Algebra - did i do this right?: X/(.20-X)^2 = 4.9 x 10^-4

[confused20]

I am having difficulty with the following problem.

X/(.20-X)^2 = 4.9 x 10^-4

X= 4.9 x 10^-4(.04 - .40X + X^2)

X = 2.0 x 10^-5 - 2.0 x 10^-4 + 4.9 x 10^-4X^2

Is this correct? Where did i go wrong?

 

[stapel]

X/(.20-X)^2 = 4.9 x 10^-4

I will guess that the instructions were to "solve", which means that you need to get the variable by itself at some point. Since you have ended up with "x = (something with 'x' in it)", this cannot be the solution. (Has your class not covered solving quadratics yet?)

Try converting to "nicer" numbers:

. . . . .x / (0.2 - x)[sup:j2c6iwng]2[/sup:j2c6iwng] = 0.00049

. . . . .x = 0.00049(0.04 - 0.4x + x[sup:j2c6iwng]2[/sup:j2c6iwng])

. . . . .x = 0.0000196 - 0.000196x + x[sup:j2c6iwng]2[/sup:j2c6iwng]

. . . . .0 = x[sup:j2c6iwng]2[/sup:j2c6iwng] - 1.000196x + 0.0000196

Now apply the Quadratic Formula, and you're done!

Eliz.

 

[soroban]

Hello, confused20!

Where did this problem come from? . Who assigned it?

I would eliminate the decimals,
. . bye it ends up with an ugly equation anyuway.

\(\displaystyle \frac{x}{(0.2 -x)^2} \;=\; 4.9\cdot10^{-4}\)

\(\displaystyle \text{Clear the denominator: }\;x \;=\;4.9\cdot10^{-4}(0.2-x)^2\)

\(\displaystyle \text{Multiply by }10^5\!:\;\;10^5x \;=\;49(0.04 - 0.4x + x^2)\)

\(\displaystyle \text{Multiply by }100\!:\;\;10^7x \;=\;49(4 - 40x + 100x^2)\)

. . . . . . . . . . . . . . . . . .\(\displaystyle 10^7x \;=\;196 - 1960x + 4900x^2\)


\(\displaystyle \text{And we have: }\;\;4900x^2 - 10,001,960x + 196 \;=\;0\)


Good luck!