# Find the value of the following expression: 4sqrt2+2(sqrt36-sqrt8)

Question
Irrational numbers
Find the value of the following expression:
$$\displaystyle{4}\sqrt{{2}}+{2}{\left(\sqrt{{36}}-\sqrt{{8}}\right)}$$

2021-02-22
$$\displaystyle{4}\sqrt{{2}}+{2}{\left(\sqrt{{36}}-\sqrt{{8}}\right)}={4}\sqrt{{2}}+{2}{\left({6}-\sqrt{{{4}\times{2}}}\right)}$$
$$\displaystyle={4}\sqrt{{2}}+{12}-{4}\sqrt{{2}}$$
$$\displaystyle={\left({4}\sqrt{{2}}-{4}\sqrt{{2}}\right)}+{12}$$
=0+12
=12
$$\displaystyle{4}\sqrt{{2}}+{2}{\left(\sqrt{{36}}-\sqrt{{8}}\right)}={12}$$

### Relevant Questions

Find the value of the following expression:
$$\displaystyle\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left(-{x}\right)}}}}}+\frac{{{\sin{{\left(-{x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}$$
Simplify the following expression:
$$\displaystyle\frac{{8}}{{{4}-\sqrt{{{6}{n}}}}}$$
$$\displaystyle{Q}:{\quad\text{if}\quad}{\left\lbrace-\frac{{3}}{{4}},\sqrt{{3}},\frac{{3}}{\sqrt{{3}}},\sqrt{{\frac{{25}}{{5}}}},{20},{1.11222},{2.5015132},\ldots,\frac{{626}}{{262}}\right\rbrace}$$,
find the following
1) Rational numbers?
2) Irrational numbers?
Find $$\displaystyle{\sqrt[{{4}}]{{{25}^{{2}}}}}$$
Consider the following statements. Select all that are always true.
The sum of a rational number and a rational number is rational.
The sum of a rational number and an irrational number is irrational.
The sum of an irrational number and an irrational number is irrational.
The product of a rational number and a rational number is rational.
The product of a rational number and an irrational number is irrational.
The product of an irrational number and an irrational number is irrational.
Find the outcome of:
$$\displaystyle{\cos{{\left(\frac{\pi}{{4}}+\frac{\pi}{{3}}\right)}}}$$
In which set(s) of numbers would you find the number $$\displaystyle\sqrt{{80}}$$
- irrational number
- whole number
- rational number
- integer
- real number
- natural number
$$\displaystyle{3}\frac{{1}}{{7}}$$ and $$\displaystyle{3}\frac{{1}}{{6}}$$
$$\frac{1}{4}$$ and $$\frac{3}{4}$$