How To Approximate The Value Of The Square Root Of 2

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• There are two common ways to simplify radical expressions, depending on the denominator.

Using the identities $\setminus {\sqrt{a}}^{2} = a$ and $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$, in fact, you can get rid of the roots at the denominator.

Case 1: the denominator consists of a single root. For example, let's say that our fraction is $\frac{3 x}{\setminus \sqrt{x + 3}}$. If we multply this fraction by $\frac{\setminus \sqrt{x + 3}}{\setminus \sqrt{x + 3}}$, we won't change its value (since of course $\frac{\setminus \sqrt{x + 3}}{\setminus \sqrt{x + 3}} = 1$, but we can rewrite it as follows:
$\frac{3 x}{\setminus \sqrt{x + 3}} \setminus \cdot \frac{\setminus \sqrt{x + 3}}{\setminus \sqrt{x + 3}} = \setminus \frac{3 x \setminus \sqrt{x + 3}}{\setminus {\sqrt{x + 3}}^{2}}$, and finally obtain
$\setminus \frac{3 x \setminus \sqrt{x + 3}}{x + 3}$

Case 2: the denominator consists of a sum/difference of roots. If we multiply by the difference/sum of the roots, we'll have the same result as above. For example, if you have

\frac{\cos(x)}{\sqrt{x}+\sqrt{\sin(x)}

You'll multiply numerator and denominator by the difference \sqrt{x}-\sqrt{\sin(x), and obtain

$\setminus \frac{\setminus \cos \left(x\right)}{\setminus \sqrt{x} + \setminus \sqrt{\setminus \sin \left(x\right)}} \setminus \cdot \setminus \frac{\setminus \sqrt{x} - \setminus \sqrt{\setminus \sin \left(x\right)}}{\setminus \sqrt{x} - \setminus \sqrt{\setminus \sin \left(x\right)}}$ which is

$\setminus \frac{\setminus \cos \left(x\right) \left(\setminus \sqrt{x} - \setminus \sqrt{\setminus \sin \left(x\right)}\right)}{\setminus {\sqrt{x}}^{2} - \setminus {\sqrt{\setminus \sin \left(x\right)}}^{2}}$

which finally equals

$\setminus \frac{\setminus \cos \left(x\right) \left(\setminus \sqrt{x} - \setminus \sqrt{\setminus \sin \left(x\right)}\right)}{x - \setminus \sin \left(x\right)}$

Of course, when working with radicals, you always need to pay attention and make sure that the argument of the root is positive, otherwise you will write things that have no meaning!

• Generally, you don't want to have radical at the denominators. So, let's say that we want to simplify the expression $\setminus \frac{\setminus \sqrt{a}}{\setminus \sqrt{b}}$, where $a$ and $b$ can be any expression you want. Since, of course, $\setminus \frac{\setminus \sqrt{b}}{\setminus \sqrt{b}} = 1$, we can multiply it without changing the value of our expression, so we have $\setminus \frac{\setminus \sqrt{a}}{\setminus \sqrt{b}} = \setminus \frac{\setminus \sqrt{a}}{\setminus \sqrt{b}} \setminus \cdot \setminus \frac{\setminus \sqrt{b}}{\setminus \sqrt{b}}$. The advantage is that now we observe that $\setminus \sqrt{b} \setminus \cdot \setminus \sqrt{b} = b$, and so our expression becomes $\setminus \frac{\setminus}{\setminus \sqrt{a b}} \left\{b\right\}$, and we got rid of the radical at the denominator.

• This is easy! If you want to multiply this are the rules: First coefficients are multiplied with each other and the sub-radical amounts each other, placing the latter product under the radical sign common and the result is simplified.

Let's go: $2 \sqrt{5}$ times $3 \sqrt{10}$

2sqrt5 × 3sqrt10 = 2 × 3sqrt(5×10)=6sqrt50

= 6sqrt(2·5^2)

$= 30 \sqrt{2}$

Now if you want to divide, then the coefficients are divided among themselves and sub-radical amounts each other, placing the latter quotient under the radical common and the result is simplified.

$2 \sqrt[3]{81 {x}^{7}}$ by $3 \sqrt[3]{3 {x}^{2}}$

$\frac{2 \sqrt[3]{81 {x}^{7}}}{3 \sqrt[3]{3 {x}^{2}}} = \frac{2}{3} \sqrt[3]{\frac{81 {x}^{7}}{3 {x}^{2}}} = \frac{2}{3} \sqrt[3]{27 {x}^{5}}$

2/3 root3 (3^3·x^3·x^2) = 2xroot3 (x^2)

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