Simplifying square root of an irrational

[erisedk]

Homework Statement

Find [(3 - 5^1/2)/2]^1/2

Homework Equations

The Attempt at a Solution

My calculator says (-1 + √5)/2
I have no idea how. Rationalising doesn't really do much good. Just tell me where to start.

 

[blue_leaf77]

Multiplying the original expression with $\sqrt{2}/\sqrt{2}$ will give you $\sqrt{6-2\sqrt{5}}$ in the numerator. Then think of the terms under the radical as having the form of $(a-b)^2 = a^2-2ab+b^2$.

 

[erisedk]

Could you please elaborate? I'm not getting anywhere. I assume you mean use something like completing the square but I can't do it.

 

[blue_leaf77]

What do you get after multiplying the original expression with $\frac{\sqrt{2}}{\sqrt{2}}$?

 

[vela]

You could also try working the other direction. Find the square of $\frac{-1 + \sqrt 5}{2}$ and see how it simplifies.

 

[erisedk]

What do you get after multiplying the original expression with $\frac{\sqrt{2}}{\sqrt{2}}$?

Oh I got that, ie √(6-2√5) ÷ 2
I don't get what I'm supposed to do after this.

 

[blue_leaf77]

√(6-√5) ÷ 2

It is √(6-2√5) ÷ 2. Then express the terms under the radical in the form I wrote in post #2, that is, write $6-2√5 = a^2+b^2-2ab$ . FInd the right pair of $a$ and $b$ such that the LHS is equal to RHS.

 

[erisedk]

Got it! Thanks!