Question

What is the solution for 2x^2-3x+2=0
Help me figure out how to solve ones like it too?

Answer 1

well you will need the general quadratic formula as the solutions are complex numbers.

Answer 2

Okay, I will look up that formula?

Answer 3

Quadratic formula:
Where you have this quadratic: \(\large ax^2+bx+c=0\)
\(\large x=\Large\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 4

Okay, I understand that much. But as my b is already negative, do I leave it that way? Or make it -(-3) thus causing a positive?

Answer 5

If you have
\(\large 2x^2-3x+2=0\)
What does a, b, and c equal?

Answer 6

the second way. make it postive

Answer 7

Okay. so x= 3 +- the sqrt of -3^2-4(2)(2)/4

Answer 8

so when you plug in you get
\(\large x=\Large\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(2)}}{2(2)}\)

Answer 9

Okay, cool. I haven't screwed up yet.

Answer 10

lol...just simplify

Answer 11

So 3+- sqrt of 9-16/4?

Answer 12

correct

Answer 13

\(\large x=\Large\frac{3\pm\sqrt{9-16}}{4}\)

Answer 14

I'll just solve it from here and you tell me if I've done it correctly. 9-16=-7 so 3+- sqrt of -7/4?

Answer 15

correct

Answer 16

that's it. unless you need to submit your answer to the nearest tenth or something like that

Answer 17

oh, wait. not it. sorry

Answer 18

you have to take out the imaginary number

Answer 19

\(\large x=\Large\frac{3\pm\sqrt{-7}}{4}\)
do you know about \(\large i\) ?

Answer 20

Well, it's either 3+-i sqrt 7/4 OR 3+-5i/4

Answer 21

I do know that i= sqrt of negative one.

Answer 22

Which is exactly what I was going to ask you lol, how do I remove the imaginary number?

Answer 23

right

Answer 24

\(\large x=\Large\frac{3\pm\sqrt{7\times-1}}{4}\rightarrow\frac{3\pm\ i\sqrt{7}}{4}\rightarrow\)

Answer 25

ignore the last arrow on teh end

Answer 26

Okay, so you turn the negative into an i?

Answer 27

Ohhh! You multiply it out! Duhh! Got it. Thank you so much!

Answer 28

:)

Answer 29

anytime, my friend \(\large\color{blue}{:)}\)