Question

Please help. Medal/Fan/Testomonial

Of the following, which is the solution to 2x2 + 3x = –5?

the quantity negative three plus or minus i times the square root of thirty one over four

the quantity negative three plus or minus i times the square root of thirty one over two

the quantity three plus or minus i times the square root of thirty one over four

the quantity negative three plus or minus i times the square root of fifty over four

Answer 1

\[-3 \pm i \sqrt{31} all \over 4\] is the first option

Answer 2

have you considered the quadratic formula?

Answer 3

whats that?

Answer 4

I know the distance formula

Answer 5

its the formulaic way to determine the solutions to a quadratic equations - its the shortcut of completing the square

Answer 6

can you help me?
Im really stuck

Answer 7

what methods does your material cover?

Answer 8

distance formula

Answer 9

I know that the answer is not B

Answer 10

as far as i can recollect, the distance formula is not useful in determining the solutions to a quadratic equation.

Answer 11

then what do I do? thats all that they taught

Answer 12

given a quadratic equation of the form:\[ax^2 + bx + c = 0\]
we can utilize the quadratic formula:\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 13

this formula should have been in your course work, or the method known as completing the square.

Answer 14

oh, the quadratic formula is the distance formula I was talking about. Sorry, I got them confused

Answer 15

distance formula:
\[d=\sqrt{x^2 + y^2}\]
yeah, not quite the same thing :)

Answer 16

oops, well then thats my fault. Im sorry. Can you help me with the quadratic formula since that is what I meant all along?

Answer 17

well, can you rewrite your problem in: ax^2 + bx + c = 0 format?

Answer 18

2x^2 + 3x +5 = 0

Answer 19

where 2x^2 = a b = 3x and c= 5

Answer 20

a,b,c are the coefficients, not the terms themselves.

[2]x^2 + [3]x +[5] = 0
a b c

Answer 21

right...okay, so what now?

Answer 22

plug and play of course ...

everywhere in the formula that you see an a at .. replace it with a 2

everywhere in the formula that you see an b at .. replace it with a 3

everywhere in the formula that you see an c at .. replace it with a 5

Answer 23

\[-3 \pm \sqrt 3^2 - 4(2)(5) \over 4\]

Answer 24

which is \[-3 \pm \sqrt -9 - 40 \over 4\]

Answer 25

I mean regular 9 not -9

Answer 26

we know its not B (assuming you tried that first and it said it was wrong)

and we need a -3 starting off, that narrows it to A or D

so the real question is: is the difference between 9 and 40 bigger or smaller than 40?

Answer 27

smaller

Answer 28

so the answer is A?

Answer 29

A is what we get to yes

\[\frac{-3\pm\sqrt{9-40}}{4}\]

Answer 30

Thank you so much! You were a great help! I gave you a medal and a fan

Answer 31

youre welcome

Answer 32

correct. since 9-40 = -31 and we have a negative inside the radical that negative becomes i so we have
\[\frac{-3\pm\sqrt{31}i}{4}\]

Answer 33

thank you too UsukiDoll...

Answer 34

:) you're welcome.

Answer 35

I have another small question

Answer 36

If something has a vertex of (-4,-1) would the formula be (x + 4)2 − 1

Answer 37

almost

Answer 38

what if that is all the information I have. If they only give me the vertex is that correct? I know that there is supposed to be an 'a' in front of the parenthesis but that is not an option

Answer 39

(x+4)^2 - 1 does have a vertex at (-4,-1)

but so does:

-3(x+4)^2 - 1 have a vertex at (-4,-1)

and:

12(x+4)^2 - 1 have a vertex at (-4,-1)

Answer 40

y = a(x+4)^2 - 1 have a vertex at (-4,-1) for any real value of 'a'

Answer 41

except maybe a=0 ... but eh

Answer 42

the first question was right and the one we did just now is right too

Answer 43

thank you both so much!

Answer 44

good luck