Question

I NEED HELP!!! MEDAL AND FAN REWARD!!!
See post for questions

Answer 1

whats the question

Answer 2

I need help with creating the following:

One function, f(x), with two real rational solutions
One function, g(x), with two real irrational solutions
One function, h(x), with two complex solutions

Explain how you know these functions meet each condition

Answer 3

I promise I'm not trying to cheat or anything I just need serious help. I can't figure this out!

@ranga
@Captain_Page_Turner
@Calliope

Answer 4

@ganeshie8

Answer 5

Pick two rational numbers say 2 and 3
Then f(x) = (x-2)(x-3) will be a function with two real rational roots because if I put x =2 or x = 3 f(x) will become 0 and therefore 2 and 3 are the roots.
You can multiply it out and f(x) = x^2 - 5x + 6

Answer 6

Okay, that makes some sense

Answer 7

For the general quadratic function ax^2 + bx + c to have irrational roots, the discriminant must NOT be a perfect square.
Discriminant is: b^2 - 4ac
Choose a, b and c such that they are not a perfect square.
Let a = 1, you choose b and c.

Answer 8

a=1, b = 7, c=2? @ranga

Answer 9

And the discriminant should be positive and not a perfect square for real, irrational roots.

Answer 10

you could use the same function for all three requirements by multiplying together two real rational roots, two real irrational roots, and ± an imaginary root.

Answer 11

Yes, a=1, b = 7, c=2 will do the job.

Answer 12

how would you suggest i go about the imaginary root part? @Peter14

Answer 13

g(x) = x^2 + 7x + 2

Answer 14

For imaginary solutions, make the discriminant negative.

Answer 15

right thats what i put @ranga

Answer 16

make b^2 - 4ac negative
let a = 1, you choose b and c

Answer 17

multiply together (x+ai) (x-ai) to make a simple function that fulfils the requirements

Answer 18

okay so the discriminant is negative for the third and i just have to decide what i want b and c to be? does it matter if they are rational or irrational?

Answer 19

choose integers but make b^2 - 4ac negative.
a= 1, b = 2, c = 3 will do the job

Answer 20

or h(x)=x^2 + 1

Answer 21

okay @ranga

Answer 22

thank you @ranga and @Peter14

Answer 23

you are welcome.

Answer 24

you're welcome

Answer 25

any other questions? or are you going to close the question now?

Answer 26

I will be closing it. but ill message you if i have any @Peter14