If I have the value of $\sqrt{1.3}$ could it be possible to find other square roots from that value? using the manipulation of surds? If I have the value of $\sqrt{1.3}$ could it be possible to find other square roots from that value? using the manipulation of surds?  
 A: You may write
$$
\sqrt{1.3}=\sqrt{\frac{1.3\times 10}{10}}=\frac{\sqrt{13}}{\sqrt{10}}=\frac{\sqrt{10}}{\sqrt{10}}\frac{\sqrt{13}}{\sqrt{10}}=\frac{\sqrt{130}}{10}.
$$
A: You can use the knowledge of $\sqrt{1.3}$ to calculate $\sqrt{x}$ only if $x = 1.3*y$, where $y$ is some other number whose square root you already know.
So $\sqrt{5.2}$ is easy, because it's just $\sqrt4*\sqrt{1.3}$, which you can calculate, assuming that you know that $\sqrt{4}=2.$
To calculate $\sqrt{2.6}$, you could argue that this is just $\sqrt2 *\sqrt{1.3}$, but that doesn't help unless you already know $\sqrt2$.
A: Let $r$ be any rational number. Then $r\sqrt {1.3}=\sqrt {1.3r^2}$, so you can find lots of other square roots, including the square roots of some integers if $r$ is an integer multiple of $10$.
It is also possible to prove that these are all the additional square roots of rational numbers which you can get in this way, even if you are able to add rationals to get numbers of the form $a+b\sqrt {1.3}$. Indeed suppose $r=(a+b\sqrt{1.3})^2=(a^2+1.3b^2)+2ab\sqrt {1.3}$. If $r$ is a rational number, then $\sqrt {1.3}$ is rational (which we know is not true) unless $ab=0$ - i.e. we must have $a=0$ or $b=0$.
Also note that every non-zero number of the form $a+b\sqrt {1.3}$ has a multiplicative inverse of the same form, because $$(a+b\sqrt{1.3})\cdot \frac {a-b\sqrt {1.3}}{a^2-1.3b^2}=1$$
These observations are also true in more general cases, and help to give a wider perspective on your particular question.
