Question

Need help on Quadratic equations please!

The equation is on the comments.

Answer 1

\[4x^{2}+9x=0\]
i know that the answer is:
\[x=\frac{ -9\pm \sqrt{81-4(4)(2)} }{ 8 }\]

Answer 2

Much easier to solve your problem would be to factor out an x,\[x(4x+9)=0\]now you have two solutions,\[x=0\]and\[4x+9=0\]\[4x=-9\]\[x=-\frac{9}{4}\]and you get

Answer 3

The equation is correct. But you put innocent "2" instead of "0" for c.

Answer 4

huh? the thing messed up, give me a sec

Answer 5

Well, then you can do the following:\[x_{1/2}=\frac{-9 \pm \sqrt{9^2-4*4*0}}{2*4}\]simplifying that you get,\[x_{1/2}=\frac{ -9 \pm 9 }{ 8 }\]Compute the rest

Answer 6

I have to use:
\[ax^{2}+bx+c=0\]
and i have to use:
\[x=\frac{ -b\pm \sqrt{b^{2}-4ac} }{ 2a }\]

Answer 7

Yes.. that is what I have used. Just make c=0, and you will get the same result as I have.

Answer 8

but c = 2... thats what the answer says.

Answer 9

Well then you have given us a wrong problem\[4x^2+9x=0\]does not have a c in it..

Answer 10

my bad... /facepalm

its:
\[4x^{2}+9x+2=0\]

Answer 11

@gezimbasha

Answer 12

Hehe, no worries. Now you have to use the formula you input\[x_{1/2}=\frac{-9 \pm \sqrt{9^2-4*4*2}}{2*4}\]compute the values inside the square root first\[x_{1/2}=\frac{-9 \pm \sqrt{81- 32}}{8}\]and complete calculation\[x_{1/2}=\frac{-9 \pm 7}{8}\] and you get,\[x_1= -2\]\[x_2=-\frac{1}{4}\]

Answer 13

where did you get \[x _{1/2} \]

Answer 14

That just means that there are two solutions we will get from the equation. Its just mathematical notation. I am used to writing it like that. Don't worry, you can write just x, it will be fine.

Answer 15

im only in high school. :/ :L

Answer 16

Yeah I figured. The quadratic equations are studied extensively in high school. You're doing good tho. Might want to reconsider your profile picture heh ;)

Answer 17

it says i don't hate math. open my profile and look at the white! everyone says that. lol do you not see the white scribble on the \[^{don't}\]
I hate math

Answer 18

Oh.. then I owe you an apology. XD.

Answer 19

Its okay. I need to make it bigger anyway. lol. Thanks tho.

Answer 20

You're welcome man. Anytime!

Answer 21

I gave you a medal. dont for get to give one back~ ;)

Answer 22

Hey, my bad, my first day over here.

Answer 23

what do you mean?

Answer 24

This is my first day in open study, so I'm still getting used to the medals and stuff.

Answer 25

how would i set up:
\[3x^{2}+11x+5=0\] and oh. i didn't know

Answer 26

Same principle, replace a=3, b=11, and c=5.

Answer 27

i mean in the fraction form