If x1, x2 are solution of a polynomial f(x),

then f(x1) and f(x2) must vanish.. i.e. f(x1)=0 and f(x2)=0. But in above example it is not so...

Thanks sir,

Sir, u rightly ponied the variation in the scales in-between different coefficients of the polynomial. I am working on it.

kindly answer one wore thing. I have following one example.

aw = 23; bw = 34; cw = 12; nc = (aw/bw) + 10/cw

How can I set Precision to evaluate "nc"

Consider an exact polynomial expression:

poly = -6 + 17 x - 32 x^2 + 26 x^3 - 20 x^4 + 14 x^5 - 8 x^6 + 3 x^7

Note that it contains no approximate numbers like 1.4: all of the coefficients are exact rational numbers (integers in this case). In Mathematica, any number containing a decimal point is an approximate number, not an exact number. You cannot expect exact results from approximate input.

Make an equation, solve it:

exact = Solve[poly == 0]
{{x -> 1/3 (1 - I Sqrt[2])}, {x -> 1/3 (1 + I Sqrt[2])}, {x -> 
   Root[-6 + 5 #1 - 4 #1^2 + 3 #1^3 - 2 #1^4 + #1^5 &, 1]}, {x -> 
   Root[-6 + 5 #1 - 4 #1^2 + 3 #1^3 - 2 #1^4 + #1^5 &, 2]}, {x -> 
   Root[-6 + 5 #1 - 4 #1^2 + 3 #1^3 - 2 #1^4 + #1^5 &, 3]}, {x -> 
   Root[-6 + 5 #1 - 4 #1^2 + 3 #1^3 - 2 #1^4 + #1^5 &, 4]}, {x -> 
   Root[-6 + 5 #1 - 4 #1^2 + 3 #1^3 - 2 #1^4 + #1^5 &, 5]}}

I get seven roots, all exact. Two are expressed as radicals, five have no such representation and are expressed using Root. These are the exact solutions, as we can verify:

Simplify[poly /. exact]
{0, 0, 0, 0, 0, 0, 0}

However, if I want numerical results:

approx = NSolve[poly == 0]
{{x -> -0.551685 - 1.25335 I}, {x -> -0.551685 + 1.25335 I}, {x -> 
   0.333333 - 0.471405 I}, {x -> 0.333333 + 0.471405 I}, {x -> 
   0.805786 - 1.2229 I}, {x -> 0.805786 + 1.2229 I}, {x -> 1.4918}}

The numerical solutions are necessarily approximate: exact numerical solutions of this equation are not possible. Most polynomial equations do not have exact numerical solutions: the solutions are irrational and thus have no finite numerical representation. Substituting the approximate solutions back into the original polynomial yields results that are approximately, but not exactly, zero:

poly /. approx
{1.84741*10^-13 + 1.91847*10^-13 I, 1.84741*10^-13 - 1.91847*10^-13 I,
  6.66134*10^-16 + 1.9984*10^-15 I, 
 6.66134*10^-16 - 1.9984*10^-15 I, -1.42109*10^-14 + 
  7.10543*10^-15 I, -1.42109*10^-14 - 
  7.10543*10^-15 I, -2.13163*10^-14}

And that is just as you should expect.

This is not making sense. You have an approximate set of values and plug them in to compute residuals. How can those residuals possibly be "exactly" zero?

This is a matter of numerical mathematics, not Mathematica.

If you want exact results, you must use exact input. You are using approximate numbers like 0.51 in your input. Any number expressed with a decimal point is approximate in Mathematica. 0.51 (approximate) is not the same thing as 51/100 (exact, rational). If you don't actually want exact results, but rather results with better precision than you're getting, you can try higher WorkingPrecision as Daniel suggests.

Try NSolve with higher than default WorkingPrecision. And be prepared to wait.

yes it worked for this polynomial...

But in my real problem the polynomial in more complex... Neither "Solve" nor "NSolve" are working well for that...

kindly suggest some another method...

If u wish I can upload my orignal equation