### Video Transcript

Suppose the two roots of the equation π₯ squared minus π plus six π₯ minus 10π minus nine equals zero are equal. Determine all possible values of π and then find the two roots.

When weβre looking to solve a problem like this where we have been given information about the roots of a quadratic equation, then what we are going to use is the discriminant. Now, first of all, what we want is an equation in the form ππ₯ squared plus ππ₯ plus π equals zero. And then what we can say is that the discriminant is π squared minus four ππ. And we have three different scenarios for our discriminant. So, what we have is when π squared minus four ππ, our discriminant is less than zero; there are no real roots. If π squared minus four ππ is equal to zero, then there is one repeated root. And if π squared minus four ππ is greater than zero, then there are two real and different roots.

Well, in this problem, weβre gonna look at the middle scenario, where π squared minus four ππ is equal to zero, because weβre told the roots are equal. So, therefore, what we have is our one repeated root. So, if we take a look at our equation from the question, then we can see that itβs already in the form ππ₯ squared plus ππ₯ plus π equals zero. So, what we can do is identify what our π, π, and π are going to be. Well, because the coefficient of our π₯ squared term is one, because itβs just a single π₯ squared, π is going to be equal to one, then π is gonna be equal to negative π plus six, and π is going to be equal to negative 10π minus nine.

So, therefore, using the discriminant, we can set up the equation negative π minus six all squared βcause thatβs our π squared. And we get that because if we look at π, it was negative π plus six. So, if we distribute across the parentheses, so we multiply each of the terms by the negative one. It gives us our negative π minus six. And then we square that minus four multiplied by one multiplied by. Then, weβve got negative 10π plus nine equals zero. And again, we got negative 10π plus nine because what we needed to do is multiply everything in the parentheses, so our 10π minus nine, by negative one.

So, the first thing we need to do is distribute across the parentheses for negative π minus six all squared. When we do that, we get π squared plus 12π plus 36. And we get that because itβs the same as multiplying negative π minus six by negative π minus six. And if we want to distribute across the parentheses here, first of all, weβd have negative π multiplied by negative π, which gives us π squared, and then negative π multiplied by negative six, which gives us positive six π. And then we add another six π βcause weβve got negative six multiplied by negative π again. And then we add 36 because weβve got negative six multiplied by negative six. And then collecting like terms gives us our π squared plus 12π plus 36.

Okay, now letβs move on and simplify the rest of our equation. Well, what weβre gonna have is plus 40π. And thatβs because we had negative four multiplied by one multiplied by negative 10π, so thatβs positive 40π. Then weβve got minus 36 equals zero. And we got that minus 36 because we had negative four multiplied by positive nine, which gives us negative 36. So when simplified, weβre left with π squared plus 52π equals zero. So, now, to solve, what we can do is factor the left-hand side. And when we do that, weβre gonna have π multiplied by π plus two equals zero. So, therefore, we can say that π is equal to zero or negative 52. And thatβs because if we had π equal to zero, weβd have zero multiplied by 52, which would just give us zero. And if π was equal to negative 52, weβd have negative 52 multiplied by and then weβd have zero because negative 52 plus 52 is zero, which once again would give us our zero.

So, have we finished there? Have we solved the problem? Well, what we have done is found the possible values of π. So, yes, weβve solved that part of the problem. However, what we then need to do is find the two roots. So, what weβre going to have to do is substitute our values of π back in to our original equation. So, to give us room to do that, what weβre going to do is clear a space on the left-hand side.

So, if we start when π is equal to zero, then weβre going to have π₯ squared minus six π₯. And thatβs because we had zero plus six, which is just six. So, minus six π₯ plus nine equals zero. And we get plus nine because what we had was negative then weβve got 10π, which would just be zero, minus nine. So negative multiplied by negative nine, which is gonna give us positive nine. So, then, if we factor once again, what weβre gonna get is π₯ minus three multiplied by π₯ minus three is equal to zero. And thatβs because negative three multiplied by negative three is positive nine and negative three added to negative three is negative six. So, therefore, our roots would just be three or three. So, we could just say that π₯ is equal to three. Because if it was three, then that would make each of parentheses equal to zero. So, therefore, the right-hand side would be zero.

Okay, great. Thatβs when π is equal to zero. Now, letβs take a look at when π is equal to negative 52. So, when π is equal to negative 52, weβre going to have π₯ squared minus negative 46π₯ minus negative 520 minus nine equals zero. So, now if we simplify this, weβre gonna have π₯ squared plus 46π₯ plus 529 equals zero, remembering that we got our positive 46π₯ and positive 529 because of the negatives in front of our parentheses. So, we had negative multiplied by negative 46, which gave us our positive, and negative multiplied by negative 529, which gave us our positive. So, once again, what we could do now is in fact factor. Well, once again, when we factor, we get repeated root because we have π₯ plus 23 multiplied by π₯ plus 23 is equal to zero. But this is exactly what we were expecting because the question wanted us to assume the roots of our equation were equal.

So, we have here in both scenarios got roots that are equal, because we can say that when π is equal to negative 52, π₯ is equal to negative 23. So, therefore, what we could say is that the possible values of π and the two roots are π equals zero with two roots that both repeated, as we said, three and three, or when π equals negative 52, the two repeated roots are negative 23 and negative 23.