Quadratic Equations Explained Simply
A quadratic equation is one in which the unknown appears squared and in no higher power. Its graph is always a parabola, and asking for the solutions is asking where that parabola crosses the x-axis. That can happen twice, once, or not at all.
Two formulas, one idea
Many students treat the quadratic formula and the reduced pq-formula as separate methods. They are not. The pq-formula is simply the quadratic formula for the special case a = 1. Divide 2x² + 6x − 8 = 0 by 2 and you get x² + 3x − 4 = 0, where the shorter version applies.
The decisive part of both is the expression under the square root, called the discriminant. If it is positive there are two distinct solutions. If it is exactly zero both solutions coincide and the parabola touches the axis instead of crossing it. If it is negative there is no real solution, because no real number squares to a negative value.
Three parabolas with two, one and no roots
D = 64 − 48 = 16
x = ( 8 ± 4 ) / 4 → x₁ = 3, x₂ = 1
x² − 6x + 9 = 0 → D = 0 → x = 3 (double root)
x² + 2x + 5 = 0 → D = −16 → no real solution
| D = b² − 4ac | Solutions | The parabola … |
|---|---|---|
| D > 0 | two distinct | crosses the x-axis twice |
| D = 0 | one double root | touches the axis at its vertex |
| D < 0 | none real | never reaches the axis |
When no formula is needed
If the constant term is missing, factor out x: x(ax + b) = 0 gives x = 0 or x = −b/a. If the linear term is missing, as in x² − 49 = 0, take the square root directly and remember both signs: x = ±7. Vieta's formulas also help — for x² + px + q the roots add up to −p and multiply to q, which cracks most integer problems in your head.
Using the pq-formula when the leading coefficient is not 1.
Losing the sign of b. If b = −8, the formula contains +8.
Giving only the positive square root. x² = 49 has two solutions.
Not moving everything to one side first. Both formulas assume the
equation equals zero.
A rectangular bed is 3 m longer than it is wide and covers 40 m². How wide is it?
Show solution
x(x + 3) = 40 → x² + 3x − 40 = 0 → x = −1.5 ± 6.5 → x₁ = 5, x₂ = −8. A width cannot be negative, so the bed is 5 m wide and 8 m long.