# Complex Numbers Quadratic. Sorry if my explanation is a bit long.

• Sensayshun
In summary, the equation z = -(3-8i) +/- \sqrt{b^2 - 4ac} can be rewritten as z = -3+8i +/- \sqrt{(-55-48i) - (4(2i(-6+7i))} which simplifies to z = -3 +/- 8i. Using this, the equation for z^3- (3-8i)z- 6+7i= 0 can be solved for z= -111-96i.

## Homework Statement

$$2iz^2 - (3-8i)z -6 + 7i$$

## Homework Equations

$$z = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}$$

## The Attempt at a Solution

Right, here goes...

a = 2i
b = -3 + 8i (is this correct? Or would it be better to leave it as 3-8i?)
c = (-6 + 7i)

so using these:

z = $$\frac{3-8i) +/- \sqrt{(-55-48i) - (4(2i(-6 + 7i))}}{4i}$$

I'll just work with the top line for now to make writing it easier.
But that last section under the square root, simplifies to:

$$\sqrt{(-55 - 48i) - (48i + 56)}$$
which is then:
$$\sqrt{-111 - 96i}$$

THIS IS WHERE I THINK I START TO GET STUCK

Do I then take the complex conjugate of the bottom to give:

$$\frac{(3 - 8i)(-4i)}{(4i)(-4i)} +/- \frac{\sqrt{111 - 96i}}{4i}$$

multiplying this out gives:

$$\frac{-3}{4}i - 2 +/- \frac{\sqrt{111 - 96i}}{4i}$$

and then I'm not sure where to go from here, or if I've headed in the correct direction

Thanks for any help given.

Last edited:
Sensayshun said:

## Homework Statement

$$2iz^2 - (3-8i)z -6 + 7i$$
Do you mean $z^3- (3-8i)z- 6+7i= 0$?

## Homework Equations

$$z = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}$$

## The Attempt at a Solution

Right, here goes...

a = 2i
b = -3 + 8i (is this correct? Or would it be better to leave it as 3-8i?)[/quote]
Yes, that is correct. You could leave it as -(3- 8i) but it is not -3+8i.

c = (-6 + 7i)

so using these:

z = $$\frac{3-8i) +/- \sqrt{(-55-48i) - (4(2i(-6 + 7i))}}{4i}$$

I'll just work with the top line for now to make writing it easier.
But that last section under the square root, simplifies to:

$$\sqrt{(-55 - 48i) - (48i + 56)}$$
4ac= 4(2i)(-6+7i)= 8i(-6+7i)= -48i- 56 so -4ac= 48i+ 56 not -(48i+ 56)

which is then:
$$\sqrt{-111 - 96i}$$
$$\sqrt{-55- 48i+ 48i+ 56}= \sqrt{1}= 1$$
That should make it much easier!

THIS IS WHERE I THINK I START TO GET STUCK

Do I then take the complex conjugate of the bottom to give:

$$\frac{(3 - 8i)(-4i)}{(4i)(-4i)} +/- \frac{\sqrt{111 - 96i}}{4i}$$

multiplying this out gives:

$$\frac{-3}{4}i - 2 +/- \frac{\sqrt{111 - 96i}}{4i}$$

and then I'm not sure where to go from here, or if I've headed in the correct direction

Thanks for any help given.

Under the problem statement, you give this:
$$2iz^2 - (3-8i)z -6 + 7i$$

What exactly are you trying to do? This is not an equation, so there is nothing to solve. It is an expression, which you can factor or otherwise rewrite in some different (but equal) form.

To mark, sorry it's meant to be = 0 on the end. Certain bits I couldn't get to format correctly, and I thought it'd be clear(ish).

HallsOfIvy - Thanks ever so much, that might make things far simpler!
Me and my friends were sat around trying to do this and we thought we'd have gone wrong with the signs somewhere, but we couldn't see it, so assumed it was fine. Someone did end up with -1, but no one got just 1. :P I'll work through it now with what you've shown.

Thanks very much, greatly appreciated :)

## What are complex numbers?

Complex numbers are numbers that have both a real and an imaginary part. They are expressed in the form a + bi, where a is the real part and bi is the imaginary part (i is the imaginary unit). Complex numbers are useful in mathematics and physics, particularly in solving quadratic equations.

## What is a quadratic equation?

A quadratic equation is a second-degree polynomial equation in one variable, usually written in the form ax² + bx + c = 0. It has two solutions, known as roots, which can be found using the quadratic formula or by factoring. Quadratic equations are commonly used to model parabolas and other curved shapes.

## How do complex numbers relate to quadratic equations?

Complex numbers are used to solve quadratic equations that have no real solutions. When the discriminant of a quadratic equation (b² - 4ac) is negative, the solutions will be complex numbers. By factoring or using the quadratic formula with complex numbers, we can find the solutions to these equations.

## What is the role of the imaginary unit (i) in complex numbers?

The imaginary unit, represented by the letter i, is a number that, when squared, gives a negative result (-1). It is used to represent the imaginary part of a complex number, which cannot be expressed as a real number. The imaginary unit is essential in solving quadratic equations with complex solutions.

## What are some applications of complex numbers in real life?

Complex numbers have numerous applications in fields such as engineering, physics, and economics. They are used in electrical engineering to represent alternating currents and in signal processing. In physics, complex numbers are used to describe the behavior of quantum systems. In economics, they are used to model financial markets and forecasting. Complex numbers are also used in computer graphics to create 3D images and animations.

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