A Nice Algebra Problem | Maths Olympiad | How to solve for X in this problem ?

Added: 2026-05-05

[00:00:00]Hello, you're welcome to solve this math problem of x^ 3 - x^ 2 is = 100 to find the value of x from this equation.

[00:00:14]Now solution in the first step we will take this 100 to the left side. So it will be x c^ 3 - x^ 2 when I take this side it will be - 100 is equal to zero.

[00:00:31]Then in the next step it will be x^ 3 - x^ 2 - 100. We split into difference of a numbers difference of a numbers which can be form of exponents. So 100 is same as 125 minus 25 bracket is = 0. Then here it will be x c^ 3 - x^ 2 we take negative inside. So here it will be -25us and minus it will be + 25 is = 0.

[00:01:16]Then here it will be x^ 3 - x^ 2 - here 125 into exponents it is 5^ 3 + 25 into exponent it is 5^ 2 is = 0.

[00:01:36]Then here with like powers of three. So here we start by this x^ of 3 then - 5^ of 3 then with light powers of 2. So here it will be - x^ 2 then + 5^ 2 is equal to zero.

[00:02:00]Then here inside the bracket so it will be x^ 3 - 5^ 3 inside the bracket. Then here we take negative outside the bracket. So - x² / it is positive x² positive 5² / is - 5² bracket is = 0.

[00:02:29]Now into here is in the form of difference of two cubes. So we'll apply a rule in this form whereas it is in the form of a^ of 3 - b^ 3 is equal to a - bracket bracket a² + a b + b² bracket and here it is in the form of difference of two squares. So we'll apply a rule in this form which is a² - b² is equal to a - bracket bracket a + b bracket.

[00:03:14]So here when we compare a^ 3 with x^ 3 a is x when we compare b^ 3 is 5^ 3 b is 5 then we apply this form a minus b it will be x - 5. So here x - 5 bracket bracket a² it will be x c² + a it will be x * 5 is 5 x + b² it will be 5 square which is 25 bracket then minus here this is in this form then we change into this form a minus b it will be x - 5 so here bracket X - 5 bracket bracket A + B it will be X + 5. So here X + 5 bracket [snorts] is equal to zero.

[00:04:10]Then into here X - 5 is common. So take X - 5 bracket outside the bracket. Now this / X - 5. It is a quadratic expression of x² + 5x + 25.

[00:04:32]Then here minus this here divide by this. It is this. So because you have negative so here we place inside the bracket x + 5 bracket then bracket is equal to zero.

[00:04:50]Then into here it will be x - 5 bracket bracket x c² + 5 x + 25 we take negative inside so it will be - x here it will be - 5 bracket is = 0 then into here it will be x - 5 bracket bracket xc C² 5 X - X is 4 X. So here + 4 X 25 - 5 here it is 20. So + 20 bracket is equal to zero.

[00:05:35]Then here we have two solutions where this first solution of x - 5 is = 0 and this second solution of x² + 4 x + 20 is = 0.

[00:05:53]Then here I'll take5 to this side. So it will be x is = 5.

[00:06:00]So this is the first value of x which is really solution.

[00:06:05]Now [snorts] let's solve this second solution which is quadratic equation. So we'll solve by using quadratic formula to find the values of x is = b + or minus square<unk> of b² - 4 a c over 2 a. So into here it will be x is equal to b. B is cofficient of X which is 4 plus or minus square<unk> of B square it will be 4² - 4 * A a it is cofficient of X² which is 1 * C is constant which is 20 then over 2 * A a it is 1 then in the next Step here it will be x is = -4 + or minus square t of 4 square 4 square it is 16 then -4 * 20 here it is - 80 then over 2 * 1 it is 2 so here it will be x is = -4 4 + or minus square<unk> of 16 - 80 here it is - 10 - 6 here it is 4 here borrow 1 to be 7 - 1 it is 6 then over this two then in the next step here it will be x is = -4 + or minus square<unk> of -64 is same as 64 * -1 then over 2. So into here it will be x is = -4 + or minus we separate so it will be square<unk> of 64 *<unk> of -1 then over 2.

[00:08:22]So here it will be x is = -4 + or minus roo<unk> of 64 it is 8<unk> of -1 it is i then over 2. So here it will be x is equal to we divide by two in this part and this part. So it will be this over this. So be -4 / 2 + or minus 8 I / 2.

[00:08:51]So here it will be x is equal to -4 / 2 it is -2 + or minus 8 I / 2 it is 4 I.

[00:09:04]So from here we have two complex solutions.

[00:09:08]Therefore our conclusion the first value of x which is real solution it is this here which is five five.

[00:09:23]The second value of x is equal to from this complex solution it is it will be -2 + 4 i. So here -2 + 4 i. The third value of x is equal to here when it is negative to be -2 - 4 i. So into here we have three solutions into this our problem. One real solution and two complex solutions.

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