Solve for x in this nice Algebra equation | Math Olympiad Mathematics

Added: 2026-05-19

[00:00:00]In this video, let us solve for x given x * x * x + x is = 10.

[00:00:22]We're given x * x * x + x is = 10.

[00:00:33]x * x * x here can be written as x raised to power 3. Then + x is = 10.

[00:00:47]Let us move this 10 to the left hand side. This will give us x rais^ 3 + x - 10 is = 0.

[00:01:02]Then 10 here is equivalent to 8 + 2 and 8 here is equivalent to 2 raised to power 3.

[00:01:18]So let us replace 10 with 2^ 3 + 2 giving us x raised to power 3 + x - remember to put a bracket 2^ 3 + 2 = 0.

[00:01:42]If we open up this bracket we get x^ 3 + x - 2 raised to power 3 - 2 is equal to zero.

[00:01:58]If we rearrange by order of power we get x raised to power 3 - 2 raised to power 3 then + x - 2 is equal to z.

[00:02:16]These first two expressions here is difference of two cubes and by identity given a raised to power 3 - b raised to power 3. We can factoriize this as a minus b into bracket a^ 2 + a b + b 2.

[00:02:42]So we use this to factoriize this. This will then give us x - 2 into bracket x^ 2 + 2 * x + 2 raised to power 2.

[00:03:02]Then + x - 2 is equal to zero.

[00:03:16]Let's tidy this up. This will give us x - 2 into bracket x^2 + 2x + 4 + x - 2 is = 0.

[00:03:38]If we introduce a bracket here and we compare these two expressions, we see x - 2 here and x - 2 here. So we can factor that as x - 2 into brackets x^2 + 2 x + 4 then + 1 is = 0.

[00:04:10]x - 2 into bracket x^2 + 2x + 5 is = 0.

[00:04:26]This will imply either x - 2 is = 0 or x^2 + 2 x + 5 is = 0.

[00:04:44]From the first linear equation, we see that x = 2.

[00:04:49]Let's call that x1.

[00:04:52]Then from this quadratic equation, we can get two more values for x.

[00:04:58]Before we do that, I'll compare this with a x^2 + bx + c = 0.

[00:05:06]This is the general quadratic equation form and we can get the coefficient of x^2 which is a 1, coent of x which is b 2 and the constant time c which is 5.

[00:05:24]Then we'll use this parameter to solve for x using the quadratic formula b + or minus square<unk> of b^ 2 - 4 a c then divided by 2 a.

[00:05:47]We have a, b and c. Therefore, x is going to be - b, which is - 2 + or -<unk> - 2^ 2 - 4 * a is 1 and c is 5.

[00:06:13]All of that divided by 2 times a a is 1 giving us x is = -2 + or minus square root of this will be 4 minus 20 square root of that / 2 * 1 here is 2.

[00:06:43]This will give us x = - 2 + or minus<unk> of 4 - 20 will give us -16.

[00:06:56]then divided by two which we can rewrite as x is = -2 + or -<unk> of -16 the same thing as saying 16 * -1 then divided by 2.

[00:07:21]So x is now -2 plus or minus we separate this radical into square<unk> of 16 *<unk> of -1 then / 2.

[00:07:38]This will give us x is = -2 + or minus roo<unk> of 16 is 4 and root of -1 is i. and then divided by 2.

[00:07:53]I'm going to separate this division to give us x is = -2 / 2 plus or minus 4 i / 2 here is 1 2 here is 2 here 1 2 here is -1 so we have x is = -1 + or - 2 i and earlier we got The value for x, we call that x1, which was 2, positive2.

[00:08:31]But in here, we have two different signs. So, we're going to have x1.

[00:08:38]x1 is 2. So this will be x2 -1 + 2 i and x3 = -1 - 2 i giving us all three possible values of this problem. Thanks for watching. Please like and share and also remember to subscribe to my channel and I'll see you in my next video.

[00:09:08]Bye.