Complex Number System?

Question

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

Answer 1

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

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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

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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

Answer 2

• 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.