Walk me through this please

[nacho]

someone a few days ago, asked me this question.. I did put it in a calc, quite complex, but could someone show me how to algebraically solve it?
solve for b in:

 a/b + b/c = d

answer: http://www.wolframalpha.com/input/?i=solve+b%3A+a%2Fb+%2B+b%2Fc+%3D+d

[vgardiy]

ac + b^2 =dbc
b^2 -dc*b +ac = 0

Then quadratic formula it so

b= (dc+sqroot(d^2c^2-4ac))/2 and
b= (dc-sqroot(d^2c^2-4ac))/2

which is what wolfram says

[iNerd]

ac + b^2 =dbc
b^2 -dc*b +ac = 0

Then quadratic formula it so

b= (dc+sqroot(d^2c^2-4ac))/2 and
b= (dc-sqroot(d^2c^2-4ac))/2

which is what wolfram says

Facepalm - so obvious. I can't believe I took the +ac to the other side to make it negative and then got stuck -.-

[Furbob]

I'm going to throw in a random question since it's a one-off (and nacho wont mind hopefully)

from Essentials 1G Q9.b

Let g: [b, 2] → R where g(x) = 1 − x2. If b is the smallest real value such that g has an
inverse function, find b and g−1(x)

[nacho]

wow essentials q's own.
anyway, here:

I'm going to throw in a random question since it's a one-off (and nacho wont mind hopefully)

from Essentials 1G Q9.b

Let g: [b, 2] → R where g(x) = 1 − x2. If b is the smallest real value such that g has an
inverse function, find b and g−1(x)

g:[b,2] ---> R
g(x) = 1-x^2
x = 1 - y^2
therefore g(x) = sqrt.(1-x)
x e [-3, 1] (domain of x) as g:[b,2]
lowest value of b is 0, as you cannot take the square root of negative numbers