Question

4. Write a program in a script file that determines the real roots of a quadratic equation ax2+bx+c 0 When the file runs, it asks the user to enter the values of the constants a, b, and c. To calculate the roots of the equation the program calculates the discriminant D, given by: D b2-4ac When D 0, the program displays message "The equation has two roots," and the roots are displayed in the next line. When D= 0, the program displays message "The equation has one root," and the root is displayed in the next line. WhenD< 0, the program displays CSC 113-Fall 2019 message "The equation has no real roots." Run the script file in the Command Window three times to obtain solutions to the following three equations: (b) -3x2+ 4x-6 0 (e) -3x2+ 7x + 5 = 0 (a) 3x2+6x+3=0 5. Write a program that generates a vector with 20 random elements between -10 and 10 and then finds the sum of the positive elements 6. Finish In-Class Activities problems (if you haven't finished them yet). Upload your solution in the related In-Class Activity (ICA) folder or type it here. If you type the ICAs here, make sure to add the ICA number and Problem number. For example: ICA 14- Problem 1
4-6 on matlab

Answer 1

%4

a=input('Enter a: ');
b=input('Enter b: ');
c=input('Enter c: ');

D=b*b-4*a*c;
if D>0
fprintf('The equation has two roots\n%f\t%f\n',(-b+sqrt(D))/(2*a),(-b-sqrt(D))/(2*a))
elseif D==0
fprintf('The equation has one root\n%f\n',(-b)/(2*a))
else
fprintf('The equation has no real roots\n')
end

%%%%%%%%%%%%%%%%%%%%

%5

vec=-10+20*rand([1,20]);

sumPositive=sum(vec(vec>0));

fprintf('Sum of positive elements: %f\n',sumPositive)

%6 is not really a matlab questions

quadEq.m > quadEq Enter a a-input'Enter a: '); b-input ('Enter b: '); c input'Enter c: '); Enter b D-b*b-4*a*c; 5 Enter c if D e -6 fprintf('The equation has two roots\n%f\t%f\n*,(-b+sqrt(D))/ (2*a), (-b-sqrt(D) ) /(2*a)) 7 The equation has two roots elseif D-e fprintf('The equation has one root\n%f\n',( -b) /(2*a)) else fprintf('The equation has no real roots\n') 1.000000 -6.000000 9 10 >> sumPos 11 Sum of positive elements: 34.916906 end 12 >> sumPos.m + vec-10+20*rand ( [1, 20] ); sumPositive=sum (vec (vec >0) ); fprintf('Sum of positive elements: %f\n',sumPositive)