Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,356 questions

17,804 answers

1,438 comments

35,964 users

Complex Number System?

+1 vote
What exactly does it mean to Solve each polynomial equation in the complex number system.

Especially the \'complex number system\' part.

x^2 - 6x + 10 = 0

x^2 + x + 1 = 0

x^2 + 6x + 11 = 0

New Problem

Ex) x^4-1=0
asked Dec 25, 2012 in ALGEBRA 2 by alg2trig Rookie

2 Answers

+1 vote

When x2 - 6x + 10 is not factorable apply quadratic formula.

Roots of ax2 + bx + c = 0, a  ≠0 then x = (- b ± √(b2 - 4 ac))/2a

Compare x2 - 6x + 10 with ax2 + bx + c and find the coefficients.

 a = 1, b = -6 and c = 10

x = (6 ± √(62 - 4 * 1 * 10))/2 *1

 x = (6 ± √(36 - 40))/2

 x = (6 ± √(-4))/2

 Substitute i2 = -1

 x = (6 ± √(4i2))/2

 x = (6 ± 2i)/2

 x = 3 ± i

Roots are   x = 3 + i and   x = 3 - i

-----------------------------------------------------------

The equation is x2+x+1=0

Using formulae for roots of quadratic equation ax2+bx+c=0,

x= (-b ±√ (b2-4ac))/2a

x= (-1 ±√ ((1)2-4(1)(1))/2(1) 

  = (-1 ±√ (1-4))/2

  = (-1 ±√-3)/2

  = (-1 ±√3i2)/2 

  = (-1 ±√3i)/2

  = (-1/2) ±( √3/2)i

Roots are (-1/2) + ( √3/2)i and  (-1/2)-( √3/2)i

-----------------------------------------------------------

The equation is x2+6x+11=0

Using formulae for roots of quadratic equation ax2+bx+c=0,

x= (-b ±√ (b2-4ac))/2a  (Substitute; a=1, b=6 and c=11)

x= (-(6) ±√ ((6)2-4(1)(11))/2(1)                                       

  = (-6 ±√ (36-44))/2

  = (-6 ±√-8)/2

  = (-6 ±√8i2)/2             

  = (-6 ±2√2i)/2

  = -3 ±√2i

Roots are -3+√2i and -3-√2i

answered Dec 28, 2012 by steve Scholar
0 votes
  • Ex).

The equation is x4 - 1 = 0.

Rewrite the equation as (x2)2 - 12 = 0.

Difference of two squares formula : a 2- b 2 = (a + b )(a - b ).

(x2 + 1)(x2 - 1) = 0

Apply zero product property.

x2 + 1 = 0 and x2 - 1 = 0

x2 = - 1 and x2 = 1

Put - 1 = i2 .

x2 = i2 and x2 = 1

⇒ x = ± i and x = ± 1.

Solution is x = ± i and x = ± 1.

answered Jun 21, 2014 by lilly Expert
edited Jun 21, 2014 by lilly

Related questions

...