How to solve for #x# , #(3.6*10^(-4)) = ((x*x)/(0.010 - x))# ?

3 Answers
Apr 10, 2017

See below.

Explanation:

#(3.6*10^(-4)) = ((x*x)/(0.010 - x))->(3.6*10^(-4))(0.010 - x)=x^2#

or

#x^2+(3.6*10^(-4))x-0.010 xx (3.6*10^(-4))=0#

so

#x = (-(3.6*10^(-4))pm sqrt((3.6*10^(-4))^2+4 xx0.010 xx (3.6*10^(-4)) ))/2#

and then two solutions:

#x = -0.00208589# and #x = 0.00172589#

Apr 11, 2017

I get #0.17#

Explanation:

I show you my derivation to help you see if 1) your procedure was correct but you typed in wrong numbers on your calculator, or 2) your approach contained an error. To make a calculation error-free clarity of visual presentation is essential. Know that a large number of mistakes occur in the last step because your brain relaxes when it shouldn't.

#3.6 /(100*100) = x^2/(1/100-x) = 100 x^2/(1-100x)#
#3.6/100 = y^2/(1-y)# where #y=100 x#
#100 y^2 + 3.6 y -3.6=0#
#y = 1/200(-3.6 pm sqrt(3.6^2 + 4*100*3.6))#
#y = 17.2588562091#
#x = 0.172588562091#

Apr 11, 2017

Given equation is
#3.6cdot10^-4=( xcdotx)/(0.010-x)#
Multiplying both sides with #(0.010-x)# we get
#(3.6cdot10^-4)(0.010-x) = xcdotx#
Rearranging as a quadratic
#x^2+3.6cdot10^-4x-3.6cdot10^-4xx0.010=0#
#=>x^2+3.6cdot10^-4x-3.6cdot10^-6=0#

Discriminant #Delta=b^2-4ac=(3.6cdot10^-4)^2+4xx1xx3.6cdot10^-6#
#=0.0000145296#
The two roots are #(-b+-sqrtDelta)/(2a)#
For the given equation we have
#x=(-b+-sqrtDelta)/(2)#
#=>x=(-3.6cdot10^-4+-sqrt0.0000145296)/(2)#
#=>x=-1.8cdot10^-4+-sqrt0.0000036324#

We get two solutions as

#x=−0.00208589 and 0.00172589#, rounded to eight decimal places
The second one matches with the given answer.