Question

Need a quadratic equation with two real irrational solutions.

Answer 1

well, how about if you give us two real irrational solutions first?

Answer 2

I have to create the equation and find the solutions.

Answer 3

yes, but is simpler if you just get the solutions first, then make the equation :)

Answer 4

Oh can you give me an example of an irrational solution?

Answer 5

hmmm well.. do you know what an irrational number is?

Answer 6

One that doesn't stop?

Answer 7

well... not quite... because that's true for 1/3 yet is a fraction -> http://www.mathsisfun.com/irrational-numbers.html

Answer 8

So i could use 1/3 and 5/6?

Answer 9

well, 1/3 can be expressed as a RATION thus is a RATIONAL number

Answer 10

I need irrational ones that are still real numbers

Answer 11

so for the sake of simplicity let's say let us use .... root of two prime like \(\bf \sqrt{17}\qquad \sqrt{23}\)

Answer 12

How would i graph that?

Answer 13

hmmmm well... those are just the solutions :)

to find the original equation, you'd just multiply them both

Answer 14

I'm not following what you mean by multiply them both. Multiply them by what? and how does that give me the equation?

Answer 15

\(\bf x=\sqrt{17}\implies (x-\sqrt{17})=0\\ \quad \\
x=\sqrt{23}\implies (x-\sqrt{23})=0\\ \quad \\ \quad \\

(x-\sqrt{23})(x-\sqrt{17})=\textit{original equation}\)

Answer 16

How would I graph that?

Answer 17

one can say that if you were to use the conjugates, will make it even simpler, say \(\bf x=\sqrt{17}\implies (x-\sqrt{17})=0\\ \quad \\
x=-\sqrt{17}\implies (x+\sqrt{17})=0\\ \quad \\ \quad \\

(x-\sqrt{23})(x+\sqrt{17})=\textit{original equation}
\)

Answer 18

once you FOIL it, you'd get an equation you can graph

Answer 19

woops darn typos I meant \(\bf x=\sqrt{17}\implies (x-\sqrt{17})=0\\ \quad \\
x=-\sqrt{17}\implies (x+\sqrt{17})=0\\ \quad \\ \quad \\

(x-\sqrt{17})(x+\sqrt{17})=\textit{original equation}\)

Answer 20

so (x-17) and x=17 and 23?

Answer 21

jdoe0001 isn't here anymore
x will equal sqroot (17)

Answer 22

so x=17 or x=\[\sqrt{17}\]

Answer 23

Why not try creating the original equation?
(x−sq root(17) (x+sq root(17)=original equation
can you multiply that ?

Answer 24

yeah (x-17)

Answer 25

No - you'd have to have an x² in there.

Answer 26

oh (\[x ^{2}\]-17)

Answer 27

jdoe0001 mentioned FOIL - do you know what that is?

Answer 28

yeah first outer inner last

Answer 29

So multiply this using FOIL
(x−sq root(17) (x+sq root(17)

Answer 30

x2-17

Answer 31

That's close but instead of the 17, what should it be?

Answer 32

No wait I was wrong - that is it !!!!!!!!!!

Answer 33

\[(x-\sqrt{17})(x+\sqrt{17})\]

Answer 34

That's how you'd factor the equation to get the 2 values of x.

Answer 35

i need two solutions though thats only 1

Answer 36

look carefully , there are 2 solutions there

Answer 37

\[\sqrt{17} \sqrt{-17}?\]

Answer 38

That is right those are the 2 solutions

Answer 39

okay i need 1 more problem. Equation with no real solutions and two imaginary

Answer 40

Okay I'll see if I can come up with something.

Answer 41

x² + 2x + 4 =0

The problem with using that is that both answers are equal

Answer 42

Need two imaginary ones.

Answer 43

BOTH of those answers are imaginary.
but the answers equal each other.
Here's a good equation:
2x² +5x +32
THe 2 answers are:
-1.25 plus 3.7997 i and
-1.25 minus 3.7997 i

Answer 44

thank you

Answer 45

u r welcome - I guess that's it right?

Answer 46

yep cya

Answer 47

okay c ya