# Video: Solving Systems of Quadratic Equations Algebraically

Find all the real solutions to the system of equations 3π¦ β π₯Β² + 3π₯ + 5 = 0, 2π¦ β π₯Β² + π₯ + 1 = 0.

04:30

### Video Transcript

Find all the real solutions to the system of equations. Three π¦ minus π₯ squared plus three π₯ plus five equals zero. Two π¦ minus π₯ squared plus π₯ plus one equals zero.

Weβve been given two equations that equal zero. One way we could solve is to set these two equations equal to each other. So we would have three π¦ minus π₯ squared plus three π₯ plus five equals two π¦ minus π₯ squared plus π₯ plus one.

What we can do now is try to solve for a variable. If we subtract two π¦ from both sides of the equation, we have π¦. But two π¦ minus two π¦ equals zero. So that goes away. And we now have π¦ minus π₯ squared plus three π₯ plus five equals negative π₯ squared plus π₯ plus one. We have negative π₯ squared on both sides of the equation. If we add π₯ squared to both sides, the negative π₯ squared plus π₯ squared equals zero on both sides of the equation. Leaving us with π¦ plus three π₯ plus five equals π₯ plus one.

From there, we can subtract three π₯ from both sides of the equation. And weβll have π¦ plus five equals negative two π₯ plus one. From there, we subtract five from both sides. And we see that when these equations are equal, π¦ is equal to negative two π₯ minus four.

We can use this information to substitute negative two π₯ minus four in for π¦ in either one of the equations. If we substitute that value in for π¦ in the first equation, weβll have three times negative two π₯ minus four minus π₯ squared plus three π₯ plus five equals zero.

First, weβll expand this multiplication over the parentheses. Three times negative two π₯ is negative six π₯. Three times negative four is negative 12. And weβll bring everything else down. If we organise our terms with descending exponents, weβll start with negative π₯ squared. And then we can combine negative six π₯ plus three π₯ to get negative three π₯. And then weβll combine negative 12 plus five, which is negative seven.

We now have negative π₯ squared minus three π₯ minus seven equals zero. We can multiply the entire equation by negative one. So that the coefficient in front of the π₯ squared term is positive. Weβll have π₯ squared plus three π₯ plus seven equals zero.

At first, it might seem like we should factor this equation to find the values of π₯. However, I know that one times seven equals seven. Or negative one times negative seven is equal to seven. Neither one of those options though will give us this middle term of positive three. This means we cannot solve this equation by factoring.

We could use the quadratic formula negative π plus or minus the square root of π squared minus four ππ all over two π. This would give us the roots of the equation. But before we do that, we remember that weβre only looking for the real solutions.

And so itβs worth checking to see if there are any real solutions for this system. And we can use the discriminant to do that. The discriminant is the part under the radical in the quadratic formula π squared minus four ππ. And if π squared minus four ππ is less than zero, there are no real solutions.

Remember, the general form is ππ₯ squared plus ππ₯ plus π. We do have an equation in that format. And so for us, π squared equals three squared minus four times one times seven. Three squared is nine. Negative four times one times seven is negative 28. Nine minus 28 is negative 19. We have a negative discriminant. And since we have a negative discriminant, we can say there are no real solutions.