how to solve quadratic equation?

Hai,

How could I solve a quadratic equation in matlab? Looking for your reply.

BSD

Answer 1

[Walter Roberson]

roots(). Or if you have the symbolic toolbox, solve()

[Naz]

More explicit: x^2-2x+1. Coefficients are 1 -2 1, therefore:

roots([1 -2 1])

[Walter Roberson]

And note that any root that is output might be complex. If you want only the real roots, filter to TheRoots(imag(TheRoots)==0)

[bsd]

why is that == 0 used?

BSD

[Walter Roberson]

You can tell whether a number has a complex part or not by testing to see if the imaginary part is 0. imag(x) gives you the imaginary part of x, so imag(x)==0 tests whether the imaginary part is 0. TheRoots(imag(TheRoots)==0) thus selects only the roots which are real-valued with no imaginary component.

Of course for a quadratic function over real coefficients, either _neither_ root is complex or _both_ roots are complex...

Answer 2

[Reina Young]

This code works for real and imaginary roots.

%%QUADRATIC FORMULA%%

%  A(x^2)+B(x)+C=0
begin_prompt = 'Quadratic Formaula (Yes/No)(1/0)';
d = input(begin_prompt)
if d  == 1    
prompt = 'Ax^2+Bx+C=0...A=';
A = input(prompt)

prompt2 = 'Ax^2+Bx+C=0...B=';
B = input(prompt2)

prompt3 = 'Ax^2+Bx+C=0...C=';
C = input(prompt3)
Answer1= ((-B)+((B^2-4*A*C))^0.5)/(2*A)
Answer2= ((-B)-((B^2-4*A*C))^0.5)/(2*A)
disp(Answer1)
disp(Answer2)
else 
disp( 'Error in Start Prompt Input, Please Pick Yes (1) or this code will not work');

end

Answer 3

[Emre Komur]

Solution Method-1

1- You can create function as below

function quadraticEquation

a = input('please enter a value :');
b = input('please enter b value :');
c = input('please enter a value :');

delta = (b.^2)-(4*a*c);

if(delta < 0)      
disp("Delta < 0 The equation does not have a real root");
elseif (delta == 0)
   disp('The equation has only one real roots');
   disp(-b./(2*a));   
else
   disp('The equation has two real roots');
   disp((-b+sqrt(delta))./(2*a));
   disp((-b-sqrt(delta))./(2*a));
    
end
end

2- You should call the function as below

quadraticEquation

Answer 4

[Rick Rosson]

Please try:

   x = zeros(2,1);

   d = sqrt(b^2 - 4*a*c);

   x(1) = ( -b + d ) / (2*a);
   x(2) = ( -b - d ) / (2*a);

HTH.

Rick

[Walter Roberson]

Which is equivalent to x = roots([a,b,c]) except that roots() does not promise any particular order.

[Walter Roberson]

Beware a = 0 !

[Rick Rosson]

If a = 0, then it is not (strictly speaking) a quadratic equation. It's a linear equation, and the solution in that case is trivial to compute.

[Walter Roberson]

Yes, but it is not an uncommon problem for people to calculate or randomly generate the coefficients and forget to double-check that the system is still of the same order.

Answer 5

[Temidayo Daniel]

a = input ('Enter the value of the coefficient of x square   ');
b = input ('Enter the value of the coefficient of x   ');
c = input ('Enter the third value  ');
t = sqrt ((b^2) - (4*a*c));
u = (-b+t)/(2*a);
v = (-b-t)/(2*a);
x1 = u
x2 = v