Solving Complex Numbers?

Question

-Solve in the complex number system:

1. 3x^2 = 2x-5

2. x^3 - 8 = 0

How? Thanks.

Answer 1

Question 1
3x² - 2x + 5 = 0
x = [ 2 ± √(- 56) ] / 6
x = [ 2 ± 2 i √(14) ] / 6
x = [ 1 ± i √(14) ] / 3

Question 2
x³ = 8
x = 2

Answer 2

For the first one, if you rewrite it as 3x^2 - 2x + 5 = 0, you can use the quadratic equation where a = 3, b = -2 and c = 5.

For the second one, you have to use an obscure factoring rule called Difference of Cubes. Note how both x^3 and 8 are perfect cubes; knowing that, the formula for difference of cubes is:

(a - b)(a^2 + ab + b^2)

where a = cuberoot(x^3) and b = cuberoot(8). Therefore, with a = x and b = 2, we get the following factorization:

(x - 2)(x^2 + 2x + 4)

The imaginary number comes in when we do the quadratic equation on the second term - the trinomial - where a = 1, b = 2 and c = 4.

Try that and see if it helps.

Answer 3

Use the quadratic formula or complete the square.

3x^2 -2x =-5

x^2 -2/3(x ) =-5/3

x^2 -2/3(x ) + 1/9 =-5/3 + 1/9

(x - 1/3)^2 = -14/9

x-1/3 = =/-i/3(sqrt14)

x = 1/3+/-i/3(sqrt14)

This has a formula to factor, then apply the quadratic formula or complete the square for the trinomial.

(x - 2 )(x^2 +2 x + 4) = 0

x=2 is a rational root,

x^2 + 2x =-4
x^2 + 2x + 1=-4 + 1
(x+1)^2 = -3
x+1 = +/- i*sqrt3

x = -1 +/- i*sqrt3


I chose this method because of the typing involved with the quadratic formula.

Answer 4

enable z = (a, b) = a + bi with a and b real numbers. Then |z|^2 = a^2 + b^2 (= the modulus squared) Your equation will become a^2 + b^2 - 2a - 2bi = a million + 2i To have an equality, you need to a minimum of one after the different make the real and imaginary aspects equivalent on the two aspects: a^2 + b^2 - 2a = a million -2b = 2 so as that b = -a million Then a^2 + a million - 2a = a million or a (a - 2) = 0 so as that a = 0 or a = 2. There are 2 solutions: (0, -a million) and (2, -a million).

Answer 5

1. 3x^2 = 2x-5 =
6x = 2x-5 =
6x - 2x = 5 =
4x = 5 =
***x = 5/4***

2. x^3 - 8 = 0 =
3x - 8 = 0 =
3x = 8 =
***x=8/3***

Answer 6

For the first one you, have 3x^2=2x-5 so move all numbers to one side of the equation. 3x^2-2x+5
Then you can factor, use quadratic formula, use vertex formula, etc