The example is as follows

A following equation is a fourth-order polynomial equation of the form

(1)

While some authors (Beyer 1987b, p. 34) use the term "biquadratic equation" as a synonym for quartic equation, others (Hazewinkel 1988, Gellert et al. 1989) reserve the term for a quartic equation having no cubic term, i.e., a quadratic equation in .

The roots of this equation satisfy Vieta's formulas:

	(2)
	(3)
	(4)
	(5)

where the denominators on the right side are all . Writing the quartic in the standard form

(6)

the properties of the symmetric polynomials appearing in Vieta's formulas then give

			(7)
			(8)
			(9)
			(10)

for quadratic equations the concept is called the discriminant.it isproved as follows.

by completion of squares concept

(ax^+bX+c) / a=0

x^2+bx/a=c/a=0

x^2+bx/2a+c/a+b^2/4a^2-b^2/4a^2=0

(x+b/2a)^2+c/a-b^2/4a^2=0

(x+b/2a)^2=b^2/4a^2-c/a

(x+b/2a)^2=b^2-4ac/4a^2

x+b/a=+-(b^2-4ac/4a^2)^1/2

x+b/a=+-(b^2-4ac)^1/2  /2a

x=-b+-(b^2-4ac)^1/2  /2a

where b^2-4ac=d(discriminant)

thus x=-b+-(d)^1/2  /2a