# Math Help - Simple Equation Help

1. ## Simple Equation Help

Hey All,

I am looking through an engineering book and it has assumed the user can solve this equation and given the answer for C, but I cant figure out how to solve it myself. Can anyone help??

$4c2+1820c+17400 = 238900$

The 2 in 4c2 means squared.

The books answer is C = 99.8

Hope you can help!

Thanks

2. $c_1 \approx 99,8$

$c_2 \approx -554.8$

u can use 2 ways ... quadratic formula .... or completing the square ...

3. Thanks yeKciM, I will look up quadratic formula... although im not sure how you have solved both of the c's? I am just trying to make C be on its own.

Thanks!

4. Use "^" to get powers in LaTex.
Although it is not necessary, I would recommend dividing each number by 4 to simplify the equation: $c^2+ 455c+ 4350= 59725$. Now, since the "standard" form for a quadratic equation is " $ax^2+ bx+ c= 0$, subtract 59725 from each side to get $c^2+ 455c- 55375= 0$

There are several ways to solve quadratic equations but most general is the "quadratic formula":
solutions to $ax^2+ bx+ c= 0$ are given by $\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. Note the " $\pm$". There are two solutions to this equation- although there may be other reasons why a negative solution is not valid.

5. I understand the first bit where we get rid of 59725 and make it look a little simpler, but then that equations just kind of blows me away. Is that the answer or is that what I need to do to get the answer. Do I change a and b to 455 and -553775? and C is still C?

Thanks!

6. $a=1 , b= 455 , c=-55375$

ur $C$is $x$ in $ax^2+bx+c=0$

here u have commands for LaTex... so u can write it little better
LaTeX:Commands - AoPSWiki

7. Woohoo! 99.8.

Thanks for helping me understand guys, that's awesome!

8. Ok, I have just tried it on another one but I don't think my answer is right. Could someone check it please?

$4c^2+1490c+9310=88889$
$c^2+1490c+9310=2222.25$
$c^2+1490c+7087.75=0$
$-1490+\sqrt{1490^2-4x1x7087.75}$
$/ 2X1$
$= -4.77$

9. heheheheh i guess that x-es are *

solution for that one .... is

$c_1\approx47.38 , c_2\approx-419.88$

$4c^2+1490c+9310=88889$

$4c^2+1490-79579=0$

so u have that $a=4, b=1490, c=-79579$

u put it in...

$C_{1/2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

10. Qudratic is:-

$c^2 + 1490c + 7087.75 = 0$

11. I see, so I need the A for it to work. But in the last example we used 1 as A. Is that right? Why does that not work in this instance. 47.38 sounds more like it!

I see we now also have:-

$C X 0.5$

Is this how you removed the C from 1490?

12. Originally Posted by Gazeranco
Ok, I have just tried it on another one but I don't think my answer is right. Could someone check it please?

$4c^2+1490c+9310=88889$
$c^2+1490c+9310=2222.25$
$c^2+1490c+7087.75=0$
$-1490+\sqrt{1490^2-4x1x7087.75}$
$/ 2X1$
$= -4.77$
dude... there's no need for u to divide with 4... but when u do that u divide all .... meaning that u can't from
$4c^2+1490c+9310=88889$ divide with 4 get what u got
$4c^2+1490c+9310=2222.25$ is wrooooong

if u are gonna divide it u divide all so u get
$c^2+372.5c+2327.5=22222.25$

P.S. basic
equation would not be changed if u divide it or multiply it with number different from zero... (but all elements of equation)

13. lol I see... Doh!

Still not sure where the half came from in your solution, could explain that part please?

14. Originally Posted by Gazeranco
I see, so I need the A for it to work. But in the last example we used 1 as A. Is that right? Why does that not work in this instance. 47.38 sounds more like it!

I see we now also have:-

$C X 0.5$

Is this how you removed the C from 1490?

wait...
let's go easy
when u got any quadratic equation... any u have
$ax^2+bx+c=0$ now in ur equations that C is actually x or unknown ? are we ok till here ?

now when u look quadratic formula to solve quadratic equation like this

$x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

u will have solution of quadratic equation .... but these a,b and c that u put in formula are : a is number that is with $x^2$, when it is let's say [x^2] means that a=1 ... means it's $1*x^2$.... b is number that stands with just x , and c is number that have no x-es

15. Originally Posted by Gazeranco
lol I see... Doh!

Still not sure where the half came from in your solution, could explain that part please?

lol that's not C*0,5 it's C one two ... meaning there are two solutions .... witch u get from $\pm$ if there exist sqrt... if there is not sqrt meaning its 0 u have only one solution

like this

$C_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$

$C_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Page 1 of 2 12 Last