Today most people don't mind knowing math. What's more, many teachers teach math how to memorize rules.
But real math involves learning how things work. And radiciation is one of those parts that are not explained correctly.
First we must understand that the exponent values, ie the numbers that we say a number is high, means how much of that number is multiplied by itself.
Two squared is twice two. So a number X raised to the number n is X times X, "n" times.
But mathematicians thought they could raise a number to a fraction. For example what would be the value of "X" raised in half. Let's say we have a site with an area of 25 square meters.
If I raise it in half, I would get 5 to the 2, times that to the half, and the answer would be 5. So with that I figured out the side of a square area. With that, it was defined that raising the half is to discover the side of the area of a square.
And so it was called Square Root. And so for the value of raised to half, they made a drawing, to represent the root.
So the actual square root shape is not the drawing, but the value raised to 1 divided by 2.
The same thing happens with for example a volume of 125 cubic meters. If I raise the value of 1 divided by 3, I find the value on the side of the cube. That's why the cubic root name.
So the real way to understand radiciation is to know that all radiciation is a number raised to a division between 0 and 1.
The difference between knowing useless math and knowing useful math is knowing what the math was made for. Therefore, in engineering, financial mathematics, physics, chemistry and other practical calculations the number raised to a fraction is used, and not drawings that hinder the calculation.
Understanding how it works, let's do some exercises.
Root of 8, plus root of 32, plus root of 72 minus root of 50, we extract the roots and see that all values have a root of 2, so we can highlight the root of 2 and add the values, which are inside the parentheses, and then we come to 7 root of 2.
In the next exercise it's the same way of solving as the first one, but now what's at the root is the value 3, and so again we put the root of 3 in evidence, and we have adding the values, 49 root of 3. Remember if you must know factoring well to solve the exercises.
The rule for this next exercise is that after removing the values from the root, add those with the same root. And so the value is 7 root of 5 minus 5 root of 6.
In the next exercise, the important thing is to know that 100 is ten squared and that 20 is 4 times 5, and then add those with equal root. In this exercise, the most important thing is to know how to factor, to get the cubic values, and add the values that in the end generates the value 0. Note that it is usually necessary to know how to factor, as factoring the rest is easier.
And again in this exercise we have the same rule, knowing how to extract the cubic root, and so the answer is 2 cubic root of 3. Remember that factoring is a basic rule for solving problems.
And now we have an exercise with just letters, the idea is to know that the root values are division values. So the "a" to the four is the same thing as the "a" to the cube times "a", and the "a" to the cube comes out of the cube root. So always remember to order the letters so you can see what the evidence will be to add. In this case, the answer to the exercise is 0.
Sometimes it may even seem boring to learn, but the knowledge of what surrounds us and how to really know how to live, is based on knowing things.
2^{2}=2.2\
X^{n}=X.X.X....\
25^{frac{1}{2}}=5^{2.frac{1}{2}}=5\
sqrt{25}=5\
sqrt[3]{125}=125^{frac{1}{3}}=5^{frac{3}{3}}=5
sqrt{8}+sqrt{32}+sqrt{72}-sqrt{50}=\
2.sqrt{2}+4.sqrt{2}+6.sqrt{2}-5.sqrt{2}=\
(2+4+6-5).sqrt{2}=7.sqrt{2}\
5.sqrt{108}+2.sqrt{243}-sqrt{27}+2sqrt{12}=\
5.6.sqrt{3}+2.9.sqrt{3}-3.sqrt{3}+2.2.sqrt{3}=\
(30+18-3+4).sqrt{3}=49.sqrt{3}\
sqrt{20}-sqrt{24}+sqrt{125}-sqrt{54}=\
sqrt{4.5}-sqrt{4.6}+sqrt{25.5}-sqrt{9.6}=\
2.sqrt{5}-2.sqrt{6}+5.sqrt{5}-3.sqrt{6}=7.sqrt{5}-5.sqrt{6}\
sqrt{2000}+sqrt{200}+sqrt{20}+sqrt{2}=\
20.sqrt{5}+10.sqrt{2}+2sqrt{5}+sqrt{2}=22.sqrt{5}+11.sqrt{2}\
sqrt[3]{128}-sqrt[3]{250}+sqrt[3]{54}-sqrt[3]{16}=\
4.sqrt[3]{2}-5.sqrt[3]{2}+3sqrt[3]{2}-2.sqrt[3]{2}=0\
sqrt[3]{375}-sqrt[3]{24}+sqrt[3]{81}-sqrt[3]{192}=\
5.sqrt[3]{3}-2sqrt[3]{3}+3.sqrt[3]{3}-4.sqrt[3]{3}=2sqrt[3]{ 3}\
a.sqrt[3]{ab^{4}}+bsqrt[3]{a^{4}b}+sqrt[3]{a^{4}b^{4}}-3ab sqrt[3]{ab}=\
ab.sqrt[3]{ab}+absqrt[3]{ab}+absqrt[3]{ab}-3absqrt[3]{ab}=0\
\
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sqrt[3]{a^{4}}=sqrt[3]{a^{3}.a}=asqrt[3]{a}
