Equation Lab #21 – Quadratic Equations: Systems

So we’re done exploring the many methods of solving quadratic equations, but there’s one more topic to go over, and that is systems using quadratic equations. We know that a quadratic equation has a solution for each time the parabola crosses the x-axis. Sometimes, it doesn’t have to be the x-axis. If it’s not, we subtract the value of $y$ from both sides.

Now what if the line that passes through the parabola is slanted? What if it’s another parabola? That’s what we’ll be covering today. This time, we aren’t looking for the y-values.

Systems of Linear and Quadratic Equations:

So let’s get started on our first example. The system of equations to solve is $y=x^{2}-4x+8$ and $y=2x+3$. For the equation $y=x^{2}-4x+8$, if $y=0$, there would be no real solution. But, by the substitution method, $x^{2}-4x+8$ isn’t being placed against $0$. It’s placed against $2x+3$. So the equation to solve is $x^{2}-4x+8=2x+3$.

The first step is to subtract both $2x$ and $3$ from both sides. Since we can’t have any terms other than $0$ when put against a quadratic trinomial, we need to nuke the other side without tipping the equation out of balance. By subtracting $2x$ from both sides, the equation is $x^{2}-6x+8=3$. By subtracting $3$ from both sides, the equation becomes $x^{2}-6x+5=0$. Now that can be solved.

We’ll use the factoring method this time. Since $5$ is a prime number, the only two factors are $1$ and $5$, which have a sum of $6$. But $6$ is negative, so the equation is $(x-1)(x-5)=0$. This means either $x-1=0$ or $x-5=0$. The results are $x=1$ and $x=5$.

1. $y=x^{2}-4x+8$
   $y=2x+3$
2. $x^{2}-4x+8=2x+3$
3. $x^{2}-6x+5=0$
4. $(x-1)(x-5)=0$
5. $x-1=0, x-5=0$
6. $x=1, x=5$

Now what if we plugged either value into either equation. Using $y=2x+3$, if $x=1$, then $2x+3=2+3$, which equals $5$. If $x=5$, then $2x+3=10+3$, which equals $13$. So our points of intersection are $(1,5)$ and $(5,13)$.

Now let’s try another one. The system of equations to solve is $y=x^{2}-14x+56$ and $y=-x^{2}+20x-84$. This time, it’s a quadratic trinomial vs another quadratic trinomial. This is tense! But we can bring this down. Using the substitution method, $x^{2}-14x+56=-x^{2}+20x-84$.

We’ll have to move one expression to the other side. We want to obliterate $-x^{2}+20x-84$, so we’ll add $x^{2}$ and $84$ to both sides, while subtracting $20x$ from both sides. We should get $2x^{2}-34x+140=0$.

It may look like another equation that requires using factors of both $a$ and $c$ to get the middle term, but since they all have a common factor, we can divide the whole equation by $2$. This should make the equation $x^{2}-17x+70=0$. Now what two factors of $70$ add up to $17$? Judging by a quick observation, $70$ is a product of $10$ and $7$. The sum of these two numbers is $17$. And since $17$ is negative, we can write $x^{2}-17x+70=0$ as $(x-7)(x-10)=0$. Using the zero-product property, either $x-7=0$ or $x-10=0$. Solving each binomial, either $x=7$ or $x=10$.

1. $y=x^{2}-14x+56$
   $y=-x^{2}+20x-84$
2. $x^{2}-14x+56=-x^{2}+20x-84$
3. $2x^{2}-34x+140=0$
4. $2(x-7)(x-10)=0$
5. $x-7=0, x-10=0$
6. $x-7, x-10$

As for evaluating the variables, if $x=7$, then $x^{2}-14x+56=(7)^{2}-14(7)+56$, which equals $7$. If $x=10$, then $x^{2}-14x+56=(10)^{2}-14(10)+56$, which equals $16$. So our points of intersection are $(7,7)$ and $(10, 16)$.

By the way, did you know that both of these points are the vertices to both parabolas? Well, $(7,7)$ is the minimum to $y=x^{2}-14x+16$, as $(10,16)$ is the maximum to $y=-x^{2}+20x-84$.

Quiz #21:

As the last quiz for the week, let’s see if you can solve for $x$. Forget about solving for $y$ this time. Just remember, I will take any answer for as long as it fits the solution. So if the answer is $a=3,-3$, you can use “$3$“, “$-3$“, “$3,-3“$, “$-\mathrm{3,3}$“, “$a=3$“, “$a=-3$“, “$a=3,-3$“, or “$a=-\mathrm{3,3}$“.