Solving Simultaneous Linear And Quadratic Equations
Learn to solve simultaneous linear and quadratic equations for Queensland Year 11 Mathematical Methods (QCAA). These find exactly where a straight line meets a parabola.
You will learn to substitute the line into the quadratic, solve the resulting equation, and pair each x with its y — then use the discriminant to count intersections and test whether the line is a tangent, touching the curve just once.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a simultaneous linear and quadratic system asks where a line meets a parabola. You substitute the line into the quadratic, solve the resulting quadratic, and use the discriminant to count the intersections. This page covers finding intersection points, counting them, and the tangency condition.
A simultaneous system here pairs a straight line \(y=mx+k\) with a parabola \(y=ax^2+bx+c\). A solution is a point \((x,y)\) that lies on both graphs — a point of intersection.
The standard method is substitution: replace \(y\) in the quadratic with the linear expression, which leaves a single quadratic in \(x\). Solving it gives the \(x\)-values of the intersections; substitute each back into the line for the matching \(y\).
The discriminant of that resulting quadratic counts the intersections: two if \(\Delta>0\), one if \(\Delta=0\) (the line is a tangent), none if \(\Delta<0\).
Substituting the line into the parabola gives a quadratic \(ax^2+bx+c=0\); its discriminant counts the intersections:
How to solve a line-and-parabola system
- Substitute: put the linear expression for \(y\) into the quadratic (or set the two right-hand sides equal).
- Solve: gather all terms to one side as \(ax^2+bx+c=0\) and solve (factorise, formula, or the discriminant if only the count is asked).
- Back-substitute: put each \(x\) into the line to get its \(y\); write the intersection points as coordinates.
Substitute — set the right-hand sides equal:
| \(x^2-1\) | \(=\) | \(x+1\) |
Gather to one side and solve:
| \(x^2-x-2\) | \(=\) | \(0\) |
| \((x-2)(x+1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(2 \ \text{or}\ -1\) |
Back-substitute into \(y=x+1\):
| \(x=2:\ y\) | \(=\) | \(2+1=3\) |
| \(x=-1:\ y\) | \(=\) | \(-1+1=0\) |
They meet at \((2,\,3)\) and \((-1,\,0)\).
Substitute and gather to one side:
| \(x^2+2x+3\) | \(=\) | \(x+1\) |
| \(x^2+x+2\) | \(=\) | \(0\) |
Only the count is asked, so use the discriminant:
| \(\Delta\) | \(=\) | \((1)^2-4(1)(2)\) |
| \(=\) | \(1-8\) | |
| \(=\) | \(-7\) |
\(\Delta=-7<0\), so the quadratic has no real solutions.
They do not meet — there are \(0\) intersections.
Substitute the line into the parabola:
| \(x^2\) | \(=\) | \(x+k\) |
| \(x^2-x-k\) | \(=\) | \(0\) |
Tangent means one solution, so \(\Delta=0\):
| \((-1)^2-4(1)(-k)\) | \(=\) | \(0\) |
| \(1+4k\) | \(=\) | \(0\) |
| \(4k\) | \(=\) | \(-1\) |
| \(k\) | \(=\) | \(-\dfrac{1}{4}\) |
The line is a tangent when \(k=-\dfrac{1}{4}\).
Substitute and gather to one side:
| \(x^2-x\) | \(=\) | \(x+1\) |
| \(x^2-2x-1\) | \(=\) | \(0\) |
Solve with the quadratic formula:
| \(x\) | \(=\) | \(\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)}\) |
| \(=\) | \(\dfrac{2\pm\sqrt{8}}{2}\) | |
| \(=\) | \(1\pm\sqrt{2}\) |
Back-substitute into \(y=x+1\):
| \(y\) | \(=\) | \((1\pm\sqrt{2})+1\) |
| \(=\) | \(2\pm\sqrt{2}\) |
They meet at \((1+\sqrt{2},\,2+\sqrt{2})\) and \((1-\sqrt{2},\,2-\sqrt{2})\).
Common pitfalls
Frequently asked questions
How do you solve a simultaneous linear and quadratic system?
Substitute the linear equation into the quadratic to get a single quadratic in \(x\), solve it, then substitute each \(x\) back into the line to find \(y\).
How many times can a line meet a parabola?
Zero, one or two times. The discriminant of the combined quadratic decides: \(\Delta<0\) none, \(\Delta=0\) one (tangent), \(\Delta>0\) two.
What does it mean for a line to be a tangent to a parabola?
The line touches the parabola at exactly one point, which happens when the substituted quadratic has \(\Delta=0\).
How do I find where two graphs intersect?
Set their equations equal, collect all terms to one side as \(ax^2+bx+c=0\), solve for \(x\), then find each \(y\) from either equation.
Do I use the discriminant of the line or the parabola?
Neither on its own — use the discriminant of the quadratic you get after substituting the line into the parabola.
How do I find a parameter for a given number of intersections?
Form the combined quadratic, then set \(\Delta>0\) for two intersections, \(\Delta=0\) for a tangent, or \(\Delta<0\) for none, and solve for the parameter.