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# Common Core Standards: Math

# Math.CCSS.Math.Content.HSA-REI.C.7

**7. Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = -3x and the circle x^{2} + y^{2} = 3.**

Students should know that this is a slight expansion on the previous standard. Our only change is that we instead of having two lines, we have a linear equation (a straight line) and a quadratic equation (a not so straight line). No big deal.

In the simplest terms, the quadratic equation is just a linear equation with a square sign over a variable. It can have more terms, but as long as the largest exponent over a variable is 2, it's a quadratic equation. Simple enough, right?

Students must know the quadratic formula:

Actually, they should have it tattooed on their faces. It'll be a great icebreaker at parties.

The variables in the quadratic equation (lovingly named *a*, *b*, and *c*) are derived from our standard quadratic equation *ax*^{2} + *bx* + *c* = 0.

Let's use the actual problem in the standard itself as an example. We have *y* = -3*x* as our linear equation, and *x*^{2} + *y*^{2} = 3 as our quadratic. (Technically, the second equation isn't a "true" quadratic, but there's no shame in torturing your students a little bit to teach them some more. We're only kidding about the torturing part.)

If we graph these bad boys, we can already see that we have two points of intersection.

We can prove why this is the case algebraically and calculate the exact points of intersection. First, we can solve for our quadratic as we did with linear equations: by substituting one equation into another.

That turns *x*^{2} + *y*^{2} = 3 into *x*^{2} + (-3*x*)^{2} = 3. We can eventually get it into the *ax*^{2} + *bx* + *c* = 0 format (which ends up being 10*x*^{2} – 3 = 0) so that we can use the quadratic formula. In our case, *a* = 10, *b* = 0, and *c* = -3.

If we plug in those values into the quadratic formula, we get

*x* ≈ ±0.55

Since we know what *x* is now, we can simply plug those values into our *y *= -3*x* equation to get our answers of ±1.64.
That means our points of intersection are (-0.55, 1.64) and (0.55, -1.64). It matches up with the graph, so we're done.

Students should also make use of the discriminant in order to figure out whether the line and quadratic intersect once, twice, or not at all. The discriminant is the part of the quadratic formula under the radical: *b*^{2} – 4*ac*.

If the value of the discriminant is less than 0, there are no points of intersection. If the answer is equal to 0, there is one point of intersection (a tangent line). If the answer is greater than 0, there are two points of intersection.

The value of our discriminant is (0)^{2} – (4 × 10 × -3) = 120. That means there are two points of intersection. Using the graph, the quadratic formula, and this little tidbit, we've triple checked our answer. Talk about thorough.