Question Video: Solving Systems of Quadratic Equations Algebraically

Video Transcript

Find all the real solutions to the system of equations 𝑦 equals negative two 𝑥 squared plus four 𝑥 plus six and 𝑦 equals four 𝑥 squared plus 22𝑥 plus 24.

So when we take a look at our system of equations, we can see that they both are 𝑦 equals and then we have an expression. So therefore, what we can do to solve the problem is equate them to each other. So therefore, what we can say is that negative two 𝑥 squared plus four 𝑥 plus six is equal to four 𝑥 squared plus 22𝑥 plus 24.

So now what we want to do is we want to actually have a quadratic in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero to help us solve it. So in order to do that, what we’re gonna do is add two 𝑥 squared, subtract four 𝑥, and subtract six from both sides of the equation. And when we do that, what we’re gonna be left with is zero is equal to six 𝑥 squared plus 18𝑥 plus 18. And as we can see that each term on the right-hand side is divisible by six, what we can do is divide through by six to make it easier to solve. So when we do that, what we’re gonna get is zero is equal to 𝑥 squared plus three 𝑥 plus three.

Or straightaway, we can identify that this is in fact not going to factor, because we could factor to solve the quadratic but in this case it won’t work because we have three. And we need to have a product that makes three. Well the only product that makes three is one multiplied by three, which equals three. Well, if you add one and three, you get four, whereas we want to have a sum of three. So we know that it’s not going to factor.

Well, we’re now at the stage where we think how can we solve our quadratic using other methods? What we could do is plug it into the quadratic equation and have a look at how we get on. And we will do that to show how we get the same result.

However, what you might notice by inspection is that the coefficient of 𝑥 and our numerical term are both positive and the same value and less than five. And also, it’s worth noting, which is important, the 𝑥 squared coefficient is also positive. And by noticing this, we can actually think, “Well, is it possible that there couldn’t be any real roots at all?”

And what we can do to check this out is use the discriminant, because what we know from the discriminant is that if we have 𝑏 squared minus four 𝑎𝑐 is less than zero, then there are no real roots. If 𝑏 squared minus four 𝑎𝑐 is equal to zero, then there’s one real root. And if we have 𝑏 squared minus four 𝑎𝑐 is greater than zero, then there are two distinct real roots.

And in our equation, 𝑎 is equal to one because that’s the coefficient of 𝑥 squared. 𝑏 is equal to three because it’s our coefficient of 𝑥. And then our numerical term 𝑐 is equal to three as well.

So now what we can do is work out our discriminant. And to do that, we’ll have three squared minus four multiplied by one multiplied by three. So our discriminant’s gonna be nine minus 12. So our discriminant is gonna be negative three. So the discriminant therefore is less than zero. So we can say that they’re gonna be no real roots.

And we managed to notice this because, as I said, we had the coefficient of 𝑥 was positive three and the numerical value was positive three. And we also had the coefficient of 𝑥 squared being positive. So therefore, what we know is that it was going to always look like they were gonna be no real roots. Because if we think about it, if we have a number squared and then we subtract four multiplied by that number and another number, it’s always gonna be a negative value.

Now this was using the discriminant. What I did mention was that we could have found this out also if we’d just gone straight into using the quadratic formula. Well, the quadratic formula is that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 over two 𝑎. Well, if we use the quadratic formula, what we’d have is 𝑥 equals negative three plus or minus the square root of three squared minus four multiplied by one multiplied by three all over two multiplied by one.

So then this would give us 𝑥 is equal to negative three plus or minus the square root of negative three over two. So now what we get is root negative three in our numerator. Well, root negative three is undefined, because we can’t have a result for the square root of a negative number if we’re using real numbers. So therefore, what we could say is that the system of equations would have no real roots.