Question

5x^2-5=120 ? please help (5x to the second power minus 5 equals 120)

Answer 1

x = 5?

Answer 2

well i think u mean (5x)^2 as apposed to 5x^2 so then u get (5x)^2 =125 when u add 5 to both sides. so then square 5x giving u 25x^2 = 125. divide both sides and u get x = square root of 5

Answer 3

if u mean 5x^2 u get x = 5 as amistre said

Answer 4

\[5x^2 - 5 = 120 \]\[\implies 5x^2 = 120 + 5\]
\[\implies x^2 = 25\]
\[\implies x = \pm 5\]

Answer 5

is that all i do ? it says solve each equation by using square root .man i guess you are right .! i have more too ! please keep helping me

Answer 6

Well, write the next one and see how much you can do on your own.

Answer 7

i dont know what to do !! :( im trying though
6r^2+5v=10

Answer 8

Well what are you supposed to be solving for?

Answer 9

i am supposed to solve the equation with the quadratic formula

Answer 10

I don't understand though you have 2 variables there, v and r.

Answer 11

it must have been a misprint. we can just use 1 variable. v? so whats my first step?

Answer 12

Well, what is the form you need for the quadratic formula?

Answer 13

the general form?

Answer 14

yeah

Answer 15

ax2 + bx + c = 0

Answer 16

Right. So make the equation you have look like that.

Answer 17

(ax2+bx+c=0)

6x2=5=0?

Answer 18

no thats not it !! uhm
im sorry i am wasting your time.

Answer 19

Wait, what's the equation you started with?

Answer 20

(6v)^2+5v=10

Answer 21

Are there really parentheses there?

Answer 22

no , i just dont know how to square it on here

Answer 23

oh, ok. So if you subtract a 10 from both sides wont the left side equal 0?

Answer 24

so it would be something like ...

6v^2+5v-10=0 ?

Answer 25

It would be exactly like that. So what is a, b, and c?

Answer 26

a = 6v or 6 ?
b = 5
c= -10

Answer 27

Well, the standard form is ax^2 + bx + c, so the a would just be 6. It's the coefficient in front of the x^2 (or in this case v^2) term.

Answer 28

ooooo , ok i think im starting to get this a little more. so what do i need to do next in this equation ?

Answer 29

Well, given the new form we have here.. You just plug in your a, b and c to the quadratic equation to find what x is.

\[ax^2 + bx + c =0 \implies x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 30

=( im going to fail this algebra 2 class !!

Answer 31

No. You're doing very very well. Have you been taught the quadratic equation?

Answer 32

yea we have but i was really sick and my teacher has not helped me really .

Answer 33

Ok, well that is what it is. If you have something in the form
\[ax^2 + bx + c = 0\] you can find what x equals by plugging in a, b, and c into that equation.

Answer 34

My algebra2 teacher taught it to us using a song.

Answer 35

x= -5+ the square root 5^2-4(6)(-10)
-
_____________________________________
2(6)

Answer 36

lol sorry for the huge spaces. i tried though . and how does the song go ?

Answer 37

Yes.

Answer 38

It's to the tune of jingle bells.

Quadratic equations / Dashing through the snow
often make you cry. / in a one horse open sleigh

But now there is a way / or' the fields we go
to wave your tears bye bye. / laughing all the way

Don't need to factor now. / Bells on bob-tails ring
Don't need complete the square. / making spirits bright

Just one simple forumla this secret I will share.

Oh x equals / Jingle bells
minus b / Jingle bells
plus or minus square root of / Jingle all the way
b squared minus 4ac / oh what fun it is to ride
all divided by 2a! / in a one horse open sleigh.

Answer 39

omg !!! ok can i please use this song , i would love to write it down and use this, we have a test thursday and this could help me extremly considering the fact i love to sing !!!

Answer 40

In anycase, it helps to take the square root part first. The part in the square root is called the discriminant. Then you can take the square root of the discriminant and add and subtract it from the -b to give you your two values for x.

Answer 41

Yes of course you can use it ;p

Answer 42

you are a life saver ! &'d you are extremely smart at this lol i know this is besides the point but how old are you genius ?

Answer 43

Heh.. I'm 33.

I found this video of people singing it. Theres a lot of other videos of the quadratic forumula sung to many different songs, but this is the only one I could find for jingle-bells.

Anyway, hope it helps.

http://www.youtube.com/watch?v=CnJT1ojHT28&feature=related

Answer 44

lol it will ! are you on here all the time ? i will most likely start getting on here daily with this work. that is if i dont have to work .

Answer 45

I'm on pretty often yeah =)

Answer 46

Or you can always find me a gmail if I'm not online.

Answer 47

err at gmail rather.

Answer 48

well i really hope each time i get on i can find you...im new to this lol i just made me a membership once my question was on here. i dont know how to really work this sight

Answer 49

my email is selenadowns@yahoo.com
or i have a facebook lol just saying

Answer 50

i meant selenadowns73@yahoo.com

Answer 51

ok well feel free to drop me a line if you have questions, or you can post them here and someone will no doubt answer eventually =)

Answer 52

one more question before i log off &'d head to work what are the steps to quadratic systems ?
the next problem is y=-x^2-3x+10
y=x+5

Answer 53

Well. If y = x+5 and y = \(x^2 - 3x + 10\)
Then can't you set those two right hand parts equal to eachother?

Answer 54

huh ? sweetie you lost me lol set what &'d what equal to eachother ?

Answer 55

When working with a system, you're finding solutions that satisfy both equations. So the x from one equals the x from the other, and same for the y.

Answer 56

So if y = x+5 and y = \(x^2 - 3x + 10\)

then you have

\(x+5 = y = x^2 - 3x + 10\)

Answer 57

So you can get rid of the y.

Answer 58

and you're left with

\(x+5 = x^2 - 3x + 10\)

Answer 59

-x^2-3x+10=x+5

Answer 60

o :/ what am i doing wrong ?

Answer 61

You did it right. I forgot the negative sign on the \(x^2\) term

Answer 62

Then just solve the same way we did the last one. move everything over so you have a 0 on one side and ax^2 + bx + c on the other.

Answer 63

from what i had...
i subtract the -x-5 on both sides?

-x^2-4x+5=0
0=x^2+4x-5 ????????????

Answer 64

Yep. Either of those two will work (and give the same result)

Answer 65

(x+5)(x-1)
x=-5 x=1

y=-5+5 y=1+5

y+0 y=6

(-5,0) (1,6)

Answer 66

HOLD ON IM VERY MAD!!!! i just had a moment........that problem i just did were my notes that were already done . lol wow im stupid

Answer 67

It's still good practice!

Answer 68

i got all excited because i thought i comprehended

Answer 69

i really hope that in the future i marry someone who has mathmatic potential as yourself because if not i dont know what i will do lol

Answer 70

Heh.. you're doing really well actually!

Answer 71

thanks sweetie. well i am off to work. hopefully i get your help soon. email me asap . i get them to my phone . i will really need more help when i get off work

Answer 72

I already did.

Answer 73

i will check my phone then . well bye =) thank you SOOOOOOOOOOOOOOO MUCH

Answer 74

hello :)

Answer 75

lol, hi.

Answer 76

are you online?