The program to find all the roots of a quadratic equation is discussed here.
The general form of a quadratic equation is (ax^2 + bx + c = 0). The highest degree in a quadratic equation is 2. Hence, a quadratic equation will have two roots.
The formula to find the roots of a quadratic equation is given as follows
x = [-b +/- sqrt(-b^2 – 4ac)]/2a
The discriminant of the quadratic equation is
k = (b^2 – 4ac).
Depending upon the nature of the discriminant, the roots can be found in different ways.
Case 1: If the discriminant is positive,
r1 = (-b +?k)/ 2a and r2 =(b +?k)/ 2a are the two roots.
Case 2: If the discriminant is zero,
r1 = r2 = (-b / 2a)are the two roots.
Case 3: If the discriminant is negative,
r1 = (-b +i ?k)/ 2a and r2 =(b + i?k)/ 2a are the two roots.
For example, consider the following equationnn2x^2 8x + 3 = 0.nna = 2, b = -8, c = 3nnDiscriminant value, k =b^2 - 4acn= 8^2 - 4*(-8)*3n = 40 nnThe discriminant value is positive. Hence, the roots are real and distinct.nnr1 = (-b +?k)/ 2an = (8 +?40) /2*2n = 2.3875nnr2 =(b +?k)/ 2an =(-8 +?40) /2*2n = -0.3875nnr1 = 2.3875andr2 = -0.3875 are the two roots.n
1. Input the value of a, b, c.
2. Calculate k = b*b – 4*a*c
3. If (d < 0)
Display “Roots are Imaginary, calculater1 = (-b +i ?k)/ 2a and r2 =(b + i?k)/ 2a.
else if (d = 0)
Display “Roots are Equal” and calculate r1 = r2 = (-b / 2*a)
else
Display “Roots are real and calculate r1 = -b + ?d / 2*a andr2 = -b – ?d / 2*a
4. Print r1 and r2.
5. End the algorithm.
@@coding::1@@
Time complexity: O(1)
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