Kannst du die Gleichung lösen? – Wurzelgleichungen

Hello everyone! Today I would like to show you step by step how to solve this equation using real numbers. If we look at our equation, we will first notice these roots on the left side, where our x is located. This means that we are dealing with a root equation here. And under this big root, more roots appear and more roots. So we have rooted roots here and we have to see how to deal with them. But maybe we should first take care of this big root here, because in root equations it always depends on the fact that you have to square this equation so that this root dissolves. And that is our first step, we square the equation. When squaring, we can already remember at the end that we still have to do a test, because as soon as you square an equation, solutions can arise that are not real solutions at all, so-called Scheinlösungen. That's why we have to enjoy our solutions carefully at the end. And only after the test, when we have used them, when we enter the equation and the equation opens, then we know that our found solutions are actually solutions. Otherwise not, then they will be rejected again. But that's why the test is super important. So, now we square our equation. That means the entire left side is squared. Because by squaring, this big root just dissolves. That means only what was under this big root then remains on the left side. You have to make sure that you don't get confused with all the roots, but that's how it should fit. But on the right side, we have to take our 2 to the power of 7 and square it. You can't forget that. Maybe we'll embellish the whole thing here on the right side. A potential that is potentiated again. There is a law of potency that says that these exponents are then simply multiplied with each other. So 7 times 2 is calculated. That means our new high number is 14. And that makes the whole thing look nicer. Then our x is still under a root here. The x has to get out of there. That means the root has to disappear. We just have to square again. It's not a big deal. We have to do the test anyway. So it doesn't matter. And we already know a little bit how it works. We first square the left side. Because there is a time in between here, we can square each part individually. So the x is squared once. That becomes x to the power of 2. And this root is squared. That's how it gets out. And only what was under the root is left. And don't forget on the right side that this potential also has to be squared here. But we already know that the exponents are simply multiplied with each other. That means 14 times 2 is 28. Then our x is still under a root. We know what to do. We square again. There is a time in between here. That means we can square each part individually. x to the power of 2 squared. We can also write it down first and then solve it. Then x is squared. Well, we know that x is squared. And the root gets out by squaring. And only the x under it remains. Here we also have to square again. We know that 2 times 28 is 56. Then we have to make the left side a bit nicer. Here, too, the exponents are multiplied with each other. 2 times 2 is 4 as a new exponent. And then we have different potentials here that are multiplied with each other. They all have the same base. Here is x to the power of 1. And then there is a law of potentials that says keep the base and add the exponents. 4 plus 2 plus 1 is 7. And then we have created a nice little equation here, which I wrote down here again. And now we just have to solve it for x. That means this 7 must disappear. And we can do that by simply pulling the 7th root on both sides. This ensures that this 7 disappears on the left side. Then there is only the x. But we also have to use the 7th root of everything that was there on the right. And now we have basically solved for x. But that's not particularly nice yet. We still have to simplify that a bit. Because you can rewrite such a root, especially if there is such a nice potential under the root. You just keep the base. And the exponent will become a fraction. The number that was up here also goes into the counter. So high 56. And in the denominator the number that was up here lands. The how manyth root you pull, that's the 7th root. 2 high this fraction. We can calculate this fraction. 56 divided by 7. That simply results in 8. So our solution is 2 high 8. Or if we want to calculate that, 2 times 2 times 2 times 2 times 2 times 2 times 2 times 2. That results in 256. And that's a nice solution. But we don't know if it's a solution. We have squared. That means we still have to do the test. We can't get around it. We have to calculate for every x here. So let's put it on the left side and see if we get 2 high 7. So on the left side we have this big root. Then there is our x. That means that's where our found number comes in. Then comes the middle root, in which there is also x. So also insert there. And then we have the small root, in which our number lands again. Then we just have to calculate. Let's start inside here. We can subtract. That's a square. That results in 16, because 16 times 16 equals 256. Then we can take care of the middle root. We don't have to calculate it in such a way that we have to multiply it. I'll write the rest down first. But we can now make two individual roots out of it. Because there is an x in between. That means we have the root of the first part, of 256, times the root of the second number, of 16. And they are easier to calculate. The root of 16 is nothing more than 4. And here we have already subtracted the root. That results in 16. The same principle. Everywhere there is an x in between. We subtract the big root into smaller roots. Then we have here the root of 256, the root of 16, and the root of 4. And we can subtract them. Here we have 16. In the second corner there is 4. The root of 4 is 2. What do we want to go beyond? We want to see that 2 to the power of 7 comes out. Well, then maybe we should see that we have 2 to the power of 2 here. The 16 is a 2 to the power of 2. It is 2 times 2 times 2 times 2. So 2 to the power of 4. The 4 is also a 2 to the power of 2. It is 2 to the power of 2. And the 2 is also a 2 to the power of 2. It is 2 to the power of 1. And then it actually says here 2 to the power of 7. Because the base is simply taken over and the high numbers are added. And then we come to our result, which was desired. This means that our found solution is not a Schein solution, but can also end up in the solution quantity. And with that we have solved the problem. Then I hope that it helped you. If you have any questions, feel free to write them in the comments. Otherwise, have a wonderful day and we'll see you in the next video. Take care!