How are used imaginary numbers in solving quadratic equations ?

neela | Student

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.

giorgiana1976 | Student

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.

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