# HARD QUESTIONS ON SURDS

In our previous post, we learn the addition and subtraction of surds, we are to be going over some hard questions on surds.

These questions are not so hard. You just need to be careful and follow the methods.

Question 1
Given that $\frac{2\sqrt{2}}{\sqrt{3}-1}-\frac{2\sqrt{3}}{\sqrt{2}+1}=x\sqrt{2}+y\sqrt{3}+z\sqrt{6}$. Find the value of $x$, $y$ and $z$.

For simplicity, we would solve the L.H.S and R.H.S separately.
Ok firstly, let's solve the L.H.S ($\frac{2\sqrt{2}}{\sqrt{3}-1}$)

By rationalizing the denominator
$\frac{2\sqrt{2}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\frac{2\sqrt{2}(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2}$
$\frac{2\sqrt{6}+2\sqrt{2}}{3-1}$
$\frac{2\sqrt{6}+2\sqrt{2}}{2}$

$2$ is common
$\frac{\require{cancel}\bcancel{2}(\sqrt{6}+\sqrt{6})}{\require{cancel}\bcancel{2}}$
$\sqrt{6}+\sqrt{2}$

Now, let solve the other part of the fraction $\frac{2\sqrt{3}}{\sqrt{2}+1}$

By rationalizing the denominator
$\frac{2\sqrt{3}}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}$
$\frac{2\sqrt{3}(\sqrt{2}-1)}{(\sqrt)^2-(1)^2}$
$\frac{2\sqrt{6}-2\sqrt{3}}{2-1}$
$\frac{2\sqrt{6}-2\sqrt{3}}{1}$
$2\sqrt{6}-2\sqrt{3}$

Having solve for both part of the fraction, Now let's subtract both answers
$\sqrt{6}+\sqrt{2}-(2\sqrt{6}-2\sqrt{3})$

Opening the bracket
$\sqrt{6}+\sqrt{2}-2\sqrt{6}+2\sqrt{3}$
$-\sqrt{6}+\sqrt{2}+2\sqrt{3}$
$\sqrt{2}+2\sqrt{3}-\sqrt{6}$

To derive the value of $x$, $y$ and $z$, ,we will compare.
$\sqrt{2}+2\sqrt{3}-\sqrt{6}=x\sqrt{2}+y\sqrt{3}+z\sqrt{6}$

Hence,
$x=1$, $y=2$, $z=-1$.

Question 2
Evaluate $\sqrt{\frac{8}{3}}+\frac{3}{2}\times\sqrt{\frac{8}{27}}$

$\frac{\sqrt{8}}{\sqrt{3}}+\frac{3}{2}\times\frac{\sqrt{8}}{\sqrt{3}}$

$\sqrt{8}$ and $\sqrt{27}$ can be further simplified. Thus
$\frac{\sqrt{4\times2}}{\sqrt{3}}+\frac{3}{2}\times\frac{\sqrt{4\times2}}{\sqrt{9\times3}}$
$\frac{\sqrt{4}\times\sqrt{2}}{\sqrt{3}}+\frac{3}{2}\times\frac{\sqrt{4}\times\sqrt{2}}{\sqrt{9}\times\sqrt{3}}$
$\frac{2\sqrt{2}}{\sqrt{3}}+\frac{3}{2}\times\frac{2\sqrt{2}}{3\sqrt{3}}$
$\frac{2\sqrt{2}}{\sqrt{3}}+\frac{6\sqrt{2}}{6\sqrt{3}}$

Taking the L.C.D
$\frac{12\sqrt{2}+6\sqrt{2}}{6\sqrt{3}}$
$\frac{18\sqrt{2}}{6\sqrt{3}}$
$\frac{3\sqrt{2}}{2\sqrt{3}}$

Rationalizing the denominator
$\frac{3\sqrt{2}}{2\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\frac{3\sqrt{6}}{\sqrt{9}}$
$\frac{3\sqrt{6}}{3}$
$\frac{\require{cancel}\bcancel{3}\sqrt{6}}{\require{cancel}\bcancel{3}}=\sqrt{6}$

Related post
Question 3
Simplify $\frac{(2+\sqrt{3})^2-(1-\sqrt{3})^2}{\sqrt{3}}$ completely.

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

By expansion
$\frac{2(2+\sqrt{3})+\sqrt{3}(2+\sqrt{3})-(1(1-\sqrt{3})-\sqrt{3}(1-\sqrt{3}))}{\sqrt{3}}$
$\frac{4+2\sqrt{3}+2\sqrt{3}+\sqrt{9}-(1-\sqrt{3}-\sqrt{3}+\sqrt{9})}{\sqrt{3}}$
$\frac{4+2\sqrt{3}+2\sqrt{3}+3-(1-\sqrt{3}-\sqrt{3}+3)}{\sqrt{3}}$
$\frac{7+4\sqrt{3}-(4-2\sqrt{3})}{\sqrt{3}}$

Opening the bracket
$\frac{7+4\sqrt{3}-4+2\sqrt{3}}{\sqrt{3}}$
$\frac{3+6\sqrt{3}}{\sqrt{3}}$

Rationalizing the denominator
$\frac{3+6\sqrt{3}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\frac{\sqrt{3}(3+6\sqrt{3})}{\sqrt{9}}$
$\frac{3\sqrt{3}+18}{3}$

3 is common to both side, hence
$\frac{\require{cancel}\bcancel{3}(\sqrt{3}+6)}{\require{cancel}\bcancel{3}}$
$\sqrt{3}+6$
$6+\sqrt{3}$

Question 4
If $k\sqrt{28}+\sqrt{63}-\sqrt{7}=0$, find k

We can further simplify these surds.
$k\sqrt{4\times7}+\sqrt{9\times7}-\sqrt{7}=0$
$2k\sqrt{7}+3\sqrt{7}-\sqrt{7}=0$
$2k\sqrt{7}+2\sqrt{7}=0$
$2k\sqrt{7}=-2\sqrt{7}$
$k(2\sqrt{7})=-1(2\sqrt{7}$

Divide both sides by $2\sqrt{7}$
$\frac{k(2\sqrt{7})}{2\sqrt{7}}=\frac{-1(2\sqrt{7})}{2\sqrt{7}}$

We have come to the end of our series on surds. However, if you are still not familiar with surds, make reference to this mega post on surd where we explain everything there is to know on surds.