How are used imaginary numbers in solving quadratic equations ?
It is vague to say imaginary numbers are used in quadratic equations. A quadratic equation is 2nd degree polynomial equation , ax^2+bx+c = 0 ,where the quadratic coefficient of x^2 a , the linear coefficient x , b and the constant term c are all real.
But in quadratic equations the solutions or the roots could be imaginary or complex.
We know ax^2+bx+c = 0 is a general quadratic equation. Then a(x^2+bx/a)+c/a = 0
x^2+bx/a +((b/2a)^2 -(b/2a)^2 + c/a = 0
(x+b/2a)^2 = (b^2-4ac)/(2a)^2
x +b/2a = +(1/2a)sqrt(b^2-4ac).
x+b/2a = -(1/2a)sqrt(b^2-4ac).
x = -(b/2a) + (1/2a)(b^2-4ac)..........(1) Or
x = -(b/2a) - (1/2a) (b^2-4ac)..........(2)
From (1) and (2) it is only when b^2-4ac is negative that an imaginary roots can occur.
Also it could be noticed that the imaginary roots can occur in conjugate pairs like A+Bi and A-Bi and never in single imagiginary root. The idea is usuful while solving the quadratic equation.
Imaginary numbers are involved in solving of the quadratic equations, when the discriminant of the equation is negative.
The quadratic equation:
ax^2 + bx + c = 0
x1 = [-b+sqrt(delta)]/2a
x2 = [-b-sqrt(delta)]/2a
Discriminant of the quadratic equation:
delta = b^2 - 4ac
If delta is negative, then sqrt -delta = i*sqrt delta.
We have used the property of imaginary unit i:
i^2 = -1
i = sqrt(-1)
We'll solve an example:
x^2 - 4x + 8 = 0
delta = (-4)^2 - 4*1*8
delta = 16 - 32
delta = -16
sqrt delta = sqrt (-1)*(16) = i*sqrt 16 = 4i
The roots of the quadratic equation are:
x1 = (4+4i)/2
x1 = 2 + 2i
x2 = 2 - 2i
We remark here the fact that if an equation has a complex root, then the equation has as root the conjugate of the complex root, also.