# Question Video: Quadratic Equations with Complex Discriminants Mathematics

Solve π§Β² + (2 β 2π) π§ β (7 + 26π) = 0.

04:08

### Video Transcript

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

Here we have a quadratic equation. This time though, it has nonreal coefficients. This doesnβt really matter though. It means only to be a little bit careful. But we can still apply the quadratic formula.

Remember, for an equation of the form ππ§ squared plus ππ§ plus π equals zero, the solutions are given by π§ equals negative π plus or minus the square root of π squared minus four ππ all over two π. And so for our equation, π, the coefficient of π§ squared, is equal to one. π is the coefficient of π§. So thatβs two minus two π. And π is the constant term, so thatβs negative seven plus 26π.

Letβs substitute everything we have into our quadratic formula. Negative π is negative two minus two π. We then have plus or minus the square root of π squared minus four ππ. Remember, π squared minus four ππ is the discriminant. And here thatβs two minus two π all squared minus four times one times negative seven plus 26π. And this is all over two π. So thatβs just two times one.

Letβs distribute our parentheses. Negative two minus two π is negative two plus two π. When we square two minus two π, we get four minus eight π plus four π squared. And negative four times one times negative seven plus 26π is just 28 plus 104π. And this is all over two.

Now we can simplify a little bit further. We know that π squared is equal to negative one. So we can replace four π squared with negative four. And then we see that four minus four is zero. And so our solution simplify to negative two plus two π plus or minus the square root of 28 plus 96π over two. But how do we find the square root of a complex number?

Well, letβs say the solution to this is π₯ equals π plus ππ. π₯ squared is π plus ππ all squared, which when we distribute the parentheses becomes π squared minus π squared plus two πππ. We know that 28 plus 96π must be equal to this value, π squared minus π squared plus two πππ. And remember of course π and π are real constants.

We can now equate coefficients. Letβs compare the coefficients of the real part. So on the left, thatβs 28. And on the right, thatβs π squared minus π squared. So 28 equals π squared minus π squared.

Weβll now compare the imaginary parts. On the left, thatβs 96. And on the right, thatβs two ππ. And so 96 equals two ππ. Letβs rearrange this second equation to make π the subject. We divide through by two π. And we find that π is equal to 48 over π. We then substitute this into our other equation. So we get 28 equals π squared minus 48 over π all squared. Next, we square 48 over π and then multiply this entire equation by π squared.

Gathering all the terms together, and we get π to the fourth power minus 28π squared minus 2304 equals zero. Factoring this expression, we get π squared minus 64 times π squared plus 36. And thatβs equal to zero. Now what this actually means is either π squared minus 64 is equal to zero. And thus π squared is equal to 64. Or π squared plus 36 is equal to zero. And π squared is equal to negative 36.

In fact though, we disregard this solution because we said that π had to be a real constant. If we find the square root of negative 36, we have a nonreal number. So weβre actually only interested in π squared equals 64, which means that π is equal to plus or minus eight. Substituting in this back into our expression for π, we get 48 over plus or minus eight, which is equal to plus or minus six.

Now what we could do is replace the square root of 28 plus 96π with plus or minus eight plus or minus six π. But actually, we really donβt need the plus or minuses now. And thatβs because weβre going to find plus or minus eight plus six π. So all options are really included here.

Letβs clear some space and perform the last steps. Weβll separate our two solutions. We have negative two plus two π plus eight plus six π over two or negative two plus two π minus eight plus six π over two. On the left, that simplifies to six plus eight π over two. And on the right, it becomes negative 10 minus four π over two. All thatβs left to do is to divide everything through by two. And we obtain our two solutions. They are π§ equals three plus four π or negative five minus two π.