Question

I need help with these two problems 4x^2+15x+10=0 and 2x^2+31=-15x

Answer 1

dp you need to solve them...?

Answer 2

Yeah. I have NO clue on how to do it. :/

Answer 3

ok... they can't be factored so you'll need the general quadratic formula

for an equation

\[ax^2 + bx + c = 0\]

the values of x thank make the equation true can be found using

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

in your 1st equation

a = 4 , b = 15 and c = 10

substitute and then evaluate

then 2nd equation needs to be rewritten

\[2x^2 + 15x +31 = 0\]

again you'll need the general quadratic formula

but this time

a = 2, b = 15 and c = 31

this will have complex roots

hope this helps...

Answer 4

Where do you go from, \[x=-15\pm \sqrt{65} \]
all over 8?

Answer 5

you solve for + and - both will be valid

Answer 6

I don't know how to do that lol...

Answer 7

make 2 equations

\[x=-15+\sqrt{65}\]

and

\[x=-15-\sqrt{65}\]

Answer 8

and just solve it?

Answer 9

well if you need answer...yes =)

Answer 10

what do I do about the 8.
When I was writing the equation it wouldn't let me put it all over 8. but its -15 + or - sqrt of 65 all divided by 8

Answer 11

ah yes divide by 8 as well as its part of the equation

Answer 12

yep....the 1st one is

\[x = \frac{-15 \pm \sqrt{65}}{8}\]

Answer 13

ok! and I just do the same problem (-15+ sqrt of 65 divided by 8) and thats my answer?

Answer 14

well perhaps you need it written as

\[x = (-15 + \sqrt{65})/8.....and.... x = (-15-\sqrt{65})/8\]

Answer 15

Yes! Idk why but that made so much more sense haha

Answer 16

I'm not sure if you need to provide and exact value... which is what has been posted or an decimal approximation... which needs to to actually calculate the value.

Answer 17

Yeah it was a long decimal.

Answer 18

ok... then you need to calculate the 2 values for x...

Answer 19

the second one I got -15 + and - the sq rt of -23 all over 4

Answer 20

thats right... but you will need to use i^2 = -1 to simplify the -23...

this is a complex number solution to the problem

and can be written

\[x = \frac{-15 \pm \sqrt{23 \times -1}}{4} = \frac{-15\pm \sqrt{23 \times i^2}}{4} = \frac{-15\pm i \sqrt{23}}{4}\]

hope it helps

Answer 21

all of those equations kind of ran together but I think I get the idea.

Answer 22

Thank you very much!

Answer 23

well to simplify


let

\[i^2 = -1\]

and then substitute it

\[-23 = 23 \times i^2\]

Answer 24

i dont think they have covered complex numbers yet - usually at that stage it is said that equation has no answer.

Answer 25

yeah we haven't. idk im not good at math at all and I missed the notes over this! lol

Answer 26

well you can use the discriminant... if you know of that

and say

\[b^2 - 4ac = -23\]

so the equation has no real roots

Answer 27

ok! thanks!

Answer 28

good luck