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

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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.

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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: *

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).*

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 :

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

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,

Each factor gives two values of z. So we get 4 solutions for z.