What kind of equation is z^4-3z^2+2=0 ?
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
The equation you have given is z^4 - 3z^2 + 2 = 0. Here you can replace z^2 = x to get a quadratic equation in x which can be solved.
z^4 - 3z^2 + 2 = 0
=> x^2 - 3x + 2 = 0
=> x^2 - 2x - x + 2 = 0
=> x(x - 2) - 1(x - 2) = 0
=> (x - 1)(x - 2) = 0
So we have x = 1 and x = 2
As x = z^2
z^2 = 1 and z^2 = 2
z = -1 and z = +1 and z = +sqrt 2 and - sqrt 2
The roots of the equation are -1 , 1 , sqrt 2 and -sqrt 2
Related Questions
- z=?What is the absolute value of z if 3z -9i = 8i + z + 4.
- 1 Educator Answer
- Find z in the unique solution of the systemx + 2y + 3z =1-x - y + 3z = 2-6x + y + z = -2
- 1 Educator Answer
- equationFind the solutions of x^4 - 3x^2 + 2 = 0
- 1 Educator Answer
- Solve the equation z+z'=z'*z^2+i*z*z' where z is a complex number.
- 1 Educator Answer
- Solve the simultaneous equations x^2+y^2=10 , x^4+y^4=82
- 1 Educator Answer
This equation is called biquadratic equation and it is easy to solve.
This equation is reduced to a quadratic equation when doing the substitution z^2 = x.
We'll re-write the equation in x:
x^2 - 3x + 2 = 0
We'll apply quadratic formula:
x1 = [3+ sqrt(9-8)]/2
x1 = (3+1)/2
x1 = 2
x2 = (3-1)/2
x2 = 1
But, we'll have to find z1,z2,z3,z4.
z^2 = x1
z^2 = 2
z1 = sqrt 2 and z2 = -sqrt 2
z^2 = x2
z^2 = 1
z3 = -1 and z4 = 1
The solutions of the biquadraticÂ equation are: {-sqrt2 ; -1 ; 1 ; sqrt2}.
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Student Answers