Johnny the math 'tard needs help! :o

Johnny · Post by **Johnny** » 10/24/2005, 2:24 pm

Well normally I do not like asking for help with my studies but, I currently have no other alternative. You see, I've been attempting to do all the work on my own, save the help from Loru and Josh, but I find that I'm making no progress at all. As of now, I'm pretty much at a standstill, so I am asking for your help.

So, what I plan on doing is posting the problems that I am having and hoping some of you will be able to help me though. And please, don't spam this thread.

happening fish · Post by **happening fish** » 10/24/2005, 2:31 pm

Woo, math

Let's see 'em.

Johnny · Post by **Johnny** » 10/24/2005, 2:38 pm

I don't know how to solve quadratic equations.
My book here shows me what to do but I don't know how to do it. I don't understand how to work the question out.

x2 + 3x = 10

superrgirll · Post by **superrgirll** » 10/24/2005, 2:42 pm

i thought the x always went after the number :freak:

happening fish · Post by **happening fish** » 10/24/2005, 2:50 pm

it's been a while but i believe the first step is to make the equation equal to zero. so,

x^2 + 3x = 10
x^2 + 3x - 10 = 0

edit: geeze, there we go

Johnny · Post by **Johnny** » 10/24/2005, 2:51 pm

its x squared. :heythere:

happening fish · Post by **happening fish** » 10/24/2005, 2:55 pm

ok now you apply the quadratic formula, ax^2 + bx + c = 0

so:

we have

ax^2 + bx + c = 0
compared to
1x^2 + 3x - 10 = 0

therefore a = 1, b = 3, c = -10

questions so far?

Johnny · Post by **Johnny** » 10/24/2005, 3:02 pm

You islolated the 10 right?

happening fish · Post by **happening fish** » 10/24/2005, 3:06 pm

yes i moved it from the right side of the equation to the left side, which is why it became negative

superrgirll · Post by **superrgirll** » 10/24/2005, 3:09 pm

Chanandler Bong wrote:its x squared.

no wonder i was so confused

Johnny · Post by **Johnny** » 10/24/2005, 3:16 pm

happening fish wrote:ok now you apply the quadratic formula, ax^2 + bx + c = 0

so:

we have

ax^2 + bx + c = 0
compared to
1x^2 + 3x - 10 = 0

therefore a = 1, b = 3, c = -10

questions so far?

So, what does one have to do now?

happening fish · Post by **happening fish** » 10/24/2005, 3:25 pm

okay so you take those numbers and plug them in to the quadratic equation.

edit: frickin' site. your book should have it, but here's my best shot:
..................__________
x = -b +/- \/ b^2 - 4ac
......-------------------------
..................2a

so wherever there is an "a" put in 1, wherever there's "b" put in 3, and wherever there's "c" put in -10. and such.

happening fish · Post by **happening fish** » 10/24/2005, 3:30 pm

Good god... I'm an ARTIST.

happening fish · Post by **happening fish** » 10/24/2005, 3:35 pm

thus we have

..................______________
x = -3 +/- \/ 3^2 - 4(1)(-10)
......-----------------------------
..................2(1)

Post by **saman** » 10/24/2005, 4:42 pm

which gives you x = -5 and 2

there's another way of doing quadratic equations that i've always preferred to the quadratic formula.

1. you've got your equation, x^2 + 3x = 10.

2. move the 10 over so there's a zero at one end: x^2 + 3x - 10 = 0.

3. now, you think of two numbers which, when added together, give you +3 (the second number in step 2), and when multiplied give you -10 (the third number in step 2).

by doing simple trial and error, you can tell that the two numbers will be +5 and -2, because +5+(-2) = +5-2 = +3, and +5 x -2 = -10.

4. these two numbers, +5 and -2 can be put in a factored equation: (x + 5)(x - 2) = 0 (it doesn't matter which order)(if you do the math on this you will get your first equation: x^2 + 3x -10 = 0)

5. now, we know that (x + 5)(x - 2) will equal to 0, so x will be either -5 or +2. just use the two numbers you got before and flip their signs.

see, we need either (x + 5) to be zero, or (x - 2) to be zero, because if either one of those is zero, and if you multiply anything by zero, the answer is zero. if we use x = -5, then (-5 + 5)(-5 - 2) = (0)(-7) = 0.

the same way, if we use x = +2, (-2 + 5)(2 - 2) = (3)(0)= 0

6. so, x = -5 and x = +2

starseed_10 · Post by **starseed_10** » 10/24/2005, 7:32 pm

omg i have everyone!

starseed_10 · Post by **starseed_10** » 10/24/2005, 7:40 pm

by the way, i'd be helping if i had time :uh:

half jill · Post by **half jill** » 10/24/2005, 7:43 pm

i have a test on this tomorrow and thursday. :GASP:

don't ask why · Post by **don't ask why** » 10/24/2005, 7:45 pm

Chanandler Bong wrote:I don't know how to solve quadratic equations.
My book here shows me what to do but I don't know how to do it. I don't understand how to work the question out.

x2 + 3x = 10

i read that as 2x + 3x = 10. and then i was like ohhh, x^2.

happening fish · Post by **happening fish** » 10/24/2005, 11:51 pm

saman, that would be factoring :lol:

but he specifically asked for quaddy... and it's a good fallback if the factoring don't work out