Solving Quadratic-Quadratic Systems Algebraically

Solve quadratic-quadratic systems of equations. Grade 11 Math 20-1 Alberta.

Lesson 2.4 of Linear Functions & Systems of Equations in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• The Method. Learn the two algebraic strategies for quadratic-quadratic systems — substitution and elimination — and use the discriminant to predict how many solutions exist.
• Substitution and Elimination. Solve two full quadratic-quadratic systems: one by elimination (setting equal), and one by substitution — including the tangent case with a repeated root.
• Special Cases. Handle the infinitely-many-solutions case (same parabola) and the no-solution case by checking the discriminant before doing any heavy algebra.
• Finding Parameters and Word Problems. Find unknown coefficients given a solution point, solve a real-world projectile problem, and use a complete quick-reference table for all quadratic-quadratic question types.

Inside the lesson — a free preview

The Big Idea

What is a Quadratic-Quadratic System?. A quadratic-quadratic system pairs two quadratic equations (two parabolas). Solving algebraically means finding every point where the parabolas intersect.

The same two strategies from linear-quadratic systems still apply:

Strategy 1 — Substitution. Solve one equation for y, then substitute that expression into the other equation. The result is a single-variable quadratic you can solve.

Strategy 2 — Elimination. Add or subtract the equations to eliminate y. You must eliminate y — not x — because the x^2 terms do not cancel cleanly when you try to remove x.

Special shortcut: when both equations are already solved for y, you can set the right-hand sides equal to each other. This is equivalent to subtracting one equation from the other to eliminate y.

Warning — Always Eliminate y. When using elimination, always eliminate y, never x. Eliminating x would require cancelling x^2 terms, which only works in special cases and is almost never the right move.

Possible Outcomes

After reducing the system to a single quadratic equation, the number of solutions depends on its discriminant.

General Steps

Follow these six steps for any quadratic-quadratic system.

Worked example: Solve by Elimination

System: y = 4x^2 + 8x + 4 and y = 3x^2 - 2x - 5. Both equations are solved for y, so we can set the right-hand sides equal — this eliminates y in one move.

Step 1 — Set the Right Sides Equal. 4x^2 + 8x + 4 = 3x^2 - 2x - 5

Step 2 — Move Everything to One Side. 4x^2 - 3x^2 + 8x + 2x + 4 + 5 = 0

x^2 + 10x + 9 = 0

Step 3 — Factor and Solve. Need two numbers multiplying to 9 and adding to 10: those are 9 and 1.

(x + 9)(x + 1) = 0

x = -9 or x = -1

Step 4 — Find y for Each x (Use y = 4x^2 + 8x + 4). For x = -9:

y = 4(-9)^2 + 8(-9) + 4

y = 4(81) - 72 + 4

y = 324 - 72 + 4

y = 256 → (-9,\ 256)

For x = -1:

y = 4(-1)^2 + 8(-1) + 4

y = 4 - 8 + 4

y = 0 → (-1,\ 0)

Solutions: (-9,\ 256) and (-1,\ 0)

Verification — Check (-1, 0). Equation 1: 0 = 4(1) + 8(-1) + 4 = 4 - 8 + 4 = 0 ✓

Equation 2: 0 = 3(1) - 2(-1) - 5 = 3 + 2 - 5 = 0 ✓

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