Solving the quadratic equationx2=1±i3x2=−2ω,−2ω2where ω is a cube root of unityThen x=±i2ω2,±i2ωNow |x|=2, so the points should lie on a circle of radius 2, but how can we tell if it’s a square, rectangle or rhombus ?

Let x1=+i2ω2,x2=+i2ω,x3=−i2ω2,x4=−i2ωAs, x1=−x3 and x2=−x4, so they must lie opposite to each other.Now, x1−x2=x4−x3=i2(ω2−ω)=6, thus parallel to real axis.Also, x1−x3=x2−x4=i2(ω2+ω)=−i2, thus parallel to imaginary axis.Therefore the figure is a rectangle as all angles are 90deg and length of height and width are different.

"Solving the quadratic equationx2=1±i3x2=−2ω,−2ω2where ω is a cube..." solved and explained

High School
Algebra
Algebra I
Quadratic function and equation

palmantkf4u

Answered question

2022-04-02

Solving the quadratic equation

x^{2} = 1 \pm i \sqrt{3}

x^{2} = - 2 ω, - 2 ω^{2}

where

ω

is a cube root of unity
Then

x = \pm i \sqrt{2} ω^{2}, \pm i \sqrt{2} ω

Now

| x | = \sqrt{2}

, so the points should lie on a circle of radius

\sqrt{2}

, but how can we tell if it’s a square, rectangle or rhombus ?

Answer & Explanation

Alannah Farmer

Beginner2022-04-03Added 11 answers

Let

x_{1} = + i \sqrt{2} ω^{2}, x_{2} = + i \sqrt{2} ω, x_{3} = - i \sqrt{2} ω^{2}, x_{4} = - i \sqrt{2} ω

As,

x_{1} = - x_{3}

and

x_{2} = - x_{4}

, so they must lie opposite to each other.
Now,

x_{1} - x_{2} = x_{4} - x_{3} = i \sqrt{2} (ω^{2} - ω) = \sqrt{6}

, thus parallel to real axis.
Also,

x_{1} - x_{3} = x_{2} - x_{4} = i \sqrt{2} (ω^{2} + ω) = - i \sqrt{2}

, thus parallel to imaginary axis.
Therefore the figure is a rectangle as all angles are 90deg and length of height and width are different.

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