Question

What are the solutions (real or imaginary ) to -5 = 2x^2 -2x - 3?

Answer 1

use the pq thing and synthetic division. i can't find it

Answer 2

uhh i never learned that.. keep in mind i am only a junior lol haven't taken calc yet

Answer 3

This is algebra II.

Answer 4

no

Answer 5

Hold on.

Answer 6

my teacher said do the pq to find the possible roots then try them all w synthetic division

Answer 7

BUT NONE OF THEM FING WORK AND I CANT DO THIS I HAVE A FINAL TOMORROW I AM GOING TO CRY THEY DONT FING WORK

Answer 8

Use the discriminant.
b^2 - 4ac
If the answer is a perfect square, then they're integers.
If the answer is a positive number that isn't a perfect square, then they're irrational roots.
If the answer is 0, then it's a double root meaning that it crosses the x-axis only once.
If the answer is any negative number, then the roots are imaginary.
a = 2
b = -2
c = 2
If you haven't learned the discriminant, then use the rational root theorem. All the possible roots are: ±1, ±2, ±1/2. Plug it into synthetic division. If none of them work, then the roots are imaginary.

Answer 9

Did you make sure to bring the -5 to the other side first?

Answer 10

i really don't understand what your explanation is saying

Answer 11

and yes

Answer 12

And you sure you tried ±1, ±2, ±1/2?

Answer 13

ya

Answer 14

Then your roots are imaginary buddy. :) THat's all there is to it.

Answer 15

all of them?

Answer 16

Yup :)

Answer 17

i thought u have to have ate last one real one

Answer 18

They're both imaginary.

Answer 19

how do i find em

Answer 20

Plug them into the quadratic formula. You have to get it to a quadratic (degree of 2) to use the quadratic formula. Therefore, the degree has to be 2 to use it. However, since you already have it as 2, you can just use it :)

Answer 21

wow i forgot about the quadratic formula. lol please don't leave i need more help if u don't mind

Answer 22

Alright. How much more?

Answer 23

not much. u don't have to help if u don't want its ok

Answer 24

No. It's fine, was just wondering how much more.

Answer 25

ur right they're both imaginary :) and not that many

Answer 26

so if i have a polynomial thats not degree 2 so cant use quad formula....how do i found the real or imaginary solutions

Answer 27

and how' do i know how many are real and how many are imaginary

Answer 28

i know i have to do the pq thing and test them w synthetic...then get to a quadratic formula to find the rest...but sometimes the pq thing doesnt work

Answer 29

for example take x^4 + 4x^2 = -4

Answer 30

You have to use synthetic division. When you find a root that works, and plug it into synthetic division, you're quotient is a depressed equation.

Answer 31

theres no other way besides synthetic? it takes forever and it doesn't even work sometimes

Answer 32

Have you learned Descartes Rule of Signs?

Answer 33

no lol please don't confuse me i have a final tomorrow. i think I'm gonna pull an allnighter :(

Answer 34

Alright. Unfortunately, that's the only way... Let me show you with your example.

Answer 35

Actually, your example can't be factored that way...
Your example is a special case...
x^4 + 4x^2 = -4
x^4 + 4x^2 + 4
(x^2 + 2)(x^2 + 2)
From there, there are only imaginary roots. The problems that are on the test won't be undoable like this one. They're gonna be doable using the rational root theorem and synthetic division.

Answer 36

so i will be able to do them? phew. because this one made me cry lol

Answer 37

Let's try:
x^4 + 2x^3 – 7x^2 – 8x + 12

Answer 38

lol no way these things take forever..

Answer 39

how about we change the 12 to a 1 :)

Answer 40

The rational roots possible are ±1, ±2, ±3, ±4, ±6, ±12

Answer 41

If you do that, you can't do it then :)

Answer 42

its ok i know how to do it lol i just get frustrated easily

Answer 43

Here's a rule. Start at the beginning. Start with 1, -1, 2, -2, in that kind of order.

Answer 44

FOr example, I know for a fact that 1 will work in this one.

Answer 45

ok

Answer 46

Here's another tip. just plug in the possible roots in for x. If it equals 0, then it works and you know to plug it into synthetic division. You don't have to do it every single time.

Answer 47

oh wow nice tip!

Answer 48

For example:
x^4 + 2x^3 – 7x^2 – 8x + 12
(1)^4 + 2(1)^3 – 7(1)^2 – 8(1) + 12
1 + 2 – 7 – 8 + 12 = 0
Since it's 0, plug it into synthetic division.

Answer 49

to make it easier

Answer 50

The synthetic division will work if it equals 0. Yes it does. It's a theorem. Maybe they didn't teach it for some reason?

Answer 51

it says consider the following function f(x) = 2x^3 - 5x^2 +18x -45 ........ it says find all the roots real and imaginary. use polynomial long division and the quadratic formula. i understand how to do it...except what do we divide it by for long division?!?!?

Answer 52

If it's long division and you plugged in the roots, you just set it equal to 0 again.
This is not an actual root, but for instance, if 2 works, then do this.
x = 2
x - 2 = 0
Divide by x - 2 if you have to use polynomial long division.

Answer 53

OH!!!! so first synthetic divide. and take that and subratct it from x!!!!!!!! you are so smart dang

Answer 54

Yeah. Sorry, but I've gotta go...

Answer 55

no problem. thank you so muchhhh girl

Answer 56

lol. I'm a guy. Anyway, bye :)