Only 3% correctly figured out this math problem

Hello everyone! Happy to see you here. Welcome back to my channel Higher Mathematics. Today we have very interesting and in the same way this very tricky algebra question. We have a to the fourth equal to a minus one raised to the fourth power and we need to find our a. If you have your solution, your approach, you can also write your solution in the comments below and then we will check our answers. So right now let's start. A lot of students might be thinking okay let's just cancel this fourth power from both sides but that's not how it works. Right now I'm going to show you how to solve this question correctly step by step. First of all let's write this expression from right side to left side with the negative sign. As a result what do we have? We have a to the fourth power. So we have a to the fourth minus this one from the right side. So we have minus a minus one to the fourth power and equal to zero. So this is our first step. Second step. Instead of this a to the fourth power we can easily write a square raised to the second power. So I hope you understand it. According to a basic power rule we can easily multiply it so we will have the same a to the fourth power. Okay. So this is our first step. And according to the same property we can easily write this one as a minus one, a minus one right here, raised to the second power and of course all of this, I'm going to write these type of brackets, we're going to raise to the second power equal to zero. So according to this property I split it like second power to the second power and second power to the second power. But right now if you look closely we can easily consider it as a difference of two squares and everyone should know this school formula. I'm going to write it in terms of x and y. So when we have x square minus y square we can write it as x plus y. So we have x plus y and times x minus y. This is our basic school identity and right now let's apply this identity to our question. So we can easily consider it, we can easily consider this one as x and this one, this is our y. So we can consider it as difference of two squares. So addition and right here subtraction. So let's do this. Let's split it. Okay. As a result what do we have? We have in the first parenthesis we have a square plus y so plus a minus one to the second power. So we have a square minus and right here we have a minus one to the second power. So we can easily write minus, then plus, doesn't matter. And the second parenthesis we write as a square plus a minus one, a minus one to the second power and equal to zero. Right now we can easily simplify this a little bit. First of all we can easily raise this a minus one to the second power in both parenthesis. So let's do this right now. As a result what do we have? We have a square minus and right now let's raise this. So a minus one to the second power according to a school identity. This is equal to a square minus two a and plus one, plus one. This is our first parenthesis. And in the second parenthesis we have a square plus the same thing. A square minus two a and plus one. So a square minus two a and plus one. Equal to, equal to zero. Right now let's simplify this a little bit. We have two parenthesis. So let's open our parenthesis on our left side. As a result what do we have? We have a square, a square minus a square. We need to change all this sign in parenthesis. So minus a square plus two a and minus minus one. This is our first parenthesis. And the same thing in the second one but we need to like get rid of our parenthesis because we have addition so we don't need to change all the signs. So we have a square plus a square minus two a and plus one. Equal to, equal to zero. So we have a product of two parenthesis but we can easily simplify this real quick because we can easily cancel this a square. We can easily add this a square and a square. We have two a square and as a result what do we get from here? We're going to get two parenthesis. The first one, two a minus one, two a minus one because we cancel our a square. And the second parenthesis we will have two a square minus two a plus one. So we have two a square minus two a and plus, plus one. Equal to, equal to zero. And right now, really basic thing because we have a product of two parenthesis. A product of two parenthesis equal to zero when first parenthesis equal to zero or the second one equal to zero. So let's start with the first one. So we have two a, two a minus one is equal to, is equal to zero. From here we have two a equal to one and from here we have that our a is equal to one, is a is equal to one half. Okay, this is our first, first root a. First we're going to check it a little bit later but right now let's solve our second parenthesis. So we have two a square minus two a and plus one equal to zero. First of all we can find our discriminant real quick. Of course with the method of coefficients we have a equal to, a equal to two, b equal to minus two and c equal to, equal to one. Right now we can easily find our discriminant real quick. So we have d equal to b square minus four a c. All known formula. From here we have b square minus two square minus four times a times two and times c times, times one. And from here our discriminant is equal to, so we have minus two square equal to, equal to four. Yeah, we have right here four. And we have right here minus, minus four times two, we have eight. So our discriminant is equal to minus four. Our discriminant is negative so it means that in this parenthesis we're going to get two complex roots. So let's find it real quick. So our a second and a third, so a second and third equal to, we have minus b plus minus square root of d and all over two a. All known formula and right now let's plug in all the things. First of all we have minus b, so we have minus minus two plus minus square root of discriminant square root of minus four and all over two times a, two times two times two. Let's simplify this real quick. So first of all minus minus is plus so we have two plus minus square root of minus four. We can easily split it like square root of minus one times four. Okay, we can easily write it as a product. Okay, all over two times two equal to, equal to four. Right now in terms of our square root property we can easily split it. We can write it as two plus minus. This one can be easily written as square root of minus one and times square root of square root of four and we divide it by four. But this square root of minus one, this is our complex unit. This is our i. So as a result we can write it as two plus minus square root of four equal to two. So we have two i and all over, all over, all over four. And the final tricky move, we're gonna divide both parts. So we can easily like, we can easily split it by real part and imaginary part. We can write it as two over four plus minus two i over, over four. We can easily write it like that. As a result what do we get from here? We have two over four equal to one