Lesson Video: Solving Quadratic Equations with Complex Coefficients | Nagwa Lesson Video: Solving Quadratic Equations with Complex Coefficients | Nagwa

# Lesson Video: Solving Quadratic Equations with Complex Coefficients Mathematics

In this video, we will learn how to solve quadratic equations with complex coefficients using the quadratic formula.

15:09

### Video Transcript

In this video, weβre going to see how to solve quadratic equations with complex coefficients. So how do we solve something of the form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are allowed to be complex numbers?

Recall that the discriminant of a quadratic equation ππ₯ squared plus ππ₯ plus π equals zero is π squared minus four ππ. The discriminant is often denoted by capital Ξ. For a quadratic equation with real coefficients, the discriminant allows us to determine the nature of its roots.

There are three possibilities. If the discriminant is positive, then we have two real roots. If itβs zero, then we have one repeated real root. And if itβs negative, then we have a complex conjugate pair of complex roots. But this is only true if the quadratic equation has real coefficients, that is, if π, π, and π are all real numbers.

Weβre going to see what we can say if π, π, and π are complex numbers. But first, letβs see how to solve a quadratic equation with complex coefficients. We have the familiar ππ§ squared plus ππ§ plus π equals zero. But whatβs new is that the coefficients π, π, and π can be complex numbers.

Now if π, π, and π were real, then we could just apply the quadratic formula. But it might not be obvious that this quadratic formula still applies when π, π, and π are complex numbers. So letβs rederive the quadratic formula by completing the square, checking that each step we use is still valid when π, π, and π are complex numbers and not just real numbers.

We can divide both sides by π to make the coefficient of π§ squared one. And now we want to complete the square, writing π§ squared plus π over ππ§ as a single square. We get π§ plus something squared. And that something is half the coefficient of π§, so half of π of π, which is π over two π. And if we distribute, we see that we do indeed get π§ squared plus π over ππ§. But we also get a term of π squared over four π squared. We need to subtract this term to get π§ squared plus π over ππ§. We make this substitution then safe in the knowledge that all the answer that weβve done holds just as true for complex numbers as it does for real numbers.

Now we can subtract the constant terms from both sides. And we can combine the two fractions on the right-hand side, getting π squared minus four ππ over four π squared. Now comes a tricky part, we need to take square roots. It turns out that the answer is what we expect from our experience with real numbers. We can check by squaring the last line and observing that we get the line above it. And by the fundamental theorem of algebra, there canβt be any solutions that weβve missed. We have a degree two polynomial and so we expect two roots. And these two roots are given by the plus and minus options on the right-hand side. There canβt be any more options.

All thatβs left to do is to subtract π over two π from both sides. And then we get the quadratic formula that we know and love. π§ equals negative π plus or minus root π squared minus four ππ all over two π. Letβs see some examples of applying this formula.

Solve three π§ squared plus five ππ§ minus two equals zero.

We solve this using the quadratic formula. The formula tells us that the quadratic equation ππ§ squared plus ππ§ plus π equals zero has solution π§ equals negative π plus or minus root π squared minus four ππ all over two π. And this formula applies even if the coefficients π, π, and π are complex numbers and not just real numbers. So we can apply it to our problem.

What is negative π? Well, the coefficient of π§ is five π. So itβs negative five π. We then have plus or minus the square root of π squared, which of course is five π squared. And from this, we subtract four times π, which is three, times π, which is negative two. And finally, we divide through by two π, π of course being three.

Now we just need to simplify. We canβt see much with negative five π at the moment. But five π squared is negative 25. And four times three times negative two is negative 24. Two times three is of course six. And now we can simplify further under the radical. Negative 25 minus negative 24 is negative 25 plus 24, which is negative one. And so we can see that we have a negative discriminant here. And we know what the square root of negative one is. Itβs π. The two solutions then are π§ equals negative four π over six and π§ equals negative six π over six. We can simplify the solutions to π§ equals negative two-thirds π and π§ equals negative π.

Note that although we have a negative discriminant here, our two solutions are not complex conjugates. If a quadratic equation with real coefficients has a negative discriminant, then we get two complex conjugate solutions. However, there are no guarantees when we have a nonreal coefficient as we do here.

Solve π§ squared plus two plus π times π§ plus π equals zero.

We use the quadratic formula, substituting the coefficients for π, π, and π. We can simplify under the radical. Two plus π squared is two squared, which is four, plus two times- two times π, which is four π, plus π squared, which is negative one. And from this, we subtract four times one times π, which is four π. We see that the terms involving π cancel. And weβre left with just four minus one, which is three under the radical.

Notice that we have a positive discriminant here. Thereβs a bit more simplification to do. Two times one on the denominator is just two. And we can distribute the minus sign over the parentheses. Doing this and rearranging the terms, we get negative two plus or minus root three minus π over two. Our solutions are therefore π§ equals negative two plus root three over two minus π over two and π§ equals negative two minus root three over two minus π over two.

Notice that although our equation had a positive discriminant, we donβt get two real solutions. If a quadratic equation with real coefficients has positive discriminant, then we get two real solutions. However, our quadratic equation has some nonreal coefficients. And so as weβve seen, we shouldnβt expect it to have two real roots.

Solve two plus three π times π§ squared plus four π§ minus six π plus four equals zero.

We use the quadratic formula. We substitute the values of the coefficients for π, π, and π. And now we simplify. In the denominator, we get four plus six π. In the numerator, we get negative four plus or minus a big radical. And inside that radical, four squared is 16. And from that, we subtract four times the product of π and π. When we distribute, we see that the terms involving π cancel. And weβre left with just eight plus 18, which is 26. 16 minus four times 26 is negative 88.

