Question

Which is the easy method to solve the polynomial equation of degree 4 i.e., bi-quadratic? Please explain it... Very urgent....!

Answer 1

Put x=1,-1,2,-2,3,-3 n in the option in which f(x)=0, choose it n make factor like if putting x =1, u get f(x)=0, the factor will be (x-1).. then divide the equ with x-1 n u ll get a cubic equ, apply same putting on it n again divide the factor you came across second time wid cubic equ, u ll get a quadratic equ n simply do its middle splitting.. Hope it helps..

Answer 2

to solve:    f(x) = x⁴ + a x³ + b x² + c x + d = 0    --- (1)

  one way is to check f(x = -1), f(x = -2), f (x = -3) or f(0), or f(x=1) or f(x=2) etc if that is zero.  Then we have a factor like (x- k), if  f (x = k) = 0.

  Then we can write f(x) = (x - k) [ x³ + (a+k) x² + (b+ak+k²) x - d / k ]
  Now, the cubic equation could be again checked if, you could find some factors.  Or, one could try the way a cubic equation is solved or reduced into a quadratic equation.  There are standard formulas for solving cubic equations.
====================
We could try other means of reducing the quartic equation into a cubic equation and then solve that ..

f (x) = x⁴  + a x³ + b x² + c x + d = 0    -- (1)

let z = x + a/4,  then    z⁴ = x⁴ + 4 x³ a/4 + 6 x² a²/4²  + 4 x a³/4³ + a⁴/4⁴
         x⁴ + a x³ =  z⁴ - 3 x²a²/8 -- xa³/16 - a⁴/256
 Substitute   x = z - a/4

f(x) = (z - a/4)⁴ + a(z - a/4)³ + b (z - a/4)² + c (z - a/4) + d = 0
     =  z⁴ - z³ a + 3 z² a²/8 - z a³/16 + a⁴/256 + a z³ - 3a² z²/4 + 3 a³ z/16 - a⁴ z/64
           + b z² - ab z/2 + a²b/16 + c z - ac/4 + d = 0

F(z) = z⁴ + z² (b -3a²/8) + z (c+ a³/8 - a⁴/64 - ab/2) + (a⁴/256 +a²b/16 -ac/4 +d) = 0    -- (2)
       Let  A = b - 3a²/8      B = c+a³/8 - a⁴/64 -ab/2
              C = d + a⁴/256 + a²b/16 - ac/4

F(z) = z⁴ + A z² + B z + C = 0          rewrite it as,        --- (3)
       =>  (z² + A)² = A z² + A² - B z - C

  Then  (z² + A + y)² = (z²+A)²  + y² + 2 y (z² + A)            --- (4)
                        = A z² + A² - Bz - C + y² + 2 y z² + 2 A y
                        = z² (A + 2y) - B z + (y² + 2Ay + A² - C)      -- (5)

 2 Roots of this equation are equal if the discriminant on the RHS is 0.  So RHS becomes a perfect square.  LEt us choose y such that Discriminant = 0.
      Discr = B² - 4 (A+2y) (y² + 2Ay + A² - C)  = 0        --- (6)
              = B² - 4 [ 2 y³ + 5A y² + 2 y (2A² - C) + A³ - AC ] = 0
              = B² - 8 y³ - 20 A y² - 8 (2A²-C) y - 4(A³- AC)  = 0    
     =>  y³ + (A²+4A)/2 y² + (2A² -C) y + [(A³-AC) /2 - B²/8] = 0      -- (7)
 
  Call  (A²+4A)/2  = P   ;   2A²-C = Q  ;    (A³-AC)/2 - B²/8 = R

This is a cubic equation.  Rewrite it as :

   G(y) = y³ + P y² + Q y + R = 0                       -- (8)
   now   substitute u = y + P/3 ,  so  y =  u - P/3
 F(y) = T(u) = (u-P/3)³ + P(u- P/3)² + Q (u - P/3) + R = 0
  T(u) = u³ - 3 u² P/3 + 3 u P²/9 + P u² - 2P²u/3 +P³/9 + Q u -PQ/3 + R = 0
         = u³ + u (Q - P²/3) + (P³/9 - PQ/3 +R)  = 0
        
   Call   (Q - P²/3) = - 3p       and  (P³/9 - PQ/3 + R) = - 2 q

   T(u) = u³ - 3p u - 2q = 0              -- (9)
Three Solutions for this are: 
   $u_k = 2 \sqrt{p}\ Cos [ \frac{1}{3} Cos^{-1} (\frac{q}{p^\frac{3}{2}}) - \frac{2\pi k}{3}],\ \ for\ k=0,\ 1,\ 2$

   From the solutions u_0, u_1, u_2, we find y = u - P/3.
   For these values of y we have the equation (7) satisfied.  Hence, the discriminant of RHS of equation (5) is zero. The solution for equation (5) is:

          (z²+A+y)² = (A+2y) [ z - B/(2A+4y) ]²
                         = [ √(A+2y) z - B / {2√(A+2y) }]²
Hence,
  [z²+A+y+√(A+2y) z - B/ {2√(A+2y)} ] [z²+A+y-√(A+2y) z +B/ {2√(A+2y)}] = 0

Each factor gives two values of z.  So we get 4 solutions for z.
 Then substitute x = z - a/4  to get the 4 solutions of equation (1).