y = x 2 − 3 x + 2and draw its graph. The parabola crosses the x axis at x = 1 and x = 2. This means that the solutions of the quadratic equation x 2 − 3 x + 2 = 0 are x = 1 and x = 2.
(x − x1)(x − x2) = 0,then the solutions are x = x1 and x = x2. When you can spot the factors, this is probably the easiest of the four methods.
x 2 − 4 x + 3 = 0.The left-hand-side can be factored:
(x − 1)(x − 3) = 0.Therefore the solutions of the quadratic equation are x = 1 and x = 3.
Warning: A common error is to think that the solutions are −1 and −3. This is not correct; the solutions are the values of x that make the factors vanish (become equal to zero). |
a x 2 + b x = −c
and combine the fractions on the right hand side:
This equation has a nice geometric interpretation. The two values of x are indicated by the red dots. The first term of the equation locates the midpoint (the point halfway between them) and the second term (the one after the ± symbol) gives the distance from the midpoint to either one of them:
x 2 − 4 x = −3.
x 2 − 4 x + 4 = −3 + 4.
(x − 2) 2 = 1.
x − 2 = ±1.
Note:
with a = 1, b = −4 and c = 3:
The + sign gives the solution x = 3 and the − sign gives the solution x = 1.
Algebra Coach Exercises |