was not sure where to put this sorry.
I have managed to get an answer to this but looking at my book I can not understand how I got there and part of the mark is good mathmatical comunication.
Here is the surd i have to simplify
7√5
----- + √63 (that was very hard to put on a computer hope i did it ok)
√35
my answer is 4 √7
please could someone help me with each step as to why we do it as I strugle with the good mathmatical comunication part.
Because you have the answer and just want to understand the logic, let's do it step by step with the general process justifying each step.
[MATH]\sqrt{ab} \equiv \sqrt{a} * \sqrt{b}[/MATH] is a general rule.
[MATH]\therefore \dfrac{7 \sqrt{5}}{\sqrt{35}} + \sqrt{63} = \dfrac{7 \sqrt{5}}{\sqrt{7 * 5}} + \sqrt{7 * 9} = \dfrac{7 \sqrt{5}}{\sqrt{7} * \sqrt{5}} + \sqrt{9} * \sqrt{7}.[/MATH]
[MATH]\dfrac{pq}{rq} \equiv \dfrac{p}{r}[/MATH] is a general rule.
[MATH]\therefore \dfrac{7 \sqrt{5}}{\sqrt{35}} + \sqrt{63} = \dfrac{7 \sqrt{5}}{\sqrt{7} * \sqrt{5}} + \sqrt{9} * \sqrt{7} =\dfrac{7}{\sqrt{7}} + \sqrt{9} * \sqrt{7}.[/MATH]
[MATH]\sqrt{a^2} = |\ a\ |[/MATH] is a general rule.
[MATH]\therefore \dfrac{7 \sqrt{5}}{\sqrt{35}} + \sqrt{63} = \dfrac{7}{\sqrt{7}} + \sqrt{9} * \sqrt{7} = \dfrac{7}{\sqrt{7}} + \sqrt{(+3)^2} * \sqrt{7} = \dfrac{7}{\sqrt{7}} + 3\sqrt{7}.[/MATH]
It is sometimes useful to remove surds from the denominator of a fraction. (It is called rationalizing the denominator.)
[MATH]v > 0 \dfrac{u}{\sqrt{v}} \equiv \dfrac{u}{\sqrt{v}} * 1 \equiv \dfrac{u}{\sqrt{v}} * \dfrac{u}{\sqrt{v}} * \dfrac{\sqrt{v}}{\sqrt{v}} \equiv[/MATH]
[MATH]\dfrac{u\sqrt{v}}{\sqrt{v} * \sqrt{v}} \equiv \dfrac{u\sqrt{v}}{\sqrt{v * v}} \equiv \dfrac{u\sqrt{v}}{\sqrt{v^2}} \equiv \dfrac{u\sqrt{v}}{v} .[/MATH]
In short, [MATH]v > 0 \implies \dfrac{u}{\sqrt{v}} \equiv \dfrac{u\sqrt{v}}{v} [/MATH] is a general rule.
[MATH]\therefore \dfrac{7 \sqrt{5}}{\sqrt{35}} + \sqrt{63} = \dfrac{7}{\sqrt{7}} + 3\sqrt{7} = \dfrac{7\sqrt{7}}{7} + 3\sqrt{7}.[/MATH]
And it is then obvious that
[MATH]\dfrac{7 \sqrt{5}}{\sqrt{35}} + \sqrt{63} = \dfrac{\cancel 7\sqrt{7}}{\cancel 7} + 3\sqrt{7} = \sqrt{7} + 3\sqrt{7} = 4\sqrt{7}.[/MATH]
It is the systematic application of general rules that you already know (except perhaps for techniques of rationalizing a denominator).
