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diff --git a/lessons/matematika/1-exponents/2.md b/lessons/matematika/1-exponents/2.md new file mode 100644 index 0000000..8a15ee3 --- /dev/null +++ b/lessons/matematika/1-exponents/2.md @@ -0,0 +1,176 @@ +# Definition +A root is an inverse of an exponent. For example: +* \\( 4^2 = 16; \sqrt{16} = 4 \\) +* \\( 5^2 = 25; \sqrt{25} = 5 \\) +* \\( 3^3 = 27; \sqrt[3]{27} = 3 \\) + +# Simplifying Square Roots +You can simplify a root by factoring the root number with another number that +is rational. In other words, divide the root number by the perfect squares. +After that, root the perfect square. + +For example: **Simplify** \\( \sqrt{60} \\)! +\\( \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \\) +!---( +#--( +## 💡 Detailed explanation +)--# +1. We factorise 60. We choose the smallest factor with a perfect square (e.g. 4, 9, 16, 25). + We will get \\(4 \times 15 \\). To verify, \\( 4 \times 15 \\) must be equal to 60. And it is! +2. Then, root the perfect square, which in this case, is 4. + \\( \sqrt{4 \times 15} = 2\sqrt{15} \\) +)---! +More examples: +* \\( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \\) +* \\( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \\) + +!---( +#--( +## ❕ There are some "leftover" roots! +)--# +Some roots are **irrational numbers**. Irrational numbers are numbers that +cannot be represented by a ratio, aka a fraction. + +We cannot simplify the roots that is irrational. So, it is **already in its +simplest form**. +)---! + +# Root Operations +Firstly, some terminology! +1. Coefficient is the multiplier of a root. For example: \\( 3\sqrt{2} \\); 3 is the coefficient here, and 2 is the root. + +## Addition +Root numbers with the same root can be summed together. + +\\( a\sqrt{c} + b\sqrt{c} = (a + b)\sqrt{c} \\) + +* \\( 3\sqrt{2} + 4\sqrt{2} = (3 + 4)\sqrt{2} = 7\sqrt2 \\) +* \\( \sqrt5 + 3\sqrt5 = (1 + 3)\sqrt5 = 4\sqrt5 \\) + +## Subtraction +Root numbers with the same root can be subtracted together. + +\\( a\sqrt{c} - b\sqrt{c} = (a - b)\sqrt{c} \\) + +* \\( 7\sqrt3 - 4\sqrt3 = (7-4)\sqrt3 = 3\sqrt2 \\) + +## Multiplication +Root numbers can be multiplied by any other root numbers. + +\\( a\sqrt{c} \times b\sqrt{d} = (a \times b)\sqrt{c \times d} \\) + +!---( +#--( +## 💡 Tip +)--# +If you multiply a root number with another root number that has the same root, +you can delete the root symbol, aka it becomes a regular number. +For example: + +\\( \sqrt5 \times \sqrt5 = 5 \\) + +!---( +#--( +### ❓ Why? +)--# +Root is the inverse of exponent. If you multiply two numbers together with the +same value, it is an exponent. If two of them meet together, they will fight +and both of them dies (they will cancel out), which means it will be a regular +ol' number. + +In this example: + +\\( \sqrt5 \times \sqrt5 = \sqrt{5^2} = 5 \\) +* Here, \\( \sqrt{5} \\) was powered to 2, which cancels it out. + * **Why do they cancel out?** We can expand this equation even more. + \\( \sqrt5 \times \sqrt5 = \sqrt{5^2} = \sqrt{25} = 5 \\) + * \\( \sqrt{25} \\) is equal to 5, so the answer is 5. + +--- + +This also works for every other number. + +--- + +### In conclusion of these tips: +* If a root is exponented, they will both cancel out. \\( \sqrt[x]{a^x} = a \\) + * Because of that, \\( \sqrt{a} \times \sqrt{a} = a \\) + +)---! +)---! + +## Division +Roots can be divided with any other roots. + +\\( a\sqrt{c} \div b\sqrt{d} = (a \div b)\sqrt{c \div d} \\) + +* \\( 8\sqrt6 \div 4\sqrt3 = (8 \div 4)\sqrt{6 \div 3} = 2\sqrt2 \\) + +# Rationalising Roots +If a fraction has an **irrational** root as the denominator, we should rationalise it. + +!---( +#--( +## ❓ What? Why? +)--# +Usually in mathemathics, a fraction should have a rational number as the +denominator. If you remember, some roots are **irrational**, like \\( \sqrt2 +\\), \\( \sqrt3 \\), \\( \sqrt 5 \\). They cannot be represented as integers or +a valid fraction. + +You don't have to rationalise a root, but it is widely accepted that the +denominator is a rational number. +)---! + +Let's learn how to rationalise a root. + +## Simple Irrational Fraction +If you have a simple irrational fraction like this: + +\\( \dfrac{1}{\sqrt2} \\) + +You can rationalise it by multiplying it with its own denominator. If you +remember correctly, when you multiply a square root with itself, it will return +its own number, therefore a rational number! + +\\( \dfrac{1}{\sqrt2} \cdot \dfrac{\sqrt2}{\sqrt2} = \sqrt{1\sqrt2}{2} \\) + +The value is still (and must be!) the same, rationalising roots are basically +representing numbers in another way. + +### More Examples +* \\( \dfrac{3}{\sqrt5} \cdot \dfrac{\sqrt5}{\sqrt5} = \dfrac{3\sqrt5}{5} \\) +* \\( \dfrac{2}{3\sqrt6} \cdot \dfrac{\sqrt6}{\sqrt6} = \dfrac{2\sqrt6}{3 \cdot 6} = \dfrac{2\sqrt6}{18} \\) (you don't need to multiply it with \\( 3\sqrt6\\)) + +## A little bit more complex fractions +If you have a fraction like this: + +\\( \dfrac{5}{3 + \sqrt2} \\) + +You can rationalise it by multiplying it with itself, but invert the signs. Like this: + +\\( \dfrac{5}{3 + \sqrt2} \cdot \dfrac{3 - \sqrt2}{3 - \sqrt2} \\) +\\( \dfrac{5 \cdot (3 - \sqrt2)}{(3 + \sqrt2)(3 - \sqrt2)} \\) + +!---( +#--( +### ❓ Looks familiar? +)--# +If you look closely, \\( (3 + \sqrt2)(3 - \sqrt2) \\) is the same pattern as +\\( (a + b)(a - b) \\)! This pattern is one of the **special binominal +products**. We use this a lot in algebra. + +So, remember this pattern: +\\( (a + b)(a - b) = a^2 - b^2 \\) +)---! + +Let's continue. + +\\( \dfrac{5 \cdot (3 - \sqrt2)}{(3 + \sqrt2)(3 - \sqrt2)} = \dfrac{5(3 - \sqrt2)}{3^2 - \sqrt{2^2}} \\) + +And if you remember correctly, \\( \sqrt{a^2} = a \\). So, + +\\( \dfrac{5(3 - \sqrt2)}{3^2 - \sqrt{2^2}} = \dfrac{5(3 - \sqrt2)}{9 - 2}\\) +\\( = \dfrac{5(3 - \sqrt2)}{6}\\) + +And that's how you rationalise complex fractions. |