converting to ax2 + bx + c = 0

[sc44sc44]

Hi I need to know all the steps to put a equation like 9=(x)(0.625+x)/(1.125-x)(0.5-x) into a standard form like: ax2 + bx + c = 0 to solve as a quadratic equation. (for a science class)

I need a internet address or something where I could learn the steps or rules and order of the steps to understand it.I know the answer already I just dont know how to get from one equation to the other, so knowing the rules to do it would be great cause I got a couple of these equation converstion to do.

And is it possible to buy a calculator who do these equation convertion? thank you

 

[wjm11]

Hi I need to know all the steps to put a equation like 9=(x)(0.625+x)/(1.125-x)(0.5-x) into a standard form like: ax2 + bx + c = 0 to solve as a quadratic equation. (for a science class)

I need a internet address or something where I could learn the steps or rules and order of the steps to understand it.

Unfortunately, I can’t give you “steps or rules” that are applicable to every problem. It will depend on haw each problem is stated. All the steps are going to be just basic algebra, though. Let’s examine the problem at hand as an example.

9=(x)(0.625+x)/(1.125-x)(0.5-x)

We notice right away that there is a denominator with x terms in it. We’d like to get x out of the denominator and even get rid of the denominator entirely if possible. We can do this by multiplying both sides of the equation by the denominator:

(9)(1.125-x)(0.5-x) = [(x)(0.625+x)/(1.125-x)(0.5-x)] (1.125-x)(0.5-x)

So on the right side of the equation, the (1.125-x)(0.5-x) on top and bottom cancel out. Our equation simplifies to

(9)(1.125-x)(0.5-x) = (x)(0.625+x)

Next, distribute (multiply everything out):

9(.5626 – 1.125x - .5x + x^2) = .625x + x^2
5.0625 – 14.625x + 9x^2 = .625x + x^2

Now we want to get all our terms on one side of the equation and zero on the other, so subtract .625x and x^2 from both sides:

5.0625 – 14.625x + 9x^2 - .625x - x^2 = 0

Finally, simply (combine like terms) and rearrange:

8x^2 – 15.25x +5.0625 = 0

Regarding calculators: some calculators can solve the problem in its original form, but I don’t know of one that will convert a problem into the form you desire.

 

[Deleted member 4993]

Hi I need to know all the steps to put a equation like 9=(x)(0.625+x)/(1.125-x)(0.5-x) into a standard form like: ax2 + bx + c = 0 to solve as a quadratic equation. (for a science class)

I need a internet address or something where I could learn the steps or rules and order of the steps to understand it.I know the answer already I just dont know how to get from one equation to the other, so knowing the rules to do it would be great cause I got a couple of these equation converstion to do.

And is it possible to buy a calculator who do these equation convertion? thank you

if your function is:

\(\displaystyle x\cdot\frac{0.625+x}{1.125-x}\cdot(0.5-x) = 9\)

start by multiplying both sides by (1.125-x)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[sc44sc44]

Regarding calculators: some calculators can solve the problem in its original form, but I don’t know of one that will convert a problem into the form you desire.

So I could just enter 9=(x)(0.625+x)/(1.125-x)(0.5-x) and it would find x ? Probably referring to some TI (Texas Instrument) calculator if im right? ( Ive just check a couple of calculator online)


Thanks for the explanation of the problem I just don't know where the .5626 come from in this part 9(.5626 – 1.125x - .5x + x^2) = .625x + x^2



To Khan:

the Kc constante for SO2+NO2 =SO3 + NO equal 9.0 at 700 Celcius. In a 4 liter recipient we put 4,5mol of SO2, 2mol of NO2 and 2,5mol of NO.

so I need to find the final concentration when the solution is balance.

Formula: SO2 + NO2 = SO3 + NO
At Start : 1,125 + 0,5 = 0 + 0,625
Reaction: -x -x +x +x
---------------------------------------------------------------------
Balanced: 1,125-x 0,5-x x 0,625+x


the formula to find it is:
Kc = (SO3)(NO)/ (SO2)(NO2) so: 9,0= (x)(0,625+x) / (1,125-x)(0,5-x)


And im stuck there where I need to put this in this form ax2 + bx + c = 0 to solve it

Its just that I don't know in this kind of problem how to have my starting equation like 9=(x)(0.625+x)/(1.125-x)(0.5-x) into ax2 + bx + c = 0

For the rest they show how to do the quadratic thing to find the X value.

So I lack (or doesn't remember) the knowledge for this kind of part: 9=(x)(0.625+x)/(1.125-x)(0.5-x) into ax2 + bx + c = 0
So each time I get a problem like this I can complete it cause of this single part.

Having a calculator who could find the X value would be good event if it would not show how it does it, as long it helps me completing my problem.

 

[soroban]

Hello, sc44sc44!

I read it the "other way".

\(\displaystyle \text{Put into standard form: }\,ax^2 + bx + c \:=\:0\)

. . . . \(\displaystyle 9 \;=\;\frac{x(0.625 + x)}{(1.125 - x)(0.5-x)}\)

\(\displaystyle \text{First, I got rid of the decimals: }\; 9 \;=\; \frac{x\left(\frac{5}{8} + x\right)} {\left(\frac{9}{8} - x\right) \left(\frac{1}{2}-x)}\)

\(\displaystyle \text{then: }\;9\left(\tfrac{9}{8}-x\right)\left(\tfrac{1}{2}-x\right) \;=\;x\left(\tfrac{5}{8}+x\right)\)

. . . . \(\displaystyle 9\left(\tfrac{9}{16} - \tfrac{13}{8}x + x^2\right) \;=\;\tfrac{5}{8}x + x^2\)

. . . . . \(\displaystyle \tfrac{81}{16} - \tfrac{117}{8}x + 9x^2 \;=\;\tfrac{5}{8}x + x^2\)


\(\displaystyle \text{Multiply by 16: }\:10x + 16x^2 \;=\;81 - 234x + 144x^2\)


\(\displaystyle \text{Therefore: }\;128x^2 - 244x + 81 \;=\;0\)

 

[chrisr]

\(\displaystyle Or\ multiply\ by\ \frac{1000}{1000}\ and\ divide\ both\ sides\ by\ 2.\)

\(\displaystyle \frac{9}{2}=\frac{x(625+1000x)}{(1125-1000x)(1-2x)}\)

\(\displaystyle \frac{9}{2}=\frac{x(125+200x)}{(225-200x)(1-2x)}\)

\(\displaystyle \frac{9}{2}=\frac{x(25+40x)}{(45-40x)(1-2x)}\)

\(\displaystyle \frac{9}{2}=\frac{x(5+8x)}{(9-8x)(1-2x)}\)

\(\displaystyle continue\)

 

[Denis]

So I could just enter 9=(x)(0.625+x)/(1.125-x)(0.5-x) and it would find x ? Probably referring to some TI (Texas Instrument) calculator if im right? ( Ive just check a couple of calculator online)

Thanks for the explanation of the problem I just don't know where the .5626 come from in this part 9(.5626 – 1.125x - .5x + x^2) = .625x + x^2

1st part: FORGET about calculators doing YOUR work: BAD way to learn :shock:

2nd part: should be .5625 (a typo, nothing else!) and comes from multiplying: (1.125-x)(.5-x) ;
1.125 times .5 = .5625

 

[sc44sc44]

Thank you,

Yeah I know the calculator is a bad way to learn but im taking my advanced math class only after this science class, so beside this kind of math problem in this science class math are not really present or kinda basic. If it could help me pass my exam (if some equation like this are present) for now since I haven't attend my math class yet to really learn this it would be welcome.