Linear-Quadratic Systems

Solve systems of one linear and one quadratic equation for the Digital SAT. Find intersection points between lines and parabolas.

A linear-quadratic system consists of one linear equation and one quadratic equation. Graphically, this is a line intersecting a parabola. Such a system can have 0, 1, or 2 solutions depending on the relative positions of the line and parabola. The Digital SAT tests this by asking you to solve the system, find the number of solutions, or determine a parameter value.

Core Concepts

Three Cases

Case Graphically Solutions
Line intersects parabola twice 2 intersection points 2 solutions
Line is tangent to parabola 1 intersection point 1 solution
Line misses parabola No intersection 0 solutions

Solving by Substitution

  1. The linear equation is usually already solved for yy.
  2. Substitute into the quadratic equation.
  3. Solve the resulting quadratic (by factoring or quadratic formula).
  4. Find the corresponding yy-values.

Example: y=x+1y = x + 1 and y=x22x+3y = x^2 - 2x + 3

Set equal: x+1=x22x+3x + 1 = x^2 - 2x + 3

0=x23x+2=(x1)(x2)0 = x^2 - 3x + 2 = (x - 1)(x - 2)

x=1x = 1y=2y = 2. x=2x = 2y=3y = 3.

Solutions: (1,2)(1, 2) and (2,3)(2, 3).

Using the Discriminant

After setting the equations equal, you get a quadratic. Apply the discriminant to determine the number of solutions without actually solving:

  • Δ>0\Delta > 0: 2 intersection points
  • Δ=0\Delta = 0: 1 intersection point (tangent)
  • Δ<0\Delta < 0: 0 intersection points

Finding Parameters

Example: For what value of kk is y=x+ky = x + k tangent to y=x2y = x^2?

x+k=x2x + k = x^2x2xk=0x^2 - x - k = 0

For tangency: Δ=1+4k=0\Delta = 1 + 4k = 0k=14k = -\frac{1}{4}

Strategy Tips

Tip 1: Substitute to Eliminate yy

Always substitute the linear equation into the quadratic. This gives a single-variable quadratic.

Tip 2: Use the Discriminant for "How Many Solutions"

If the question only asks for the number of solutions, don't solve — just compute the discriminant.

Tip 3: Remember to Find Both Coordinates

After finding xx-values, plug back to get yy-values. The solution is an ordered pair.

Tip 4: Graph on Desmos

Quickly visualise the intersection by graphing both equations.

Tip 5: If yy Isn't Isolated

If neither equation is solved for a variable, solve the simpler one (usually the linear equation) first.

Worked Example: Example 1

Problem

Solve y=2xy = 2x and y=x23y = x^2 - 3.

2x=x232x = x^2 - 3x22x3=0x^2 - 2x - 3 = 0(x3)(x+1)=0(x-3)(x+1) = 0

x=3,y=6x = 3, y = 6 and x=1,y=2x = -1, y = -2

Solution

Worked Example: Example 2

Problem

How many solutions? y=x+5y = x + 5 and y=x2+2x+3y = x^2 + 2x + 3

x+5=x2+2x+3x + 5 = x^2 + 2x + 3x2+x2=0x^2 + x - 2 = 0

Δ=1+8=9>0\Delta = 1 + 8 = 9 > 02 solutions

Solution

Worked Example: SAT-Style

Problem

The line y=kxy = kx is tangent to the parabola y=x2+1y = x^2 + 1. What is kk?

kx=x2+1kx = x^2 + 1x2kx+1=0x^2 - kx + 1 = 0

Δ=k24=0\Delta = k^2 - 4 = 0k=±2k = \pm 2

Solution

Worked Example: Example 4

Problem

Solve y=x+4y = -x + 4 and y=x24x+6y = x^2 - 4x + 6.

x+4=x24x+6-x + 4 = x^2 - 4x + 6x23x+2=0x^2 - 3x + 2 = 0(x1)(x2)=0(x-1)(x-2) = 0

x=1,y=3x = 1, y = 3 and x=2,y=2x = 2, y = 2

Solutions: (1,3)(1, 3) and (2,2)(2, 2).

Solution

Worked Example: Example 5

Problem

Does y=3y = 3 intersect y=x2+4y = x^2 + 4?

3=x2+43 = x^2 + 4x2=1x^2 = -1 — impossible over reals.

No intersection.

Solution

Practice Problems

  1. Problem 1

    Solve y=x+2y = x + 2 and y=x2y = x^2.

    Problem 2

    How many solutions? y=3x1y = 3x - 1 and y=x2+x+1y = x^2 + x + 1.

    Problem 3

    For what value of cc does y=2x+cy = 2x + c intersect y=x2y = x^2 at exactly one point?

    Problem 4

    Solve y=2x+1y = -2x + 1 and y=x2+3x5y = x^2 + 3x - 5.

    Problem 5

    A line y=mxy = mx passes through two points on y=x24y = x^2 - 4. Find the points when m=3m = 3.

    Problem 6

    Does the line y=5y = 5 intersect the parabola y=x2+4y = -x^2 + 4?

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Common Mistakes

  • Solving the wrong equation. Always substitute the linear into the quadratic (or set them equal).
  • Forgetting to find yy-values. The solutions are points (x,y)(x, y), not just xx-values.
  • Misidentifying the number of solutions. Use the discriminant of the resulting quadratic, not the original equations.
  • Errors in rearranging. When setting mx+b=ax2+bx+cmx + b = ax^2 + bx + c equal, be careful moving all terms to one side.
  • Not checking solutions. Plug back into both original equations.

Frequently Asked Questions

Can a line intersect a parabola more than twice?

No. A line can intersect a parabola at most 2 times.

What if the system has a quadratic and another quadratic?

The SAT primarily tests linear-quadratic systems. Quadratic-quadratic systems are rare.

How is tangency related to the discriminant?

A line is tangent to a parabola when the resulting quadratic has Δ=0\Delta = 0 (one repeated solution).

Can I graph this on Desmos?

Absolutely — graph both equations and visually see the intersection points.

Is this under Algebra or Advanced Math on the SAT?

Advanced Math. It's one of the harder topics in the section.

Key Takeaways

  • Substitute the linear equation into the quadratic to get a single-variable quadratic.

  • Use the discriminant to find the number of solutions without solving.

  • A line can intersect a parabola at 0, 1, or 2 points.

  • Δ=0\Delta = 0 means tangent (exactly one solution).

  • Always find both coordinates of each solution.

  • For parameter questions, set Δ=0\Delta = 0 for tangency conditions.

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