Surds: solving equations containing numbers in square roots

[~*alphAOmega*~]

I just started "IB higher level math", for anyone who knows what the IB is. On page fourteen, we're doing surds and rational numbers, etc. There is a question here which i just can't do. :? It goes like this:

"Solve each of the following equations, expressiong each answer in simplest surd form"

a) 2x - sqrt[3] +2(1 - Xsqrt[3]) = 5 + 2sqrt[3]

b) x - sqrt[3] / 2x + sqrt[3] = sqrt[3]

thanks, your help will be greatly appreciated

 

[soroban]

Re: Surds..? help plz =)

Hello, ~*alphAOmega*~!

Here's the first one . . . with baby-steps.

Solve for \(\displaystyle x:\;\;\;2x\,-\,\sqrt{3} \,+\,2\left( 1\,-\,x\sqrt{3}\right)\;=\;5\,+\,2\sqrt{3}\)

Expand: \(\displaystyle \,2x\,-\,\sqrt{3}\,+\,2-\,2x\sqrt{3}\;=\;5\,+\,2\sqrt{3}\)

Add \(\displaystyle \sqrt{3}-2\) to both sides: \(\displaystyle \,2x \,-\,2x\sqrt{3}\;=\;3\,+\,3\sqrt{3}\)

Factor: \(\displaystyle \,2x\left(1\,-\,\sqrt{3}\right)\;=\;3(1\,+\,\sqrt{3})\)

Then: \(\displaystyle \,x\;=\;\frac{3(1\,+\,\sqrt{3})}{2(1\,-\,\sqrt{3})}\)

Rationalize: \(\displaystyle \,\frac{3(1\,+\,\sqrt{3})}{2(1\,-\,\sqrt{3})}\,\cdot\,\frac{1\,+\,\sqrt{3}}{1\,+\,\sqrt{3}} \;=\;\frac{3(1\,+\,\sqrt{3}\,+\,\sqrt{3} + 3)}{2(1\,+\,\sqrt{3}\,+\,\sqrt{3}\,-\,3)} \;=\;\frac{3(4\,+\,2\sqrt{3})}{2(1\,-\,3)} \;=\;\frac{3\cdot2(2\,+\,\sqrt{3})}{2(-2)\)

Therefore: \(\displaystyle x\;=\;-\frac{3(2\,+\,\sqrt{3})}{2}\)

 

[Denis]

Re: Surds: solving equations containing numbers in square ro

b) x - sqrt[3] / 2x + sqrt[3] = sqrt[3]

The way you show that means:
x - sqrt[3] / 2x = sqrt[3] - sqrt[3]
so x - sqrt(3) = 0
x = sqrt(3)

Did you mean: x - sqrt[3] / (2x + sqrt[3]) = sqrt[3] ?