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POsolve In Matlab a quadratic equation with very small coefficients
 

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  1. POsolve In Matlab a quadratic equation with very small coefficients
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    <p>I'm implementing a code in matlab to solve quadratic equations, using the resolvent formula:</p> <p><img src="https://i.stack.imgur.com/JgsUN.gif" alt="enter image description here"></p> <p>Here´s the code:</p> <pre><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) </code></pre> <p>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:</p> <pre><code>Roots of the polynomial 1.000 x^2 + 30000000.001 x+0.250 x1= -9.313226e-009 x2= -3.000000e+007 </code></pre> <p>Knowing that the exact value of the roots of the polynomial are:</p> <pre><code>x1= -8.3333e-009 x2= -2.6844e+007 </code></pre> <p>Which gives me the following errors in the absolute and relative precision of the calculations:</p> <pre><code> 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 </code></pre> <p>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?</p>
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    1. USAndrey Rubshtein
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      1. POsolve In Matlab a quadratic equation with very small coefficients
      1. USCastilho
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    1. COjust one thing: when you think you have the "real" solutions, check them with the girard relations (we call them like that on the latin speaking world, I think in anglo-saxonic countries they are called Newton Identity or sth like that). It seems that woodchips answered your question nicely, but just recall also that in the double precision world, your actual mantissa precision is 2^53, mess with differences greater than that, or smaller than your exponent, and you end up with garbage
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      1. POsolve In Matlab a quadratic equation with very small coefficients
      1. USCastilho
    2. CO2+2^54 == 2+2^54 +1 try this in Matlab, or this 2^54 - (2^54 - 1) == 0, and see what's the return
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      1. POsolve In Matlab a quadratic equation with very small coefficients
      1. USCastilho
 

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