Notice that we have a negative discriminant here. The square root of negative 88 is π times the square root of 88 or two π times the square root of 22. Itβs tempting to say that this is our final answer here. But weβd like our denominators to be real if possible. We do that by multiplying the numerator and denominator by the complex conjugate of this denominator.

Letβs do this for the first root. We distribute in the numerator and in the denominator too. And in the denominator, we notice that the terms involving π cancel, leaving just the modulus of our complex number in the denominator. Thatβs four squared plus six squared. We evaluate the denominator and group the real and imaginary parts together in the numerator. And we notice that we can cancel a factor of four. We can therefore write our first root in simplest form as shown. And we can use exactly the same procedure to simplify the second root.

Notice that although our quadratic had a negative discriminant, these two roots are not complex conjugates. They donβt even have the same real parts.

Solve π§ squared minus four plus four π π§ plus eight π equals zero.

We use the quadratic formula. We substitute the values of the coefficients for π, π, and π. Under the radical sign, negative four plus four π squared becomes 16 plus 32π minus 16, which is just 32π. And from this, we subtract four times one times eight π, which is 32π, meaning that, under the radical, we have zero. On the denominator, we have two. Writing this again without all the crossings out, we see that our discriminant is zero. The square root of zero is zero. And so we add or subtract nothing, meaning that thereβs only one root. π§ equals four plus four π over two, which is two plus two π.

By the fundamental theorem of algebra, this must be a repeated root. And you can check that it is. So we see that this quadratic has a single repeated nonreal root. It turns out that if the discriminant is zero, then weβre guaranteed a repeated root. If the coefficients of the quadratic are real, then this root must be real itself. But if they are complex, then the roots could be complex.

We can see this by looking at the quadratic formula. If we make the discriminant zero, then weβre left with just π§ equals negative π over two π. This is the value of our repeated root. If the coefficients π and π are real, then the repeated root, negative π over two π, must be real as well. But if π or π or both are nonreal, then negative π over two π could be nonreal as well.

In the examples weβve seen in this video so far, the discriminant has always been real. But the numbers had to be carefully chosen to make this the case. In general, the discriminants can be nonreal. Letβs see an example.

Solve one plus two π times π§ squared minus three plus π equals zero. Round your answers to three significant figures.

As thereβs no term involving π§, it seems a bit silly to break out the quadratic formula. What we can do instead is to subtract negative three plus π from both sides and then divide both sides by one plus two π to find π§ squared equals three minus π over one plus two π.

Now we just need to find the square root of this complex number. But first, we multiply numerator and denominator by the complex conjugate of the denominator. We can distribute and then simplify. And we get a fifth minus seven-fifths π.

Now how do we find the square root of this number? Well, we can write it in polar form and then apply de Moivreβs theorem for roots. Whatβs the modulus of this complex number? We find that it is root two. As our complex number is in the fourth quadrant, its argument is the arctan of its imaginary part over its real part. We find that its argument is arctan of negative seven. And so we can write our complex number in polar form.

De Moivreβs theorem for roots gives us the πth roots of a complex number in polar form. Weβre looking for square roots, and so π is two. Applying this formula to our example, where π is root two and π is arctan negative seven, we get the following. Putting these into a calculator, we get π§ equals 0.898 minus 0.779π and π§ equals negative 0.898 plus 0.77π to three significant figures.

Solve π§ squared plus two minus two π π§ minus seven plus 26π equals zero.

We use the quadratic formula. We substitute the values of the coefficients for π, π, and π. We simplify under the radical sign, noticing thereβs some cancellation. And negative eight π plus 28 plus 104π is 28 plus 96π. We notice that, under the radical, we have two multiples of four. So we can take out that multiple of four in each case and bring out a two in front of the radical sign. This allows us to cancel the two in the denominator. So we get negative one plus π plus or minus the square root of seven plus 24π.

We use de Moivreβs theorem to find the square roots. But first, we need to write seven plus 24π in polar form. Its modulus is 25 and its argument is arctan 24 over seven. And de Moivreβs theorem for roots tells us how to find the square roots.

The modulus of the square root is the square root of the modulus. So thatβs the square root of 25, which is five. We have to halve the argument. The argument of the square root is arctan 24 over seven divided by two. Putting these into a calculator, we find that cos of this value is 0.8 and sin of this value is 0.6. And so upon multiplying by the modulus five, we add or subtract four plus three π. Therefore, our two roots are π§ equals three plus four π and π§ equals negative five minus two π.

Here, as the discriminant was complex, we had to use de Moivreβs theorem to find its square roots. Actually, we only found one of its square roots using de Moivreβs theorem. But we knew that the other must be its opposite. And we have this plus or minus sign here.

Here are the key points that weβve covered in this video. We can solve quadratic equations with complex coefficients using the quadratic formula. If the discriminant is zero, the equation has a single repeated root. Unlike for quadratic equations with real coefficients, this repeated root needs not to be real. If the discriminant is nonzero, then there are two distinct roots. So we see that the discriminant tells us whether the quadratic has a repeated root or two distinct roots.

We saw that, for quadratic equations with real coefficients, the sign of the discriminant told us whether the roots were real or nonreal. If it was positive, we had two distinct real roots. If it was zero, we had a single repeated real root. And if it was negative, we had two complex conjugate roots. This is not the case for quadratic equations with complex coefficients. We canβt use the sign of the discriminant to determine whether the roots are real or nonreal. All we can do is tell whether thereβs a repeated root or whether there are two distinct roots. And in fact, in general, the discriminant need not be a real number.

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