(4.44e-2 - x) (8.06e-2 - x) / (0.331 + 2x)^2 = 1.80e-2

Question

I need to rearrange this equation so that I can get a,b, and c, for the quadratic equation but I am not sure how to do this. Thank you for the help.

Math(Please answer, thank you) - bobpursley, Sunday, March 25, 2012 at 7:43am
multpily both sides by (.331+2x)^2

now multiply out the left side, getting four terms. On the right, mupltiply all that square term out

Now, collect terms. Put them all on one side, you will have the quadratic form.

It is mostly algebra.

So I have to multiply (4.44e-2 -x )(8.06e-2 - x) by (0.331 + 2x)^2 or am I suppose to multiply 1.80e-2 by it?

Answer 1

(4.44e-2 - x) (8.06e-2 - x) / (0.331 + 2x)^2 = 1.80e-2
Transpose the denominator (in bold) to the right hand side:
(4.44e-2 - x) (8.06e-2 - x) = (0.331 + 2x)^2 * 1.80e-2
Expand both sides and you will get the required quadratic equation.

Answer 2

(4.44e-2 - x) (8.06e-2 - x) / (0.331 + 2x)^2 = 1.80e-2
as bobpursley suggested, multiply both sides by
(0.331 + 2x)^2 , which will cancel that expression from the denominator on the left side

35.864e^-4 - 4.44e^-2 x - 8.06e^-2 x + x^2 = 1.80e^-2 * (.109561 + 1.324x + 4x^2)

Use you calculator to expand the right side.
That will give you a square term , an x term, and a constant
Bring all the terms to the left side
now factor the x^2 terms, the x terms and write the constants down.
you will have something looking like
x^2(.......) + x(.......) + ....... = 0
a = first bracket
b= 2nd bracket
c = 3rd bracket

I will leave the messy button-pushing up to you
(remember that e^-2 and e^-4 are just constants)

Answer 3

From the context, there is a chance that e-2 is meant to be scientific notation for 10^(-2) and so on.