How do you solve #1.7xx10^(-1)=(x^2)/(0.60-x)# for #x#?

2 Answers
Aug 20, 2017

Answer:

See below.

Explanation:

You have:

#1.7xx10^-1=x^2/(0.60-x)#

  1. Multiply both sides by the denominator of the right:

#(0.60-x)(1.7xx10^-1)=(x^2/cancelcolor(blue)((0.60-x)))cancelcolor(blue)((0.60-x))#

  1. Simplify:

#(1.7xx10^-1*0.60)-(1.7xx10^-1)x=x^2#

#=>0.102-(1.7xx10^-1)x=x^2#

Now set the left side equal to zero by adding/subtracting those terms from both sides.

#color(green)(0.102)color(blue)(-(1.7xx10^-1)x)color(blue)(+(1.7xx10^-1)x)color(green)(-0.102)=x^2+(1.7xx10^-1)x-0.102#

#=>0=x^2+(1.7xx10^-1)x-0.102#

You can now solve for #x# using the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

For equations of the form #ax+by+c=0#.

#x=(-(1.7xx10^-1)+-sqrt((1.7xx10^-1)^2-4(1)(-0.102)))/(2*1)#

#=>x=(-(1.7xx10^-1)+-0.8130)/2#

#=>x=0.245# OR #x=-0.415#

Which of these answers you use will depend on which constant it is that you are calculating. For example, if you were calculating the solubility constant, you would choose the positive answer.

Aug 20, 2017

Answer:

#x=0.245 #

or

#x = -0.415#

Explanation:

#1.7 xx 10^-1 = x^2/(0.60-x)#

#1.7 xx 10^-1# can be written as #1.7/10 or 0.17#

I would choose #0.17# to avoid involving a fraction.

#0.17 = x^2/(0.60-x)" "larr# cross multiply

#0.17(0.60 -x)=x^2#

#0.102-0.17x= x^2" "larr# make a quadratic equal to #0#

#x^2 +0.17x -0.102 =0" "larr# does not factorise!

Solve the equation using the quadratic formula;

#x = (-b+-sqrt(b^2-4ac))/(2a)" "a =1, b=0.17,c =-0.102#

#x = (-0.17+-sqrt(0.17^2-4(1)(-0.102)))/(2(1))#

#x = (-0.17 +-sqrt(0.0289+0.408))/2#

#x = (-0.17 +sqrt(0.4369))/2 =0.245 #

or
#x =(-0.17 -sqrt(0.4369))/2 = -0.415#