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Exercises 17.6 Additional Exercises: Solving the Cubic and Quartic Equations

1.

Complete the square to solve the general quadratic equation
ax2+bx+c=0
to obtain
x=b±b24ac2a.
The discriminant of the quadratic equation Δ=b24ac determines the nature of the solutions of the equation. If Δ>0, the equation has two distinct real solutions. If Δ=0, the equation has a single repeated real root. If Δ<0, there are two distinct imaginary solutions.

2.

Show that any cubic equation of the form
x3+bx2+cx+d=0
can be reduced to the form y3+py+q=0 by making the substitution x=yb/3.

3.

Prove that the cube roots of 1 are given by
ω=1+i32ω2=1i32ω3=1.

4.

Make the substitution
y=zp3z
for y in the equation y3+py+q=0 and obtain two solutions A and B for z3.

5.

Show that the product of the solutions obtained in (4) is p3/27, deducing that AB3=p/3.

6.

Prove that the possible solutions for z in (4) are given by
A3,ωA3,ω2A3,B3,ωB3,ω2B3
and use this result to show that the three possible solutions for y are
ωiq2+ p327+q243+ω2iq2 p327+q243,
where i=0,1,2.

7.

The discriminant of the cubic equation is
Δ=p327+q24.
Show that y3+py+q=0
  1. has three real roots, at least two of which are equal, if Δ=0.
  2. has one real root and two conjugate imaginary roots if Δ>0.
  3. has three distinct real roots if Δ<0.

8.

Solve the following cubic equations.
  1. x34x2+11x+30=0
  2. x33x+5=0
  3. x33x+2=0
  4. x3+x+3=0

9.

Show that the general quartic equation
x4+ax3+bx2+cx+d=0
can be reduced to
y4+py2+qy+r=0
by using the substitution x=ya/4.

10.

Show that
(y2+12z)2=(zp)y2qy+(14z2r).

12.

From Exercise 17.6.11 obtain the resolvent cubic equation
z3pz24rz+(4prq2)=0.
Solving the resolvent cubic equation, put the equation found in Exercise 17.6.10 in the form
(y2+12z)2=(my+k)2
to obtain the solution of the quartic equation.

13.

Use this method to solve the following quartic equations.
  1. x4x23x+2=0
  2. x4+x37x2x+6=0
  3. x42x2+4x3=0
  4. x44x3+3x25x+2=0