Coding a quadratic root finder

Question

Will Murphy 2020-2-17

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编辑： James Tursa 2020-2-17

Hi I'm trying to code a quadratic root finder that works in the cases that b^2 - 4ac is positive, negative and zero as well as non-existent in the case a=b=0 & c isnt, with a given a,b & c such ax^2+bx+c=0

So far, I have:

a=input('a=')
b=input('b=')
c=input('c=')
d=sqrt(b*b-4*a*c)
if d>0
    fprintf('two real roots exist:\n');
    x1=(d-b)/(2*a)
    x2=(-d-b)/(2*a)
    elseif d==0 
    fprintf('one real roots exist:\n');
    x=-b/(2*a)
elseif d<0
    fprintf('two complex roots exist:\n')

I dont know what to do to compute these roots into the form a+bi and a-bi

Furthermore I dont know how to compute a=b=0 and c is nonzero.

How do I finish off this code, any help would be greatly appreciated as I only started coding the other day. Furthermore if I am wrong elswhere please let me know where so I know how to do this in the future.

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Answer 1

Bhaskar R 2020-2-17

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You no need to do program to find roots of the quadratic quation, MATLAB has built in command roots

a=input('a=')
b=input('b=')
c=input('c=')
r = roots([a,b,c]) %% coeffients

to get real and and imaginary parts from the roots you can apply real and imag after that you can perform conditioning to state which kind they are.

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Answer 2

Sindar 2020-2-17

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One error: you want to check the sign of (b*b-4*a*c), not sqrt(b*b-4*a*c)

The case of two complex roots works exactly like two real roots. Since (b*b-4*a*c) is negative, the sqrt will be complex and thus so will your x's. If you want to write x=A+Bi:

A=real(x);
B=imag(x);

For cases where there are no roots ( a=b=0 & c , second root of d=0), fill them with NaN in case you use them later:

elseif d==0
    if %check  a=b=0 & c isn't
        fprintf('no roots exist:\n');
        x1=NaN;
        x2=NaN;
    else
        fprintf('one real roots exist:\n');
        x1=-b/(2*a);
        x2=NaN;
    end

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Sindar 2020-2-17

assuming this is a code exercise; if you just want the functionality, use Bhaskar's root solution

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Answer 3

James Tursa 2020-2-17

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编辑：James Tursa 2020-2-17

I would advise having your logic figure out what situation you have before you start solving things. E.g., something like

if( a == 0 ) % Not Quadratic
    if( b == 0 ) % Constant case
        if( c == 0 ) % OK
        else % not OK
        end
    else % Linear case
    end
else % Quadratic
    d2 = b^2 - 4*a*c;
    if( d2 == 0 )
    elseif( d2 > 0 )
    else
    end
end

Hint: The MATLAB sqrt( ) function will automatically produce a complex result if the input argument is negative. You don't have to write special code to figure out how to get that result.