complex quadratics. (1 Viewer)

[sikeveo]

i was doing some from cambridge but i dont get how they do it.

You know how there are two different forms of answers depending on whether the discriminant is larger than 0 or smaller than zero.....But in the book they always convert the answer to a complex form. E.g. -16 = 16i² and then they use the answer form for as if the discriminant is larger than 0. Anyone care to explain?

 

[Riviet]

i was doing some from cambridge but i dont get how they do it.

You know how there are two different forms of answers depending on whether the discriminant is larger than 0 or smaller than zero.....But in the book they always convert the answer to a complex form. E.g. -16 = 16i² and then they use the answer form for as if the discriminant is larger than 0. Anyone care to explain?

When ∆<0, the roots are complex, e.g 2+i, which also includes imagnary numbers such as 3i. The book only shows examples for when ∆<0, and shows how to solve equations when this occurs. If ∆>0, then this goes straight back to 2 unit maths.

 

[YBK]

i was doing some from cambridge but i dont get how they do it.

You know how there are two different forms of answers depending on whether the discriminant is larger than 0 or smaller than zero.....But in the book they always convert the answer to a complex form. E.g. -16 = 16i² and then they use the answer form for as if the discriminant is larger than 0. Anyone care to explain?

That's because it's less than zero.

You have to then find the roots of, in your example, -16.

16i^2 is the same as -16, since i^2 = -1

Therfore 16 * -1 = -16

From there you should be able to find the roots, using the normal quadratic formula. Remember complex numbers are very similar to real numbers, you use the same formulae on them usually. The only difference is that you'll have an "i" in the answer, or you'll need to have to find the square root of that "i", in which case you have to equate the real and imaginary parts.

 

[sikeveo]

thanks riv, forgot about that, however it states if /_\ is bigger than 0 then the roots are (i sqrt /_\) and - (i sqrt /_\)

 

[Riviet]

however it states if /_\ is bigger than 0 then the roots are (i sqrt /_\) and - (i sqrt /_\)

If ∆>0, this equation has two real roots given by x= (-b+sqrt∆)/2a.

If ∆<0, etc. . .

Hence the given quadratic equation has two non-real roots given by x= (-b+i.sqrt|∆|)/2a.

The part in bold is referring to when ∆<0, NOT when ∆>0.

I hope that helps.

 

[YBK]

The part in bold is referring to when ∆<0, NOT when ∆>0.

I hope that helps.

sweet, I got confused by what sikeveo said...