Solving Linear-Quadratic Systems
When a line meets a parabola, substitution turns the system into a single quadratic — which can have two, one, or no real solutions.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
Substitute the linear expression into the quadratic, move everything to one side, and solve. Then find the matching $y$-values. Equal-and-opposite cases can mean the line misses the parabola entirely.
How to solve one, step by step
Example: solve $y = x^2$ and $y = x + 2$.
- Set equal: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0$.
- Factor: $(x - 2)(x + 1) = 0$, so $x = 2$ or $x = -1$.
- Points: $(2, 4)$ and $(-1, 1)$.
The mistakes that cost points
- Substituting into the wrong equation. Replace $y$ consistently so you end up with one equation in $x$.
- Forgetting the y-values. Each solution is a point — find $y$ for every $x$.
- Missing the no-solution case. A line can miss the parabola, giving no real intersection.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
Consider the system of equations: $y = x^2 - 4x + 7$ and $y = 2x - 1$. Which of the following gives the $y$-coordinates of all intersection points of the two equations in the $xy$-plane?
- A$y = 1$ and $y = 3$
- B$x = 2$ and $x = 4$
- C$y = 3$ and $y = 23$
- D$y = 3$ and $y = 7$
Show solution
Answer: D, $y = 3$ and $y = 7$. Set the expressions equal: $x^2 - 4x + 7 = 2x - 1$. Rearrange: $x^2 - 6x + 8 = 0$. Factor: $(x-2)(x-4) = 0$, so $x = 2$ or $x = 4$. Substitute back into $y = 2x - 1$: when $x = 2$, $y = 2(2) - 1 = 3$; when $x = 4$, $y = 2(4) - 1 = 7$. The $y$-coordinates are $3$ and $7$.
Medium
The graph shows a quadratic function $f(x) = -(x - 4)^2 - 2$ and a linear function $g(x) = 4x + k$. Point $A$ is at the vertex of $f$, and point $B$ is on the graph of $g$. For what value of $k$ would the length of $AB = 22$ units, with $B$ below $A$?
- A$k = -40$
- B$k = 6$
- C$k = \dfrac{11}{2}$
- D$k = 22$
Show solution
Answer: A, $k = -40$. Vertex $A$ is at $(4, -2)$. Point $B$ has the same x-coordinate, so $B = (4, -24)$. $|AB| = |-24 - (-2)| = 22$, giving $k = -40$.
Hard
Consider the system of equations: $y = x^2 - 6x + c$ and $y = 2x - 4$, where $c$ is a positive integer. What is the smallest value of $c$ for which the system has no real solutions?
- A$c = 12$
- B$c = 16$
- C$c = 13$
- D$c = 9$
Show solution
Answer: C, $c = 13$. Set equal: $x^2 - 6x + c = 2x - 4$. Rearrange: $x^2 - 8x + (c + 4) = 0$. For no real solutions, the discriminant must be negative: $(-8)^2 - 4(1)(c+4) < 0 \Rightarrow 64 - 4c - 16 < 0 \Rightarrow 48 - 4c < 0 \Rightarrow c > 12$. The smallest positive integer greater than $12$ is $c = 13$.
Find your exact gaps
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