Solve the equation x^6-4x^3+3=0
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to solve x^6 - 4x^3 + 3 = 0
Let y = x^3
x^6 - 4x^3 + 3 = 0
=> y^2 - 4y + 3 = 0
=> y^2 - 3y - y + 3 = 0
=> y(y - 3) - 1(y - 3) = 0
=> (y - 1)(y-3) = 0
We get y = 1 and y = 3
Now y = x^3
Therefore we have
x^3 = 1 => x = 1^(1/3) = 1
x^3 = 3 => x = 3^(1/3) = cube root 3
The values are:
x = 1
x = cube root 3
Related Questions
- Solve for x log2 x + log4 x + log8 x =11/6
- 1 Educator Answer
- Solve the system x+y-14=0 4x-y-11=0
- 1 Educator Answer
- Solve the system of equations algebraically x^2+y^2=100 x-y=2
- 1 Educator Answer
- Solve for x and y x^3-y^3=7 x^2+xy+y^2=7
- 2 Educator Answers
- How to solve the equation x^8 - 64 = 0?
- 1 Educator Answer
The equation could be solvd using substitution technique. We'll reduce the equatin to a quadratic.
We'll note x^3 = t
t^2 - 4t + 3 = 0
We'll apply the quadratic formula:
t1 = [4+sqrt(16 - 12)]/2
t1 = (4+2)/2
t1 = 3
t2 = (4-2)/2
t2 = 1
But x^3 = t
x^3 = 3
x = 3^(1/3)
x^3 = 1
x = 1
The solutions of the equation are: {3^(1/3); 1 }.
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