一元二次方程式

Ex. 1
x^2 + 6x + 9 = 0
By factorization, we get (x + 3)^2 = 0, so x + 3 = 0, x = - 3.
We call (x + 3)^2 is a perfect square of x^2 + 6x + 9.
What happen if the equation is not a perfect square ? Our job will be to make a perfect square like in example 2.
Ex. 2
x^2 + 6x - 1 = 0
x^2 + 6x + 9 - 10 = 0
(x + 3)^2 - 10 = 0
(x + 3)^2 = 10
x + 3 = +/- sqrt 10
x = - 3 + sqrt 10 or - 3 - sqrt 10.
Another more complicated example,
2x^2 + 16x - 11 = 0
2(x^2 + 8x) - 11 = 0
Our job is to make x^2 + 8x a perfect square, using the principle in factorization, we know a^2 + 2ab + b^2 = (a + b)^2.
so a = x, 2ab = 8x , so b = 4, that is (x + 4)^2 = x^2 + 8x + 16,
that means x^2 + 8x = ( x + 4)^2 - 16, thus succeeding to make a perfect square. So the equation becomes:
2[(x+ 4)^2 - 16] - 11 = 0
2(x + 4)^2 - 32 - 11 = 0
2(x + 4)^2 = 43
x + 4 = +/- sqrt (43/2)
x = - 4 + sqrt (43/2) or - 4 - sqrt (43/2).
For the general quadratic equation ax^2 + bx + c = 0, you can use the same perfect square method to find that
x = [- b +/- sqrt(b^2 - 4ac)]/2a. Try to do it !