The intersecting lines method solves a quadratic equation with one curve and one straight line. You draw the parabola y = x² once, then rearrange each equation so that a straight line tells you its solutions. The x values where the line crosses the parabola are the answers.
Why it works
Take the equation x² − x − 6 = 0. Add x and 6 to both sides and it becomes x² = x + 6.
Now think of each side as a graph: y = x² on the left and y = x + 6 on the right. Where the two graphs cross, they have the same y value for the same x, so x² really does equal x + 6 there. The x-coordinates of the crossing points are therefore the solutions of the original equation.
Step by step
- Make x² have a coefficient of 1. If the equation starts with 2x², 3x² and so on, divide every term by that number.
- Rearrange to x² = mx + k. Keep x² on the left and move every other term to the right, changing their signs.
- Draw y = x² from a table of values.
- Draw the line y = mx + k on the same axes, using its y-intercept k and slope m.
- Read the x-coordinates of the crossing points. These are the solutions.
- Check each answer in the original equation.
| x | −4 | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| y = x² | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
One parabola works for every equation. Before calculators, students drew y = x² carefully on graph paper once, then solved equation after equation just by laying a ruler along a different line.
Worked examples
Example 1
Solve x² − x − 6 = 0.
Rearrange: x² = x + 6. Draw the line y = x + 6, which crosses the y-axis at 6 with slope 1.
It meets the parabola at (−2, 4) and (3, 9), so x = −2 or x = 3.
Check: (−2)² − (−2) − 6 = 4 + 2 − 6 = 0 ✓ and 3² − 3 − 6 = 0 ✓.
See Example 1 in the Quadratic Solver →
Example 2: when a is not 1
Solve 2x² − 3x − 2 = 0.
Divide every term by 2: x² − 1.5x − 1 = 0. Rearrange: x² = 1.5x + 1.
The line y = 1.5x + 1 crosses the parabola at (−0.5, 0.25) and (2, 4), so x = −½ or x = 2.
Check: 2 × 0.25 + 1.5 − 2 = 0 ✓ and 2 × 4 − 6 − 2 = 0 ✓.
See Example 2 in the Quadratic Solver →
Example 3: answers that aren’t whole numbers
Solve x² − 2x − 1 = 0.
Rearrange: x² = 2x + 1. The line crosses the parabola at about x = −0.4 and x = 2.4.
A graph can only give an estimate here. The exact answers, from the quadratic formula, are x = 1 − √2 ≈ −0.414 and x = 1 + √2 ≈ 2.414.
See Example 3 in the Quadratic Solver →
Two, one or no solutions
A straight line can meet the parabola y = x² in three ways, and each one matches a type of answer.
- The line cuts through the parabola: two crossing points, so two solutions.
- The line just touches the parabola: it is a tangent, so there is one repeated solution.
- The line passes underneath the parabola: no crossing points, so no real solutions.
The discriminant b² − 4ac tells you the same thing without a graph: positive for two solutions, zero for one, negative for none. For x² + x + 2 = 0 it is 1 − 8 = −7, so the line must miss the parabola.
A special case: no x term
An equation like x² − 5 = 0 becomes x² = 5. The line is y = 5, a horizontal line, and it crosses the parabola at x = −√5 and x = √5, about −2.24 and 2.24.
Common mistakes
- Forgetting to divide by a. The method only works with exactly x² on the left, because the parabola you drew is y = x².
- Sign errors when moving terms. In x² − x − 6 = 0, the −x and −6 become +x and +6 on the other side.
- Giving the y values. The solutions are the x-coordinates of the crossing points. In Example 1 the points are (−2, 4) and (3, 9), but the answers are x = −2 and x = 3.
To find exact answers without a graph, use factorising or the quadratic formula.
This method links quadratics to the work on straight lines in the other lessons: finding a line’s slope and intercept, and finding where two graphs cross.