I have to code this in Matlab, but a written out version works. The output for t

Question

I have to code this in Matlab, but a written out version works. The output for the code is the second picture. Program 5-Roots (homework) CSCI 251 For program 5, we will revisit program 2, and rewrite it using a function and a main program instead. The two roots of a quadratic equation ax2 +bx +c can be obtained using the following formula: -b+ b2-4ac -b-vb2-4ac 2a 2a b2- 4ac is called the discriminant of the quadratic equation, and its value determines the number of roots as follows: b2-4ac >0, two real roots: rl and r2 b2-4ac-0, one real root:r b2-4ac

Explanation / Answer

roots_yourLastName.m

function [numOfRoots] = roots_yourLastName( a, b, c)
    discriminant = b*b - 4*a*c;
    if discriminant < 0
        numOfRoots = 0;
    elseif discriminant == 0
        numOfRoots = 1;
    else
        numOfRoots = 2;
    end
end

rootsTest_yourLastName.m

function [] = rootsTest_yourLastName()
    a = input('Enter the A coefficient: ');
    b = input('Enter the B coefficient: ');
    c = input('Enter the C coefficient: ');
    numOfRoots = roots_yourLastName(a,b,c);
    fprintf('%d, %d, and %d coefficients have',a,b,c);
    discriminant = b*b - 4*a*c;
    if numOfRoots==0
        fprintf(' no real roots ');
    elseif numOfRoots==1
        fprintf(' one real root: ');
        fprintf(' r=%.2f ', (-b*1.0)/2*a );
    else
        fprintf(' two real roots: ');
        fprintf(' r1 = %.2f ',(-b+sqrt(discriminant)+0.0)/2*a);
        fprintf(' r2 = %.2f ',(-b-sqrt(discriminant)+0.0)/2*a);
    end
end

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