# Thread: solving using algebra - is this correct?

1. ## solving using algebra - is this correct?

(2x)^2/(.86-x)(.72-x)^3 = 4.5 x 10^-5 x is negligible so the answer I got for x was .0018

can anybody show me how to arrive at the correct answer using a step by step explanation of the algebra. I multiplied .86 by (.72)^3 and multiplied this amount by 4.5 x 10^-5. Then I took the square root of this and 2x^2 and divided by 2 to get .0018. Can anyone show me where I went wrong? Any input would be greatly appreciated. thank you.

2. Originally Posted by confused20
(2x)^2/(.86-x)(.72-x)^3 = 4.5 x 10^-5 x is negligible so the answer I got for x was .0018

can anybody show me how to arrive at the correct answer using a step by step explanation of the algebra. I multiplied .86 by (.72)^3 and multiplied this amount by 4.5 x 10^-5. Then I took the square root of this and 2x^2 and divided by 2 to get .0018. Can anyone show me where I went wrong? Any input would be greatly appreciated. thank you.
$\displaystyle \frac{(2x)^2}{(.86-x)(.72-x)^3} = 4.5 \times 10^{-5} x$

when rearranged gives a quartic in $\displaystyle x$. What techniques do you have available for solving a quartic?

(It has two real roots and two complex roots, the real roots are close to $\displaystyle 3.8\times 10^{-6}$ and $\displaystyle 45.6$)

RonL