- Solve x2 + 3x – 4 = 0
- This quadratic happens to factor:
- x2 + 3x – 4 = (x + 4)(x – 1) = 0
Suppose you have ax2 + bx + c = y, and you are told to plug zero in for y. The corresponding x-values are the x-intercepts of the graph. So solving ax2 + bx + c = 0 for x means, among other things, that you are trying to find x-intercepts. Since there were two solutions for x2 + 3x – 4 = 0, there must then be two x-intercepts on the graph. Graphing, we get the curve below:

Note, however, that the calculator's display of the graph will probably have some pixel-related round-off error, so you'd be checking to see that the computed and graphed values were reasonably close; don't expect an exact match.
- Solve 2x2 – 4x – 3 = 0. Round your answer to two decimal places, if necessary.
- There are no factors of (2)(–3) = –6 that add up to –4, so I know that this quadratic cannot befactored. I will apply the Quadratic Formula. In this case, a = 2, b = –4, and c = –3:
Compare the solutions of 2x2 – 4x – 3 = 0 with the x-intercepts of the graph: Just as in the previous example, the x-intercepts match the zeroes from the Quadratic Formula. This is always true. The "solutions" of an equation are also the x-intercepts of the corresponding graph. |
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