Solving quadratic equations by completing the square, the quadratic formula
Solving quadratic equations by factoring, Vieta's formula

A quadratic equation is a polynomial equation of the second degree. The general or standard form is
ax2 + bx + c = 0,   a is not 0
where a, b and c are coefficients.
Solving quadratic equations by completing the square, the quadratic formula
The solutions of the equation are called roots (which may be real or complex) and are given by the quadratic formula.
We use the completing the square method to derive the quadratic formula to find the roots,
The roots can also be written
If  c = 0, the roots of the equation         ax2 + bx = 0       we can obtain by factoring
x(ax + b) = 0
or by using the quadratic formula, so we get   x1= 0  and  x2 = -b/a.
Solving quadratic equations by factoring, Vieta’s formula
A quadratic trinomial  ax2 + bx + can be factorized as
 ax2 + bx + c = a·[x2 + (b/a)·x + c/a] = a·(x - x1)(x - x2), where x1 + x2 = b/a and  x1· x2 = c/a
That means, to factor a quadratic trinomial we should find such a pair of numbers x1 and x2 whose sum equals b/a and whose product equals c/a.
So, if a > 0 and the constant term c negative, then the signs of x1 and x2 will be different while, when c is positive, their signs will be the same.
Then we use zero product principle to solve the quadratic equation, that is, a product a · b = 0 if at least one of its factors is 0.
 Examples:     Find the roots of each quadratic equation by factoring: a)  x2 - 3x -10 = 0 since   x2 - 3x -10 = x2 + (-5 + 2)·x + (-5)·(+2) = x2 - 5x + 2x -10 = x · (x - 5) + 2 · (x - 5) = (x - 5) · (x + 2), then     x - 5 = 0    =>     x1 = 5,     or      x + 2 = 0    =>     x2 = -2. b)  2x2 - 7x + 3 = 0 as  2x2 - 7x + 3 = 2[x2 - (7/2)x + 3/2] =  2[x2 - (1/2)x - 3x + 3/2] = = 2[x(x - 1/2) - 3(x - 1/2)] = 2 (x - 1/2)(x - 3) = (2x - 1) (x - 3), then     2x - 1 = 0    =>     x1 = 1/2,     or      x - 3 = 0    =>     x2 = 3 c)  3x2 - x - 2 = 0 as  3x2 - x - 2 = 3[x2 - (1/3)x - 2/3] =  3[x2 + (2/3)x - x - 2/3] = = 3[x(x + 2/3) - (x + 2/3)] = 3·(x + 2/3)(x - 1) = (3x + 2)(x - 1), then     3x + 2 = 0    =>     x1 = -2/3,     or      x - 1 = 0    =>     x2 = 1.
Example:   Given are leading coefficient a2 = -1 and the pair of conjugate complex roots,
x1 = 1 + and  x2 = 1 - i, of a second degree polynomial;
a)  find the polynomial using the above theorem,
b)  make a check of the solutions of the polynomial using the quadratic formula.
 Solution: a)  By plugging the given values into    a2x2 + a1x + a0 = a2(x - x1) (x - x2) a2x2 + a1x + a0 = -1[x - (1 + i)] · [x - (1 - i)] = -[(x - 1) - i] · [(x - 1) + i] = = -[(x - 1)2 - i2] = -(x2 - 2x + 1 + 1)  = - x2 + 2x - 2 b)  Check the polynomial for the roots,
Pre-calculus contents B