Quadratic Equations

Any equation of degree 2 (where the highest power of x is 2) of the form
y=f(x)=a*x^2 + b*x + c=0 is called a quadratic equation.

Let m and n be two roots of the above equation. m and n can be real/imaginary.

The solution for the quadratic equation is:
x = [- b ± √(b² - 4*a*c)]/2*a.
here, we have two solutions for x or two roots of x. So, we can say,
m = [- b + √(b² - 4*a*c)]/2*a ;

n = [- b - √(b² - 4*a*c)]/2*a.

m + n = -b/a, and, m*n = c/a.

(x-m) and (x-n) are factors of y=f(x).

m, n are positive if (m + n) < 0 and m*n > 0, i.e., b/a<0 and c/a>0

m, n are positive if (m + n) < 0 and m*n > 0, i.e., b/a<0 and c/a>0.

m, n are negative if (m + n) < 0 and m*n > 0, i.e., b/a<0 and c/a>0.

      

Discriminant = D = b^2 – 4*a*c.

if D<0, m & n are complex roots.

if D=0, m=n and roots are real and rational.

if D>0, then if D is a perfect square, then roots are rational and equal, otherwise, roots are irrational and they appear in conjugate pairs.

Maxima of f(x): only when a>0, x = -b/2a. f(x) = (4*a*c – b^2)/(4*a).

Minima of f(x): only when a<0, x = -b/2a. f(x) = (4*a*c *b^2)/(4*a). 

 

To form a new quadratic equation from another quadratic equation:-

Say we have a quadratic equation: a*x^2 + b*x + c = 0 with known roots m and n.
Thus, a*x^2 + b*x + c = (x-m)*(x-n) = 0.
Say we have to find the roots of quadratic equation: p*y^2 + q*y + r = 0 where
y = h(x) and the roots are s, t.

We can solve this by the following equations:
x = m.         x = n.
h(x) = s.      h(x) = t.
m + n = -b/a.           m*n = c/a.
s + t = -q/p.             m*n = r/p.
p*[h(x)]^2 + q*[h(x)] + r = 0.