MATLAB: Quadratic equation & Minus B

homeworkMATLABminus bquadraticsolve

How do I solve the quadratic equation ax^2+bx+c=0 using the minus b formula, x=-b+- sqrt b^2-4ac/2a on MATLAB

Thank you

Best Answer

The following example code snippet works for accepting the coefficients a,b and c and calculating the roots and finding if they are real or not:

a1 = 'Enter value of a ';
a = input(a1)
b1= 'Enter value of b ';
b = input(b1)
c1= 'Enter value of c ';
c = input(c1)
x1 = (-b + sqrt(b^2 - (4 * a * c))) / (2 * a);
x2 = (-b - sqrt(b^2 - (4 * a * c))) / (2 * a);
t1=isreal(x1);
t2=isreal(x2);

Related Solutions

MATLAB: The own function is not working with 3x^2-12-15=0

First, verify that your solution is correct.

roots([3 -12 -15])
ans =
     5
    -1

Homework is good. Learn to write MATLAB, as well as learning the quadratic formula. But also learn that the solution has been written for you, and written well, by experts in the field. So normally, just use roots. You can also use solve, as Birdman points out.

If I take your code, writing it like this (since all you posted was a picture, so I will need to retype it myself.)

x1 = (-b + sqrt(b^2 - 4*a*c))/2*a
x2 = (-b - sqrt(b^2 - 4*a*c))/2*a

we WILL GET THE WRONG ANSWERS. Why? The correct way to write it was:

x1 = (-b + sqrt(b^2 - 4*a*c))/(2*a)
x2 = (-b - sqrt(b^2 - 4*a*c))/(2*a)

The difference is entirely in a necessary set of parens around the denominator. For example...

6/2*3
ans =
     9

Yet,

6/(2*3)
ans =
     1

There is a difference, and it is important. :)

MATLAB: Plotting help, please

Change the following lines in your code to use ./ instead of /

    tan3a = (-E + sqrt(E.^2 - (4 .* D .* F)))./ (2 .* D);% note ./
    tan3b = (-E - sqrt(E.^2 - (4 .* D .* F)))./ (2 .* D);
    tan4a = (-B + sqrt(B.^2 - (4 .* A .* C)))./ (2 .* A);
    tan4b = (-B - sqrt(B.^2 - (4 .* A .* C)))./ (2 .* A);

Best Answer

Related Solutions

MATLAB: The own function is not working with 3x^2-12-15=0

MATLAB: Plotting help, please

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