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- Simultaneous equations with a quadratic
Simultaneous equations with a quadratic
When a line meets a curve, they can cross twice. Substitution finds both crossing points.
The method
1. Rearrange the linear equation to make x or y the subject.
2. Substitute it into the quadratic.
3. Solve the quadratic.
4. Substitute each answer back into the linear equation.
2. Substitute it into the quadratic.
3. Solve the quadratic.
4. Substitute each answer back into the linear equation.
Example
y = x + 1 and x2 + y2 = 13
x2 + (x + 1)2 = 13
2x2 + 2x − 12 = 0, so x2 + x − 6 = 0
(x + 3)(x − 2) = 0, so x = −3 or x = 2
Pairs: (−3, −2) and (2, 3)
x2 + (x + 1)2 = 13
2x2 + 2x − 12 = 0, so x2 + x − 6 = 0
(x + 3)(x − 2) = 0, so x = −3 or x = 2
Pairs: (−3, −2) and (2, 3)
What the answers mean
Each solution is a point where the line crosses the curve.
Two solutions: crosses twice. One: the line is a tangent. None: they never meet.
Two solutions: crosses twice. One: the line is a tangent. None: they never meet.
Solve y = 2x and y = x2 − 3.
- x2 − 3 = 2x
- x2 − 2x − 3 = 0
- (x − 3)(x + 1) = 0
- x = 3, y = 6 or x = −1, y = −2
Answer: (3, 6) and (−1, −2)
Make one letter the subject of the linear equation, substitute into the quadratic, solve, then use the linear equation to find the partner values. Answers come in pairs.
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