Video: Solving Systems of Quadratic Equations Algebraically

Find all the real solutions to the system of equations 𝑦 = 3π‘₯Β² + 5π‘₯ + 2, 𝑦 = βˆ’2π‘₯Β² βˆ’ 2π‘₯ + 2.

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Video Transcript

Find all the real solutions to the system of equations 𝑦 equals three π‘₯ squared plus five π‘₯ plus two and 𝑦 equals negative two π‘₯ squared minus two π‘₯ plus two.

Well, straightaway, we can notice that each question is 𝑦 is equal to. So therefore, what we can do is equate the right-hand sides of each equation. So, when we do this, what we’re gonna get is three π‘₯ squared plus five π‘₯ plus two is equal to negative two π‘₯ squared minus two π‘₯ plus two. So now, to enable us to solve for π‘₯, what we want to do is get a quadratic in the form π‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero. So, to enable us to do that, what we’re gonna do is add two π‘₯ squared, add two π‘₯, and subtract two from each side of the equation.

And when we do that, what we’re gonna have is five π‘₯ squared plus seven π‘₯ is equal to zero, where in fact we can actually see that we haven’t got the third term because what we have is π‘₯ squared plus 𝑏π‘₯. But there is no 𝑐 because we had two minus two, which is zero. So, this is in fact gonna be even easier to factor. So, we take a look at five π‘₯ squared plus seven π‘₯. And we see a common factor. And that common factor is in fact π‘₯. So, we take out the common factor. And then inside the parentheses, we have five π‘₯ plus seven. So therefore, we can say that π‘₯ is equal zero or π‘₯ is equal to negative seven over five. We’ll keep in fractional form for the moment.

Well, the reason we get zero or negative seven over five is because what we need to do is find the values that make the answer on the left-hand side equal to zero. Well, if π‘₯ is equal to zero, then zero multiplied by anything is zero. And then, if we look inside the parentheses, if we set this equal to zero, then five π‘₯ plus seven equals zero. Then, next, what we do is subtract seven from each side of the equation. And we get five π‘₯ equals negative seven. Then, we divide by five. We get π‘₯ is equal to negative seven over five. Okay, great. So, we have our two π‘₯-values. Now, let’s find our two 𝑦-values.

Well, now to find our 𝑦-values, what we want to do is substitute π‘₯ equals zero and π‘₯ equals negative seven over five into one of our equations. Well, I’ve labeled them one and two. So, we’re gonna choose one of them to put that into. What we’re gonna do is we’re gonna substitute them into equation one, remembering we could substitute them into either equation. Well, if we substitute in π‘₯ equals zero, we get 𝑦 equals three multiplied by zero squared plus five multiplied by zero plus two. So, when we do that, what we get is 𝑦 is equal to two. So then, what we have is our first pair of solutions. So, we’ve got π‘₯ equals zero and 𝑦 equals two.

Well, next, what we can do substitute in π‘₯ equals negative 1.4. So, when we substitute this in, we’re gonna get 𝑦 is equal to three multiplied by negative 1.4 all squared plus five multiplied by negative 1.4 plus two. And if we put this into calculator, we get 𝑦 is equal to 0.88. So therefore, we’ve got our second set of solutions. So, we can say that the real solutions to this system of equations are π‘₯ equals zero and 𝑦 equals two or π‘₯ equals negative 1.4 and 𝑦 equals 0.88.

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