I have been unable to arrive at any calculus oriented methodology apart from the solutions in terms of the previous comments. Accordingly, two proposed methods which uses the quadratic formula, but do not utilise any calculus in their respective proceedures, are furnished below.

Method 1:

bx + c = 0 ..........(i)

Multiplying both sides of (i) by (x-t), we have:
(bx + c)(x-t) = 0
Or, b*x^2 + (c-bt)x - ct = 0 .........(ii)

Now this equation has one additional root, x=t when compared to (i)

In terms of the quadratic formula, we have:

x = [(bt - c) +/- V((c-bt)^2 + 4bct)]/(2b)
Or, x = [(bt - c) +/- (c + bt)]/(2b)
Or, x = t, -c/b

Disregarding the root x=t, we obtain the required solution to the
given equation as x = -c/b

Method 2:

Substituting x = 1/(z^2), the given equation yields:
b/(z^2) + c = 0
or, c*z^2 + b = 0 ( for non zero z)

Applying the quadratic formula, this gives:
z = +/- V(-4bc)/(2c)
Or, z = +/- V(-b/c)
Or, z^2 = - b/c
Or, x = - c/b, which is the required solution 

Edited on November 4, 2007, 12:14 am

Posted by K Sengupta on 2007-11-04 00:08:00