The below given is the Quartic equation tutorial which provides you the definition, formula, and example for a quartic equation. Calculations are explained in a through step by step instructions in this Fourth degree equation tutorial.
In algebra, a quartic function is defined as a function of the form ax4+ bx3 + cx2 + dx + e = 0, where 'a' is non zero, which is defined by a fourth degree polynomial, called a quartic polynomial.
Calculate the roots (x1, x2, x3, x4) of the quartic equation, 3X4 + 6X3 - 123X2 - 126X + 1080 = 0
From the above equation, the value of a = 3, b = 6, c = -123, d = -126, e = 1080.
To find x : Substitute the values in the formulas below. f = c - ( 3b ² / 8 ) g = d + ( b ³ / 8 ) - ( b x c / 2 ) h = e - ( 3 x b4 / 256 ) + ( b ² x c / 16 ) - ( b x d / 4 )
Form as Cubic Equation : y ³ + ( f / 2 ) y ² + (( f ² - 4 x h ) / 16 ) y - g ² / 64 = 0 where, a = coefficient of y ³ b = coefficient of y² c = coefficient of y d = constant
From the above equation, the values of a, b, c, and d are, a = 1, b = f / 2, c = (( f ² - 4 x h ) / 16 ), and d = - g² / 64.
To Find y: Substitute the values in the formula's below to find the roots. The variable disc is nothing but the discriminant, denoted generally as delta(Δ) discriminant(Δ) = q3 + r2 q = (3c - b2) / 9 r = -27d + b(9c - 2b2) s = r +√ (discriminant) t = r - √(discriminant) term1 = √(3.0) * ((-t + s) / 2) r13 = 2 * √(q) y1 = (- term1 + r13*cos(q3/3) ) y2 = (- term1 + r13*cos(q3+(2∏)/3) ) y3 = (- term1 + r13*cos(q3+(4∏)/3) )
We get the roots, y1 = 20.25 , y2 = 0 and y3 = 1.
After finding cubic equation solve quartic equation Substitue y1, y2, y3 in p, q, r, s. NOTE : Let p and q be the square root of any 2 non-zero roots. p = sqrt(y1) = 4.5 q = sqrt(y3) = 1 r = -g / (8pq) = 0 s = b / (4a) = 0.5
We get the roots, x1 = 5, x2 = 3, x3 = -4 and x4 = -6.
This tutorial will help you dynamically to find the roots of a Quartic Equation.