Quadratic Surds

A quadratic surd is an irrational number that is the square root of a rational number. Quadratic surd expressions are algebraic expressions containing \sqrt{a} , where a > 0 is not a perfect square number, such as 1-\sqrt{2} , \displaystyle{\frac{1}{2-\sqrt{3}}}  etc.

Basic operational rules for \sqrt{a}  :

1) (\sqrt{a})^2 = a , where a \geq 0 ,

2) \sqrt{a^2} = |a| = a , for a>0

\sqrt{a^2} = |a| = a , for a=0

\sqrt{a^2} = |a| = a , for a<0

3) \sqrt{ab} = \sqrt{|a|} \cdot \sqrt{|b|} , if ab \geq 0

4) \displaystyle{\sqrt{\frac{a}{b}} = \frac{\sqrt{|a|}}{\sqrt{|b|}}} , if ab \geq 0 , b \neq 0

5) (\sqrt{a})^n = \sqrt{a^n} , if a \geq 0

6) a\sqrt{c} + b\sqrt{c} = (a+b)\sqrt{c} , if c \geq 0

Conjugate surds are  used to rationalize the denominators, as it more easily gives you a feel of the number.

\displaystyle{\frac{1}{a\sqrt{b}+c\sqrt{d}} = \frac{a\sqrt{b}-c\sqrt{d}}{a^2b-c^2d}} , where a,b,c,d \in Q .

In this expression, a\sqrt{b}-c\sqrt{d} is the conjugate surd.

Question:  Find the number of integers x which satisfies the inequality  \displaystyle{\frac{3}{1+\sqrt{3}} < x < \frac{3}{\sqrt{5}-\sqrt{3}}} .

(A) 2                            (B) 3                         (C) 4                         (D) 5                          (E) 6

Firstly, we need to rationalize the equations.

\displaystyle{\frac{3}{1+\sqrt{3}} = \frac{3(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{3(\sqrt{3}-1)}{2}}

\displaystyle{\frac{3}{\sqrt{5}-\sqrt{3}} = \frac{3(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{3(\sqrt{5}+\sqrt{3})}{2}}

Thus, the inequality becomes :

\displaystyle{\frac{3(\sqrt{3}-1)}{2} < x < \frac{3(\sqrt{5}+\sqrt{3})}{2}}

Simplify the equation into:

\displaystyle{\sqrt{3}-1 < \frac{2}{3}x < \sqrt{5}+\sqrt{3}}

Substitute x to be 1,2,5,6 , you will find that 1<x<6 , hence, the answer is (C) 4.

However, you may feel that the previous question is too difficult to understand. No need to fear, as today my CCA friends asked me a question about surds, and the question might be seen in lower secondary math papers. I would like to share with you all the question now.

Question: Simplify \displaystyle{4\sqrt{2} - \sqrt{50} - \frac{3}{\sqrt{2}} + \frac{7}{\sqrt{72}}}

This question might seem complicated at the start, but it is actually easy to simplify. Notice that the most difficult part to simplify is the denominators that consists of surds. Therefore, naturally we need to rationalize the denominator. After rationalisation and a little bit of tidying up, the expression becomes:

\displaystyle{4\sqrt{2} - 5\sqrt{2} - \frac{3\sqrt{2}}{2} + \frac{7\sqrt{2}}{12}}

Notice that every term has a \sqrt{2} factor. Therefore, by factorisation, the expression becomes:

\displaystyle{\sqrt{2}(4-5-\frac{3}{2}+\frac{7}{12}) = \sqrt{2}(-1-\frac{11}{12}) = -\frac{23\sqrt{2}}{12}}

Therefore, the simplified form is \displaystyle{-\frac{23\sqrt{2}}{12}} .

Cheers,

etzhky =)