Question

I’m implementing a code in matlab to solve quadratic equations, using the resolvent formula:

Here´s the code:

clear all 
format short 
a=1; b=30000000.001; c=1/4; 
rdelta=sqrt(b^2-4*a*c); 
x1=(-b+rdelta)/(2*a);
x2=(-b-rdelta)/(2*a);
fprintf(' Roots of the polynomial %5.3f x^2 + %5.3f x+%5.3f \n',a,b,c)  
fprintf ('x1= %e\n',x1)
fprintf ('x2= %e\n\n',x2)
valor_real_x1= -8.3333e-009;
valor_real_x2= -2.6844e+007;

error_abs_x1 = abs (valor_real_x1-x1);
error_abs_x2 = abs (valor_real_x2-x2);

error_rel_x1 = abs (error_abs_x1/valor_real_x1);
error_rel_x2 = abs (error_abs_x2/valor_real_x2);

fprintf(' absolute_errorx1 = |real value - obtained value| = |%e - %e| = %e \n',valor_real_x1,x1,error_abs_x1)  
fprintf(' absolute_errorx2 = |real value - obtained value| = |%e - %e| = %e \n\n',valor_real_x2,x2,error_abs_x2) 

fprintf(' relative error_x1 = |absolut error / real value| = |%e / %e| = %e \n',error_abs_x1,valor_real_x1,error_rel_x1 ) 
 fprintf(' relative_error_x2 = |absolut error / real value|  = |%e / %e| = %e \n',error_abs_x2,valor_real_x2,error_rel_x2) 

The problem I have is that it gives me an exact solution, ie for values ​​a = 1, b = 30000000,001 c = 1/4, the values ​​of the roots are:

Roots of the polynomial 1.000 x^2 + 30000000.001 x+0.250 
 x1= -9.313226e-009
 x2= -3.000000e+007

Knowing that the exact value of the roots of the polynomial are:

x1= -8.3333e-009
x2= -2.6844e+007

Which gives me the following errors in the absolute and relative precision of the calculations:

 absolute_errorx1 = |real value - obtained value| = |-8.333300e-009 - -9.313226e-009| = 9.799257e-010 
 absolute_errorx2 = |real value - obtained value| = |-2.684400e+007 - -3.000000e+007| = 3.156000e+006 

 relative error_x1 = |absolut error / real value| = |9.799257e-010 / -8.333300e-009| = 1.175916e-001 
 relative_error_x2 = |absolut error / real value|  = |3.156000e+006 / -2.684400e+007| = 1.175682e-001

My question is: Is there an optimum method to obtain the roots of a quadratic equation?, ie I can make changes to my code to reduce the relative error between the expected solution and the resulting solution?

Answer 1

Using the quadratic formula directly in this cases results in a large loss of numerical precision from subtracting two values of very similar magnitude. This is because the expression

sqrt(b*b - 4*a*c)

is nearly the same as b. So you should use only one of these two roots, the one that does not involve subtracting two very close values, and for the other root you can use (for instance) the fact that the product of roots of a quadratic is c/a. I’ll let you fill in the gaps.