Most Students Cube Both Sides and Get Stuck!

Added: 2026-06-01

[00:00:00]Hello and welcome. In this math tutorial, our task is to find all the real values of x that satisfy this equation.

[00:00:08]Now, usually when faced with such problems, we start off by trying to get rid of this radical on the left-hand side of the equation. And you know that for us to do that, we raise both sides of the equation to power three.

[00:00:21]But notice that if we do that, on the right-hand side of the equation, we will generate a polynomial of power six, and that will be very hard to solve. Now, there's a very easy way of solving this equation without starting off by squaring both sides. We will start off with a substitution.

[00:00:42]Let the cube root of 2x - 1 be equal to the letter d.

[00:00:51]Now, you know that if we raise both sides of this equation to power three, then we have that d cubed is equal to 2x 1.

[00:01:05]Now, from the original equation, you can see that d is equal to 1/2 * x cubed + 1.

[00:01:18]Now, to get rid of this 1/2 on the right-hand side, we multiply both sides of this equation by two. When we do that, we have that 2d is equal to x cubed + 1. And of course, when we rearrange this equation, we have that x cubed is equal to 2d - 1.

[00:01:45]Now, you can see that we have generated two simultaneous equations in x and d.

[00:01:52]The next thing we are going to do is to try and see if we will get a relationship between X and D.

[00:02:00]Now, to do this, we subtract the second equation from the first equation.

[00:02:06]When we do that, on the left-hand side, we have D cubed minus X cubed.

[00:02:13]And of course, on the right-hand side, we have 2 X minus 2 D. Now, you know that minus 1 minus minus 1 is equal to 0.

[00:02:29]Now, rearranging this equation, we have D cubed minus X cubed plus 2 D minus 2 X is equal to 0.

[00:02:47]Now, notice that here we have the difference of two cubes, and we can factorize. When we factorize this, we have D minus X multiplied by D squared plus X times D plus X squared.

[00:03:10]And of course, we have plus Between these two terms, we have a common factor, which is 2.

[00:03:18]Now, 2 D divided by 2 is D, and minus 2 X divided by 2 is minus X, and of course, this is equal to 0.

[00:03:31]Now, when you look at these two terms, you will see a common denominator, which is D minus X. So, we can still factorize. We have D minus X into When we divide this first term by D minus X, of course, we are left with d squared + x d + x squared.

[00:04:00]And when we divide this second term by d - x, we are left with + 2. And this is equal to 0.

[00:04:11]And of course, from here, it's very easy to see that either d - x is equal to 0, from where d is equal to x, or d squared + x d + x squared + 2 is equal to 0.

[00:04:34]Now, we can rewrite the left-hand side of this equation.

[00:04:38]What we are going to do is that we will complete the squares. We will add and also subtract the square of half the coefficient of d. Now, the coefficient of d is x. So, we are going to add and subtract x over 2 squared. When we do that, we have d squared + x d + x over 2 squared and x over 2 squared. Then, we have + x squared + 2 and of course, this is still equal to 0.

[00:05:21]Now, look at these three terms. We have a perfect square. So, we take one d.

[00:05:27]And we take one x over 2.

[00:05:30]And we square.

[00:05:33]Of course, you know that if we expand this, we are going to get exactly these three terms, because we have d squared, which is this. Then, we have 2 * d * x over 2, which is x d. And finally, we have x over 2 squared, which is what we have here.

[00:05:49]Now, we still have minus x squared over 2 squared is 4. Then, we have plus.

[00:05:57]Instead of writing x squared, let us write it as 4 x squared over 4. Of course, you know that 4 x squared over 4 is still equal to x squared. Then, we have plus 2, and this is equal to 0.

[00:06:14]Now, from here, we have d plus x over 2 squared.

[00:06:22]Now, 4 x squared minus x squared is going to give us 3 x squared divided by 4.

[00:06:30]Then, we have plus 2, and this is equal to 0.

[00:06:35]Now, notice that this quantity is a square, so it must be positive.

[00:06:41]Now, 3 x squared over 4 is also a positive quantity. Then, we have 2, which is also positive. Now, this second equation is telling us that the sum of three positive numbers is equal to zero, and you know that this is impossible for real numbers. So, this second equation is of no use to us in solving this equation.

[00:07:04]So, ultimately, this is the only relationship we have between d and x.

[00:07:12]Now, remember that the cube root of 2 x minus 1 is equal to d.

[00:07:18]So, with this, we have that the cube root of 2 x minus 1 is equal to x.

[00:07:28]And now, we have a much simpler equation. And of course, to solve this equation, we start by raising both sides of the equation to power 3. When we do that, we have that 2 x minus 1 is equal to x cubed.

[00:07:43]Now, rearranging this equation, we have x cubed - 2x + 1 is equal to 0.

[00:07:54]Now, notice that when x is equal to 1, the left-hand side of this equation is equal to the right-hand side because we have 1 cubed, which is 1.

[00:08:06]- 2 * 1, which is 2 + 1. 1 + 1 is equal to 2 and 2 - 2 is equal to 0. This simply means that x 1 is a factor of this cubic. So, we can factorize using the synthetic division method.

[00:08:30]The coefficient of x cubed is 1.

[00:08:33]The coefficient of x squared is 0 because there is no term in x squared.

[00:08:39]The coefficient of x is - 2 and of course the constant term is 1 and we are dividing by 1. We bring down 1. 1 * 1 is 1. 0 + 1 is 1. 1 * 1 is 1. - 2 + 1 is - 1. 1 * - 1 is - 1. 1 - 1 is equal to 0.

[00:09:05]So, that means that when we factorize this cubic we have x - 1 * x squared + x - 1 and of course this is equal to 0.

[00:09:29]Now, already from this factor, we have that x is equal to 1. The last part of this solution is to solve the quadratic equation x squared plus x minus one is equal to zero to find the other two values of x that satisfy this equation.

[00:09:49]Now we cannot solve this quadratic equation by factorization. So let us use the quadratic formula. We have that x is equal to minus b plus or minus the square root of b squared minus four ac divided by two a.

[00:10:08]x is equal to minus one plus or minus the square root of one squared which is one minus four times one times minus one divided by two times one.

[00:10:31]x is equal to minus one plus or minus minus four times minus one is plus four.

[00:10:40]One plus four is five. So this is root five divided by two.

[00:10:48]So finally we have all the three real values of x that satisfy this equation.

[00:10:55]We have x equal to one.

[00:10:58]We have x equal to minus one plus root five over two and we have x equal to minus one minus root five over two.

[00:11:16]And with that we come to the end of this tutorial. I hope you learned something.

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