# Thread: Factoring to find a solution to what x is

1. ## Factoring to find a solution to what x is

Find all solutions of the equation $x^2+3x+8=0$ and express them in the form a+bi: (I have no idea what 'i' stands for)

I have to find 2 sets of numbers in which the first set b < 0, then another set with b > 0. A and B are both real numbers.

I can't find an easy way to factor it, and when I put it into the quadratic formula it isn't possible because you cant squareroot -26.

2. Originally Posted by Lolcats
Find all solutions of the equation $x^2+3x+8=0$ and express them in the form a+bi: (I have no idea what 'i' stands for)

I have to find 2 sets of numbers in which the first set b < 0, then another set with b > 0. A and B are both real numbers.

I can't find an easy way to factor it, and when I put it into the quadratic formula it isn't possible because you cant squareroot -26.
(actually, it's $\sqrt {-23}$ that you have to deal with) well, that's the point of the $i$. you have complex roots here. $i = \sqrt{-1}$. so, $\sqrt {-23} = \sqrt{(-1)(23)} = \sqrt {-1} \sqrt {23}= i \sqrt {23}$

3. Originally Posted by Jhevon
(actually, it's $\sqrt {-23}$ that you have to deal with) well, that's the point of the $i$. you have complex roots here. $i = \sqrt{-1}$. so, $\sqrt {-23} = \sqrt{(-1)(23)} = \sqrt {-1} \sqrt {23}= i \sqrt {23}$
Ohay, so that means I know what my i value is now? So should i just multiply my i value by my b value in the quadratic formula to get the values?

4. Originally Posted by Lolcats
Ohay, so that means I know what my i value is now? So should i just multiply my i value by my b value in the quadratic formula to get the values?
uh...no. do the quadratic formula, simplify the negative square root as i showed you, and split the answer in two terms, one with $i$ in it, one without $i$ in it.

5. Originally Posted by Jhevon
uh...no. do the quadratic formula, simplify the negative square root as i showed you, and split the answer in two terms, one with $i$ in it, one without $i$ in it.

If one solution is complex, then the other solution is complex. This is a major theorem.

Both solutions will have $i$ in it if one does (the roots are complex conjugates)

6. Ahhh, alright I'll try that. Thank you.

7. One last thing... when their asking me to express it in a + bi... is that like in terms of the variables from the quadric formula? Or am I suppose to let a + bi = the two x's found, and then solve for a and b?

8. Originally Posted by Lolcats
One last thing... when their asking me to express it in a + bi... is that like in terms of the variables from the quadric formula? Or am I suppose to let a + bi = the two x's found, and then solve for a and b?
a+bi is the general form a complex number, i know it might be confusing, but has nothing to do with the a,b from the quadratic formula, it just means that when you run through the quadratic formula you will get an answer, part of it will have i and part of it wont

write it as (real number)+(real number)i

9. Originally Posted by emathinstruction
a+bi is the general form a complex number, i know it might be confusing, but has nothing to do with the a,b from the quadratic formula, it just means that when you run through the quadratic formula you will get an answer, part of it will have i and part of it wont

write it as (real number)+(real number)i
haha, oh, now i see where the confusion came from i was wondering why the OP said to multiply i by b. yes, the b in a + ib is not the same b as the one in the quadratic formula.

10. Uhh k..

First input the solution with b< 0 here:
the real number a equals_______ and the real number b equals_______
Then input the solution with b0 here:
the real number a equals_______ and the real number b equals _______

Thats the format I'm suppose to answer it in, I'm just really confused now how to get through this...

Sorry for dragging this out, I've never had a question like this or delt with 'complex conjugates'

11. Originally Posted by Lolcats
Uhh k..

First input the solution with b< 0 here:
the real number a equals_______ and the real number b equals_______
Then input the solution with b>0 here:
the real number a equals_______ and the real number b equals _______

Thats the format I'm suppose to answer it in, I'm just really confused now how to get through this...

Sorry for dragging this out, I've never had a question like this or delt with 'complex conjugates'
example: lets say you did all i said and you ended up with something like $\frac {1 \pm i \sqrt 2}2$

first step: $\frac 12 \pm i \frac { \sqrt 2}2$

so $a = \frac 12$, and $b = \pm \frac {\sqrt 2}2$. got it?