# Question Video: Solving Systems of Quadratic Equations Algebraically Mathematics • 9th Grade

Find all the real solutions to the system of equations 𝑦 = 𝑥² − 3𝑥 − 4, 𝑦 = 2𝑥² − 4𝑥 − 6.

03:55

### Video Transcript

Find all the real solutions to the system of equations 𝑦 equals 𝑥 squared minus three 𝑥 minus four, and 𝑦 equals two 𝑥 squared minus four 𝑥 minus six.

Well, the first thing I’m gonna do is I’m gonna label our equations. So we’ve got equation one and equation two. I do this so it’s easy to see what we’re going to do them. Now, we can see at first glance that both of our equations are 𝑦 equals. So therefore, what we can do is make them equal to each other. So, as we’ve said, because they both have 𝑦 is equal to, we can set them equal to each other. So when we do that, we get two 𝑥 squared minus four 𝑥 minus six equals 𝑥 squared minus three 𝑥 minus four, with the second equation on the left-hand side. And we’ve just done that because I prefer to have the greater number of 𝑥 squared terms on the left-hand side.

So now, what we want to do is set our quadratic equal to zero because that’s what we like to do when we want to solve it. And to do that, what we’re gonna do is we’re gonna subtract 𝑥 squared add three 𝑥 and add four to each side of our equation. And when we do that, we’re gonna be left with 𝑥 squared minus 𝑥 minus two is equal to zero. And that’s because if you have two 𝑥 squared minus 𝑥 squared, you get 𝑥 squared. And if you have negative four 𝑥 add three 𝑥, you get negative 𝑥. And if you have negative six add four, you get negative two.

Okay, great. So now, what we need to do is solve this to find our two 𝑥-values. And the way to do that is by using factoring. So in order to factor, what we need to do is find a pair of factors whose products is negative two and whose sum is negative one. And that’s because we’ve got negative 𝑥. So the coefficient will be negative one. So now, there are only two pairs of factors that’s gonna give us the product negative two. And that’s negative one and two or negative two and one. Well, if we look at the sum, the sum needs to be negative one. So therefore, this tells us that the greatest value is going to be the negative. And the other one is gonna be positive. So we’re gonna have negative two and positive one. So therefore, we know that 𝑥 minus two multiplied by 𝑥 plus one is gonna be equal to zero.

So now, what we need to do is set each of our parentheses equal to zero because one of these needs to be zero if the answer to be zero. So therefore, we can do that to find out our two 𝑥 values. And when we do that, we get 𝑥 is equal to two or 𝑥 is equal to negative one. So now, what we’ve done is found the 𝑥-values or 𝑥-coordinates of our solutions. So to wrap this up, we need to now find the corresponding 𝑦-values. And to enable us to do that, what we need to do is substitute our values for 𝑥 into one of our equations. I’ve chosen equation one. So I’m gonna just first substitute 𝑥 equals two into equation one.

So if I substitute in 𝑥 equals two, I’m gonna get 𝑦 is equal to two squared minus three multiplied by two minus four, which is gonna give me four minus six minus four. So therefore, I’m gonna get 𝑦 is equal to negative six. So that’s great. We now need to find the other 𝑦-value. So now, to find the other 𝑦-value, what I’m gonna do is substitute 𝑥 equals negative one into equation one again. So when I substitute in 𝑥 equals negative one, I’m gonna get 𝑦 is equal to negative one squared minus three multiplied by negative one minus four. Which is gonna give us 𝑦 is equal to one plus three minus four. And that’s because we had negative one squared. And negative multiplied by negative is positive. And then, we had negative three multiplied by negative one, which gives us positive three. So this is gonna give us a 𝑦-value of zero.

So therefore, we can say that the solutions to the system equations are 𝑥 equals two and 𝑦 equals negative six and 𝑥 equals negative one and 𝑦 equals zero.