Question

A quadratic equation in form ax^2+bx+c=0 cannot have:
one real solution one imaginary solution
two real solutions two imaginary solutions

Answer 1

a and b are answers

Answer 2

which ones srry I didn't label them

Answer 3

one real //////////one imginary

Answer 4

its one or the other

Answer 5

It can have one real solution if you ignore "repeated" roots.

Answer 6

so one inaginary

Answer 7

Yup, if you twist my arm.

Answer 8

?

Answer 9

That was correct thanks

Answer 10

I do have a certain amount of sympathy with html1618's position.

Answer 11

It cannot have 1 imaginary solution as imaginary solutions come from cubics.
\[ax^3 + bx^2 + cx + d\]

Answer 12

@Twill What are the solutions of x^2 +2?

Answer 13

2 of 'em, right?

Answer 14

@estudier i >.>

Answer 15

well,

x^2 + 2 = 0

x^2 = -2

So x = i

Answer 16

Well, not quite; try using the quadratic formula, see what you get...

Answer 17

+/- 2!

Answer 18

I thought there was a square root sign in the quadratic formula?

Answer 19

Sorry, Pie_Boi, didn't mean to hijack your question.

Answer 20

\[-b +/- \sqrt{b^2 - 4ac}\div2a\]

a =1
b = 0
c = 2

\[\pm \sqrt{(0)^2 - 4(1)(2)} \div2(1)\]

\[\pm \sqrt{-8}\div2\]

\[\pm \sqrt{-4}\]

Gives you i?

Answer 21

I can write \[\sqrt{-4}\]
as
\[\sqrt{-1}\sqrt{4}\]
Do you agree?

Answer 22

Yes, fair point. So i and \[\pm2?\]

Answer 23

Usually written as \[\pm i \sqrt{2}\]

Answer 24

Well done.

Answer 25

Thank you!

Sorry for the Hijack Pie_Boi

Answer 26

no problem at all :)