/sci/ - Science & Math

Roots are -7 and -(1/8)

If you want more info, ill lurk on this baby for a few moments.

yea i got -8 as a root, and -1/8, however on my sheet (which is posted as correct) shows the second root as -3/8

I dont know if this step is wrong , but for some reason it seems to be right, let me explain.

::
> 8x(x+7) - now taking the first term out for the second bracket 3(x+2)

to get (x+7) (8x+3)
8x+3=0
8x=-3
=-3/8

Similarily, in this question : 15a^2-3a=3-7a

>15a^2 +9a -5a -3
>(15a^2 +9a) (-5a-3)
>3a(5a+3) -5(a-3)

roots = 5a+3 = 0 , -3/5
3a-5=0
3/5 , these values are CORRECT in both answers, so the answers cant be wrong. I think im doing something wrong in that method, and you made the same mistake, unless im wrong? englighten me

>>3046387
Why don't you just use the quadratic formula for bastards like that?

If the equation is in the form

<span class="math"> \displaystyle{ Ax^2 + Bx + C } [/spoiler]

your zeros will be

<span class="math"> \displaystyle{ \frac{ -B \pm \sqrt{B^2 - 4AC}}{2A} } [/spoiler]

ok here's what you do....get a graphing calculator go to your graph in the first y= put 8x^2+21 in the second one put -59x then zoom out and trace the intersections.....and thats algebra 2 through precal in a nutshell

>>3046399

Me again im stupid as hell. I thought that your roots would be right. However i just did it myself and the roots are 7 and (3/8)

mfw i actually thought that i could trust your mathematical skills when you are asking for help on algebra...

(to the tune of pop goes the weasel)
X equals negative b
plus or minus the square root
of b squared minus four a c
all over 2 a!

alternatively
8x^2 + 21 = -59x
8x^2 + 59x + 21 = 0
(8x+3)(x+21) = 0
bam

>>3046442
whoops sorry that last part should say (x+7) not (x+21)

>>3046442

>(8x+3)(x+21)

Foil that and tell me if you get
>8x^2 + 59x + 21 = 0

>mfw people who fail at basic algebra try to help other people who are struggling

im telling you just graph intersections in the calculator...you may think im trolling but its your loss so fuck this im out

>>3046460
>can't read my typo-correcting post that came three minutes before his own, buttfucker response

>>3046469

mfw even though you did, you still made a mistake that a dolpin with downsyndrome would even make.

>>3046492

Im with this guy.

>>3046422
That formula is god mode, just learn it. It'll get you through high school and then some.

>>3046522
X EQUALS NEGATIVE B, PLUS OR MINUS SQUARE ROOT OF
B SQUARED MINUS 4 A C
ALL OVER TWO A

Dat song

>>3046492

OH U SO MAD AS FUCK CUZZ

>>3046551

DAT ALREADY POSTED ITT SONG

I'm sure the sticky has stuff on how to do this crap.

\displaystyle{ 8x^2 + 59x +21 = 0 }
\displaystyle{ x^2 +\frac {59x}{8} + \frac{21}{8} = 0 }
\displaystyle{ (x+\frac{59}{16})^2 - \frac{2809}{256} = 0 }
\displaystyle{ x = -frac{59}{16} \pm \frac{53}{16} }

testing

I'm 77
>>3047177
>>3046422
how did you do that?

>>3046422
>>3046422
>>3046422
>>3046422

HOW YOU DO THIS?!?!

>>3046422
>>3046422
>>3046422
>>3046422
>>3046422
>>3046422
>>3046422
how did you do the formula

yo I got time, I got nothing better to do, my room mate (girl) is the only one here and wont sleep with me so I got all day I'll just keep posting.

http://wwwdotmathsisfundotcom/algebra/quadratic-equation-derivation.html

>>3046422
how did you do this formula shit... I'm for serious.

>>3047211

What do you mean how do you do the formula?

He's just some dumbass who has never learned it.

<span class="math">\displaystyle a = 8, b = 59, c = 21[/spoiler]

<span class="math">\displaystyle x = \frac{-59 \pm \sqrt{59^2 - 4(8)(21)}}{2*8}[/spoiler]

<span class="math">\displaystyle x = -\frac{3}{8}[/spoiler]

<span class="math">\displaystyle x = -7[/spoiler]

><span class="math">\sqrt{x^2+1}[/spoiler]
<span class="math">\sqrt{x^2+1}[/spoiler]

I think he's asking how to write in the special math script.

Then he can learn to stop talking like a nigger and ask just that.