Video: Finding the Complex Roots of a Cubic Equation

Solve the equation π‘₯Β³ + 1 = 0, π‘₯ ∈ β„‚.

03:40

Video Transcript

Solve the equation π‘₯ cubed plus one equals zero, where π‘₯ is a complex number.

At first glance, it might look like this is a really easy equation to solve. We would begin by subtracting one from both sides, to obtain π‘₯ cubed equals negative one. We would then take the cube root of both sides. And since the cube root of negative one is negative one, we obtain a solution π‘₯ equals negative one. And when π‘₯ is equal to negative one, our equation is negative one cubed plus one, which is indeed equal to zero. So π‘₯ equals negative one is a solution of the equation π‘₯ cubed plus one equals zero. But it’s not the only one. We need to consider the fact that a cubic equation could have three distinct solutions. So how do we find the other two?

Well, if π‘₯ equals negative one is a solution to the equation, then π‘₯ plus one must itself be a factor of π‘₯ cubed plus one. This means we should be able to divide π‘₯ cubed plus one by π‘₯ plus one, without getting a remainder. Let’s use polynomial long division to do this. I’ve written π‘₯ cubed plus one as π‘₯ cubed plus zero π‘₯ squared plus zero π‘₯ plus one just to ensure we have the space for our working. π‘₯ cubed divided by π‘₯ is π‘₯ squared. And when we multiply π‘₯ squared by π‘₯ plus one, we get π‘₯ cubed plus π‘₯ squared. We now subtract π‘₯ cubed plus π‘₯ squared from π‘₯ cubed plus zero π‘₯ squared, to get negative π‘₯ squared. And then, we bring down the next term, the zero π‘₯.

This time, we divide negative π‘₯ squared by π‘₯. That gives us negative π‘₯. And when we multiply negative π‘₯ by π‘₯ plus one, we get negative π‘₯ squared minus π‘₯. Once again, we subtract. And this time, we just end up with π‘₯. We bring down the final term. And we repeat the process once again. π‘₯ divided by π‘₯ is one. Then, we check for a remainder. We multiply one by π‘₯ plus one to get π‘₯ plus one. When we subtract, we find that the remainder is zero. There is no remainder. So π‘₯ cubed plus one divided by π‘₯ plus one is π‘₯ squared minus π‘₯ plus one. We can therefore say that π‘₯ plus one times π‘₯ squared minus π‘₯ plus one must be equal to zero.

We’ve already solved this bit. So to find the remaining solutions to our equation, we’re going to make the other part of this product equal to zero. π‘₯ squared minus π‘₯ plus one equals zero. We could solve this in a number of ways. We could use the quadratic formula. Let’s look at how we might solve this by completing the square. We begin by halving the coefficient of π‘₯. Well, the coefficient of π‘₯ is negative one. And half of that is negative one-half. We subtract negative a half squared. So we subtract a quarter. And then, we add that one back on. Simplifying, we see our equation is π‘₯ minus a half all squared plus three-quarters equals zero.

Next, we subtract three-quarters from both sides. So π‘₯ minus a half squared equals negative three-quarters. We then square root both sides, remembering to take both the positive and negative square root of negative three-quarters. Now, this might look a little bit weird. But we can do something with that in a moment. Before we do, let’s add a half to both sides. And we find the other two solutions are π‘₯ equals a half plus or minus the square root of negative three-quarters. But how can we neaten up negative three-quarters?

We recall that 𝑖 is equal to the square root of negative one. And then, we look to rewrite our root a little bit. The square root of negative three-quarters is equal to the square root of negative three over the square root of four, which is of course equal to the square root of negative three over two. And since the square root of negative one is 𝑖, we can say that the square root of negative three, which must be equal to the square root of three times the square root of negative one, is root three times 𝑖. And this then means that the square root of negative three-quarters must be root three over two 𝑖.

And we have all our solutions. They are a half plus root three over two 𝑖, negative one, a half minus root three over two 𝑖.

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