### 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 doesnβt 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 ππ§ squared plus ππ§ plus π equals zero, the solutions are given by π§ equals negative π plus or minus the square root of π squared minus four ππ 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 thatβs two minus two π. And π is the constant term, so thatβs negative seven plus 26π.

Letβs 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 ππ. Remember, π squared minus four ππ is the discriminant. And here thatβs two minus two π all squared minus four times one times negative seven plus 26π. And this is all over two π. So thatβs just two times one.

Letβs 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, letβs say the solution to this is π₯ equals π plus ππ. π₯ squared is π plus ππ all squared, which when we distribute the parentheses becomes π squared minus π squared plus two πππ. We know that 28 plus 96π must be equal to this value, π squared minus π squared plus two πππ. And remember of course π and π are real constants.

We can now equate coefficients. Letβs compare the coefficients of the real part. So on the left, thatβs 28. And on the right, thatβs π squared minus π squared. So 28 equals π squared minus π squared.

Weβll now compare the imaginary parts. On the left, thatβs 96. And on the right, thatβs two ππ. And so 96 equals two ππ. Letβs 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 thatβs 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 weβre 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 donβt need the plus or minuses now. And thatβs because weβre going to find plus or minus eight plus six π. So all options are really included here.

Letβs clear some space and perform the last steps. Weβll 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 thatβs 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.