5x^25=120 ? please help (5x to the second power minus 5 equals 120)
 anonymous
5x^25=120 ? please help (5x to the second power minus 5 equals 120)
 Stacey Warren  Expert brainly.com
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 amistre64
x = 5?
 anonymous
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
 anonymous
if u mean 5x^2 u get x = 5 as amistre said
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 anonymous
\[5x^2  5 = 120 \]\[\implies 5x^2 = 120 + 5\]
\[\implies x^2 = 25\]
\[\implies x = \pm 5\]
 anonymous
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
 anonymous
Well, write the next one and see how much you can do on your own.
 anonymous
i dont know what to do !! :( im trying though
6r^2+5v=10
 anonymous
Well what are you supposed to be solving for?
 anonymous
i am supposed to solve the equation with the quadratic formula
 anonymous
I don't understand though you have 2 variables there, v and r.
 anonymous
it must have been a misprint. we can just use 1 variable. v? so whats my first step?
 anonymous
Well, what is the form you need for the quadratic formula?
 anonymous
the general form?
 anonymous
yeah
 anonymous
ax2 + bx + c = 0
 anonymous
Right. So make the equation you have look like that.
 anonymous
(ax2+bx+c=0)
6x2=5=0?
 anonymous
no thats not it !! uhm
im sorry i am wasting your time.
 anonymous
Wait, what's the equation you started with?
 anonymous
(6v)^2+5v=10
 anonymous
Are there really parentheses there?
 anonymous
no , i just dont know how to square it on here
 anonymous
oh, ok. So if you subtract a 10 from both sides wont the left side equal 0?
 anonymous
so it would be something like ...
6v^2+5v10=0 ?
 anonymous
It would be exactly like that. So what is a, b, and c?
 anonymous
a = 6v or 6 ?
b = 5
c= 10
 anonymous
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.
 anonymous
ooooo , ok i think im starting to get this a little more. so what do i need to do next in this equation ?
 anonymous
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.
\[ax^2 + bx + c =0 \implies x = \frac{b \pm \sqrt{b^24ac}}{2a}\]
 anonymous
=( im going to fail this algebra 2 class !!
 anonymous
No. You're doing very very well. Have you been taught the quadratic equation?
 anonymous
yea we have but i was really sick and my teacher has not helped me really .
 anonymous
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.
 anonymous
My algebra2 teacher taught it to us using a song.
 anonymous
x= 5+ the square root 5^24(6)(10)

_____________________________________
2(6)
 anonymous
lol sorry for the huge spaces. i tried though . and how does the song go ?
 anonymous
Yes.
 anonymous
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 bobtails 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.
 anonymous
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 !!!
 anonymous
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.
 anonymous
Yes of course you can use it ;p
 anonymous
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 ?
 anonymous
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 jinglebells.
Anyway, hope it helps.
http://www.youtube.com/watch?v=CnJT1ojHT28&feature=related
 anonymous
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 .
 anonymous
I'm on pretty often yeah =)
 anonymous
Or you can always find me a gmail if I'm not online.
 anonymous
err at gmail rather.
 anonymous
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
 anonymous
my email is selenadowns@yahoo.com
or i have a facebook lol just saying
 anonymous
i meant selenadowns73@yahoo.com
 anonymous
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 =)
 anonymous
one more question before i log off &'d head to work what are the steps to quadratic systems ?
the next problem is y=x^23x+10
y=x+5
 anonymous
Well. If y = x+5 and y = \(x^2  3x + 10\)
Then can't you set those two right hand parts equal to eachother?
 anonymous
huh ? sweetie you lost me lol set what &'d what equal to eachother ?
 anonymous
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.
 anonymous
So if y = x+5 and y = \(x^2  3x + 10\)
then you have
\(x+5 = y = x^2  3x + 10\)
 anonymous
So you can get rid of the y.
 anonymous
and you're left with
\(x+5 = x^2  3x + 10\)
 anonymous
x^23x+10=x+5
 anonymous
o :/ what am i doing wrong ?
 anonymous
You did it right. I forgot the negative sign on the \(x^2\) term
 anonymous
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.
 anonymous
from what i had...
i subtract the x5 on both sides?
x^24x+5=0
0=x^2+4x5 ????????????
 anonymous
Yep. Either of those two will work (and give the same result)
 anonymous
(x+5)(x1)
x=5 x=1
y=5+5 y=1+5
y+0 y=6
(5,0) (1,6)
 anonymous
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
 anonymous
It's still good practice!
 anonymous
i got all excited because i thought i comprehended
 anonymous
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
 anonymous
Heh.. you're doing really well actually!
 anonymous
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
 anonymous
I already did.
 anonymous
i will check my phone then . well bye =) thank you SOOOOOOOOOOOOOOO MUCH
 anonymous
hello :)
 anonymous
lol, hi.
 anonymous
are you online?
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