The Solution of quadratic equations in the complex number system
Relating to quadratic equations, the complex roots and the imaginary numbers exist when the value under radical portion of the quadratic formula is -ve. When this takes place, the equation contains zero or no roots in the real numbers’ set.
The roots are associated with complex numbers’ set and are known as the “imaginary roots” or the “complex roots”. These imaginary roots are expressed in a ± bi form.
The standard quadratic equation is in ax2 + bx + c = 0 form, where a, b, and c are the values of real numbers with a ≠ 0 & the roots of the equation are given by –
X = -b ± √b² – 4ac / 2a
Discriminant (D) = √b2−4a
The below-mentioned roots can be of 3 forms, which depends on D’s value.
Complex Roots: When the value of D is less than zero i.e. D<0, so in this case, the quadratic equation consist of 2 complex roots. This quadratic equation is given by,
x1 = −b + i√4ac – b² / 2a,
x2 = −b – i√4ac – b² / 2a
Equal Roots: When the value of D is equal to zero i.e. D = 0,then the quadratic equation contains 2 equal roots. That quadratic equation is represented by the – x1, x2 = −b / 2a
Distinct Roots: When the value of D is greater than zero; i.e. D>0, So, the quadratic equation consist of 2 real or distinct roots. This quadratic equation is given by, x1 = −b + √b²−4ac / 2a,
x2 = −b – √b²− 4ac / 2a
Notice: A polynomial having ‘n’ degree will be having ‘n’ roots. This is called the Fundamental Theorem of Algebra.
The quadratic equation has a degree of 2, thus they have 2 roots.
The Power of i –
i² = −1
i³ = i = i(−1) = −i
i⁴ = i².i² = (−1).(−1) = 1
Therefore, the general form is represented by-
i^4k + 1 = i
i^4k+1 = −1
i^4k+2 = −i
i^4k+3 = 1
Question 1: Find the roots of the given quadratic equation – √6x² + x + √6 = 0
Solution: On having the comparison of the given equation with the general form of quadratic equation,
a = √6, b = 1, c = √6
Here Discriminant (D) = b2−4ac = (1)² – 4. (√6) (√6) = -23
x = −b ± √D / 2a
= -1 ± √-23 / 2. √6
= -1 ± √23i / 2√6 < 0