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 − x_{1})(x − x_{2}) = 0,then the solutions are x = x_{1} and x = x_{2}. 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
x^{ 2} − 4 x = −3.
x^{ 2} − 4 x + 4 = −3 + 4.
(x − 2)^{ 2} = 1.
x − 2 = ±1.
Algebra Coach Exercises |