# Solving equation involving surds

• Jun 11th 2013, 12:43 PM
Seaniboy
Solving equation involving surds
Hi,
Any advice with the problem below would be greatly appreciated.

X = √a + 1/√a + 1; X>0; express X - 2X in terms of a.

btw, answer = a + 1 + 1/a; how do you derive it though?

Thank you,
Sean.
• Jun 11th 2013, 01:33 PM
Shakarri
Re: Solving equation involving surds
X-2X is just -X, please recheck the question.
• Jun 11th 2013, 01:37 PM
HallsofIvy
Re: Solving equation involving surds
Where you write "1/√a+ 1" do you mean (1/√a)+ 1 or 1/(√a+ 1)?
• Jun 11th 2013, 04:17 PM
Soroban
Re: Solving equation involving surds
Hello, Seaniboy!

Quote:

$\displaystyle x \:=\:\sqrt{a} + \tfrac{1}{\sqrt{a}} + 1,\;x > 0.$

$\displaystyle \text{Express }x{\color{red}^2}-2x\text{ in terms of }a.$

$\displaystyle \text{Answer: }\:a + 1 + \tfrac{1}{a}$

$\displaystyle x^2-2x \;=\;x(x-2)$

. . . . . . $\displaystyle =\;\left(\sqrt{a} + \tfrac{1}{\sqrt{a}} + 1\right)\left(\sqrt{a} + \tfrac{1}{\sqrt{a}} + 1 - 2\right)$

. . . . . . $\displaystyle =\;\bigg(\left[\sqrt{a} + \tfrac{1}{\sqrt{a}}\right] + 1\bigg)\bigg(\left[\sqrt{a} + \tfrac{1}{\sqrt{a}}\right] - 1\bigg)$

. . . . . . $\displaystyle =\;\left(\sqrt{a} + \tfrac{1}{\sqrt{a}}\right)^2 - 1^2$

. . . . . . $\displaystyle =\;a + 2 + \tfrac{1}{a} - 1$

. . . . . . $\displaystyle =\;a + 1 + \tfrac{1}{a}$
• Jun 12th 2013, 07:50 AM
Seaniboy
Re: Solving equation involving surds
Thanks for your help people. Apologies for the symbols in the question. I must use an 𝑥 instead of X next time!
• Jun 12th 2013, 07:53 AM
Seaniboy
Re: Solving equation involving surds
Hi, I see what you mean. I am trying to find symbols to insert to make things a little clearer. In english it is X = √a + a fraction in the form of 1 over √a + 1. Thanks for having a look at the problem anyway.