1. ## Word Problem equality

The concentration of toxic chemical in a spring fed lake is given by the equation c(x)= 50x/x^2+3x+6, where c is given in grams per litre and x is the time in days. Determine when the concentration of the chemical is 6.16 g/L.

50/x^2+3x+6= 6.16

50/x^2+3x+6 (x^2+3x+6) (6.16) = 6.16(6.16) (x^2+3x+6)

50(6.16)= 1(x^2+3x+6)

308=x^2+3x+6

0=x^2+3x+6-308
0=x^2+3x-302

Then solve using quadratic formula???? Im not sure if this is right.

2. Originally Posted by Skoz
The concentration of toxic chemical in a spring fed lake is given by the equation c(x)= 50x/x^2+3x+6, where c is given in grams per litre and x is the time in days. Determine when the concentration of the chemical is 6.16 g/L.

50/x^2+3x+6= 6.16
Are we to assume that this is 50/(x^2+ 3x+ 6)= 6.16 and not (5/x^2)+ 3x+ 6= 6.16? Please use parentheses!

[/quote]50/x^2+3x+6 (x^2+3x+6) (6.16) = 6.16(6.16) (x^2+3x+6)[/quote]
Okay, you are multiplying both sides by x^2+ 3x+ 6 to get rid of the fraction but why are you multiplying both sides by 6.16?

50(6.16)= 1(x^2+3x+6)
No, (6.16)(6.16) is NOT 1! I suppose you meant to divide both sides by 6.16

308=x^2+3x+6

0=x^2+3x+6-308
0=x^2+3x-302

Then solve using quadratic formula???? Im not sure if this is right.
And since 50(6.16)= x^2+ 3x+ 6 is not correct, the rest is wrong.

If you divide both sides by 6.16 you get x^2+ 3x+ 6= 50/6.16= 8.117, approximately so your equation is x^2+ 3x+ 6- 8.117= x^2+ 3x- -2.117= 0 and you can use the quadratic equation to solve that.
Just multiply both sides of 50/(x^2+ 3x+ 6)= 6.16 by x^2+ 3x+ 6 to get
x^2+ 3x+ 6=

3. I have this exact same question but mine states

(50x)/(x^2+3x+6)= 6.16

How would I solve this?

4. Originally Posted by willman32
I have this exact same question but mine states

(50x)/(x^2+3x+6)= 6.16

How would I solve this?
let $k = 6.16$

$\frac{50x}{x^2+3x+6} = k$

$50x = kx^2 + 3kx + 6k$

$0 = kx^2 + (3k-50)x + 6k$

$x = \frac{(50-3k) \pm \sqrt{(3k-50)^2 - 24k^2}}{2k}$