half, one half plus minus. We have i over, over two because we're gonna cancel these two and four. We have i over two. This is the first interpretation. A lot of students don't like this one. Of course, you can easily write it with a common denominator. You can write it as two and in our numerator you're gonna get a one plus minus plus minus i. Something like that. So two interpretations. This one or this one. Maybe I think this one is much easier for us, more understandable for us because right here we have, looks like a great expression. We don't have a fraction. We have only one fraction with the plus, with the plus minus. So basically, we can write our final answer with a second equal to one plus i over two. Let's write it like that. One plus i over two and a third equal to one minus i over two. So these are our two, two complex, two complex roots. And a few thoughts about this, about this question because a lot of students, as I said before, as I said in the beginning, a lot of students might be thinking I'm gonna rewrite my question real quick. So right here I'm gonna do it. So a to the fourth equal to a minus one, a minus one to the fourth, to the fourth power. And a lot of students, this is my, like, hence, this is my notes to this question. And a lot of students might be thinking, okay, let's just cancel this fourth power right here and right here. As a result, what do we get? A equal to a minus one. And then a lot of students are confused because right here we can easily cancel this a and we have zero equal to minus one. And they are confused because they think what just happened? How can we can't, how can we find one real number root? And they're confused about this, about this solution. And a few thoughts about this, about this approach because a lot of students may be thinking, of course, we can cancel this fourth power we raised on both sides. We can cancel it. Of course, this is a correct way. But a lot of students forget about the thing which is called absolute value. Of course, you have a to the fourth equal to a minus one to the fourth power. And when you cancel an even power like fourth on both sides, you need to put absolute value because this a can be positive, can be negative. So we need to put absolute value a on the left side. And of course, you need to put absolute value on the right side. And in this way, you can easily find if you solve this absolute value question. Of course, there are a lot of ways. How can we do this? The first way is like a combination method. You can easily write four equations with the first case plus this plus this minus this plus this plus this negative. And of course, the last one, both negative. And I think in this case, when this is positive, this is negative. You can find that our a is equal to one half. In this part, you can easily find that a equal to one half. So you can easily solve this question like that. You can easily cancel fourth power, but don't forget about absolute value. And with absolute value, of course, you can easily find a equal to one half. But few things about this approach. This is not a full solution because according to a fundamental theorem of algebra, whenever we have expression with the fourth power, it means that you're going to have like four roots in total. You don't know. We don't know how many real number roots, how many complex root, but we have four roots in total. And in this case, with the canceling fourth power, you avoid a lot of complex root. In our case, you avoid this part with complex root, which is extremely important because you forget about another part of solution. This is extremely important thing in terms of math. You can easily solve this question. If you, for example, if you don't have time on your exam, you can easily cancel it. You can easily find one real number root and everything is great. But if you want to solve this question completely and step by step, it's a great to factor it. It's a great to simplify this a little bit. And as a result, you can easily say, okay, I solve this question completely. I have a first equal to one half, a second equal to, and a third equal to. So right now, let's write our final answer. And in the end, we're going to check our a equal to one half. Our final answer to this question, our final answer to this question. So a first equal to one half, a second. This is our, let's start with this one, one plus i over two. And right here, we have a third, a third equal to one minus i, one minus i over two. These roots are complex root, complex root. This one is real number, real number, real number. So this is our full solution. And in the end, let's check it real quick. A first equal to one half. So let's do this. I don't want to rewrite this expression one more. Let's plug in real quick right here. So we can easily see this equation from the beginning. So let's split it real quick and let's check it. Okay, let's check our a equal to one half. Let's do this. So we have one half raised to the fourth power equal to one half minus one and to the fourth power. And right here, we're going to have like one fourth, one second to the fourth power. This is equal to one over 16. One over 16 right here, because we have one over two to the power four equal to right here, we're going to have minus one half in parentheses, but according to an even power, we can write it with the positive sign because we can cancel it. We can write it like that. We can write it minus one half to the fourth power, which is equal to this one. We can easily cancel because we can write it, if you want a full solution, full proof, we can write it as minus one times one half. And according to this property, we can raise four to the minus one to the power four, which is equal to one, so we can easily get rid of this negative sign. And we have one fourth, one second, one half to the fourth power, which is absolutely the same thing that we have on the left side. So our route is great. So this is my solution to the question. I really enjoyed this question because, you know, a lot of students getting confused when they see this question, when they try to solve it, they cancel this fourth power. They solve this question quickly, but as you can see, without easy manipulation, to be honest, because we factor it with the basic school identity. We raise this with a basic school identity. We simplify this with the basic school knowledge and we have a product of two parentheses. It's not that hard, but, you know, maybe a lot of students in this video, they find out how can we solve this question correctly. So I really enjoy and I really appreciate if you understand it. If you still have any question, write a question down into the comment section. It will be really interesting to discuss a little bit. And thank you everyone for your time. Thank you everyone for your support. Thank you everyone for your likes, for your dislikes, for your comments, for your response. I really appreciate it. And see you in the next videos. Have a great day and take care of yourself.