Stanford Math Tournament 2023 | Find All 10 Roots?

Hinzugefügt: 2026-05-26

[00:00:01]We are going to find all the 10 roots of this equation. So, let's get it started.

[00:00:07]We can take x ^ 8 common from these three terms.

[00:00:12]We can take x ^ 4 common from these three terms.

[00:00:17]So, we will get x ^ 8 in the bracket x ^ 2 + x + 1 Then, we'll take x ^ 4 common out.

[00:00:31]So, in the bracket x ^ 2 + x + 1 Then, from remaining three terms, we can take one common.

[00:00:44]So, in the bracket we'll get all the three terms x ^ 2 + x + 1 = 0.

[00:00:55]Now, we have x ^ 2 + x + 1 overall common.

[00:01:01]So, we will get x ^ 8 + x ^ 4 + 1 in one bracket * x ^ 2 + x + 1 This product = 0.

[00:01:23]Now, we are going to use Sophie Germain identity here.

[00:01:28]So, I will write x ^ 8 as it is.

[00:01:32]Then, we have x ^ 4, which we can write 2 * x ^ 4 - x ^ 4.

[00:01:43]We have one also. We will write here.

[00:01:48]* x ^ + x + 1 = 0.

[00:01:58]Now, from these three terms, we can write one perfect square of x ^ 4 + 1 whole square minus x ^ 4 times x ^ 2 + x + 1 equal to zero.

[00:02:23]Now, we can write x ^ 4 as x ^ 2 whole square.

[00:02:29]So, I will write here x ^ 2 whole square.

[00:02:33]Now, we will apply difference of two squares formula and we will write x ^ 4 + x ^ 2 + 1 times x ^ 4 - x ^ 2 + 1 times x ^ 2 + x + 1 equal to zero.

[00:03:03]Again, we will use Sophie Germain identity over here.

[00:03:09]So, let me write our equation once.

[00:03:13]Here it is. Now, we will take up this bracket first.

[00:03:18]Which we can write x ^ 4 + x ^ 2 can be written as + 2 x ^ 2 - x ^ 2 Then, we have + 1 which we will write over here.

[00:03:34]Now, from first three terms, we can write this is perfect square of x ^ 2 + 1 Then, we have - x ^ 2 Again, we can use difference of two squares formula over here.

[00:03:51]So, we will get x ^ 2 + 1 + x, which we will write x ^ 2 + x + 1.

[00:04:00]In another bracket, we'll get x ^ 2 + 1 - x, which we can write x ^ 2 - x + 1.

[00:04:10]Now, we will take up second bracket over here.

[00:04:14]We will write x ^ 4 - x ^ 2 can be written as + 2 x ^ 2 - 3 x ^ 2 Then, we have + 1.

[00:04:30]Again, we'll consider first three terms, which is one perfect square of x ^ 2 + 1.

[00:04:40]So, we can write x ^ 2 + 1 whole ^ 2 - 3 x ^ 2 Now, we can write 3 as √3.

[00:04:52]So, we will write here √3 * x whole ^ 2 Now, we can apply difference of two squares formula.

[00:05:04]So, we will get x ^ 2 + 1 + √3 x, so which we can write + √3 * x + 1 in one bracket.

[00:05:19]And in other bracket, we need to write x ^ 2 - √3 * x + 1 Let us put all the factorization here.

[00:05:34]We will get So, I can write x ^ 2 + x + 1 * x ^ 2 - x + 1 in place of x ^ 4 + x ^ + 1.

[00:05:53]Now, we have x ^ 4 - x ^ + 1, which we have factored x ^ + √3 * x + 1 * x ^ - √3 * x + 1.

[00:06:13]Then, we have last bracket x ^ + x + 1 = 0.

[00:06:24]So, we have x ^ + x + 1 here also and here also.

[00:06:31]So, we will write x ^ + x + 1 whole ^.

[00:06:38]Then, we have x ^ - x + 1.

[00:06:44]Then, we have x ^ + √3 x + 1.

[00:06:53]Then, we have x ^ - √3 x + 1 = 0. Now, we can use product zero rule.

[00:07:05]We are going to get four equations.

[00:07:08]x ^ + x + 1 whole ^ either this should be zero or x ^ - x + 1 this must be equal to zero or x ^ + √3 * x + 1 this should be zero or last bracket x ^ - √3 x + 1 this must be zero.

[00:07:42]So, we have four equations. From our first equation, we are going to get multiplicity solution or multiple solutions.

[00:07:52]So, let's begin with the first one.

[00:07:55]I will write quadratic x ^ 2 + x + 1 ^ 2 = 0.

[00:08:06]Now, we'll write here x ^ 2 + x + 1 = 0 with multiplicity two. Let's use quadratic formula.

[00:08:17]We will get x = - b. B is + 1, so - 1 plus minus square root of b ^ 2 will be 1. 1 ^ 2 is 1 - 4 * 1 * 1, which is 4 over 2 * 1 two.

[00:08:35]So, we can write here x will be equal to - 1 plus minus i square root three over two.

[00:08:45]With multiplicity two.

[00:08:51]So, we have overall four solutions. Two with plus sign in between and two with negative sign in between.

[00:08:59]Now, we will take up second equation.

[00:09:03]x ^ 2 - x + 1 = 0.

[00:09:08]We are going to use quadratic formula and we will get 1 plus minus i times square root three over two.

[00:09:19]So, we have four solutions here.

[00:09:23]And we have two solutions here. Overall, six solutions.

[00:09:29]Let me put in the box.

[00:09:32]Now, we will take up third equation.

[00:09:35]x ^ 2 + root three x plus 1 equal to 0.

[00:09:46]Again, we have to use quadratic formula.

[00:09:49]We will get minus square root 3 plus minus i square root 1 I can write square root 1 over 2 which will be equal to minus square root 3 plus minus i over 2.

[00:10:10]So, two solutions from here.

[00:10:14]Now, we have one last equation.

[00:10:17]We will write x squared minus square root 3 x plus 1 equal to 0.

[00:10:26]We'll get here x equal to square root 3 plus minus i square root 1 is 1 over 2 * 1 is 2.

[00:10:39]So, we have 10 complex solutions for our equation.

[00:10:44]Two solutions from the last equation.

[00:10:47]So, 4 + 2 + 2 overall 10 complex solutions. I hope friends, you will like this video. Thank you so very much for watching. Do not forget to like, share, subscribe. Bye-bye.