrt{5}})^2=(\sqrt{a}+\sqrt{b})^2

This gives us:

9+4\sqrt{5}=a+b+2\sqrt{ab}

Now, we can see that the left side of the equation only contains a constant and a square root, while the right side contains both constants and a square root. This suggests that we can separate the equation into two parts: one with the constant and one with the square root.

So, we can set up a system of equations:

a+b=9 (equation 1)
2\sqrt{ab}=4\sqrt{5} (equation 2)

From equation 1, we can see that a=9-b. Substituting this into equation 2, we get:

2\sqrt{(9-b)b}=4\sqrt{5}

Squaring both sides again, we get:

4(9-b)b=80

Simplifying, we get:

b^2-9b+20=0

Solving for b, we get b=4 or b=5. Substituting these values back into equation 1, we get a=5 or a=4.

Therefore, the ordered pairs (a,b) are (5,4) and (4,5).

I hope this helps and clears up any confusion. Remember, always try to simplify the problem and break it down into smaller parts. Good luck!