# write down 3 quadratic equations with the following root and prove the roots are in fact the values of x for which y is equal to zero 2 imaginary roots write down 3 quadratic equations with the following root and prove the roots are in fact the values of x for which y is equal to zero equation two imaginary roots You should remember that an equation that will have a complex root, `x_1 = a + b*i` , it will also have, as other root, the conjugate of complex root, `x_2 = a - b*i.`

Since a quadratic equation cannot have more than two roots, then it will either have two real roots, or two complex roots, but never a combination of one real and one complex.

You also need to remember how to form a quadratic equation when the roots are given, such that:

`x^2 - (x_1+x_2)*x + x_1*x_2 = 0`

Considering `x_1 = 1 + ` `i,`  then, `x_2`  needs to be its conjugate, `x_2 = 1 - i.`

You need to find the sum and the product of the roots, such that:

`x_1 + x_2 = 1 + i + 1 - i`

`x_1 + x_2 = 2`

`x_1*x_2 = (1 + i)(1 - i) => x_1*x_2 = 1 - i + i - i^2`

Since the complex number theory states that `i^2 = -1` , yields:

`x_1*x_2 = 1 - (-1) => x_1*x_2 = 2`

Hence, substituting 2 for `x_1 + x_2`  and 2 for `x_1*x_2`  in quadratic equation, yields:

`x^2 - 2x + 2 = 0`

You may check if the given roots verify the equation, either substituting, one by one, the roots in equation, or using quadratic formula.

`x_(1,2) = (2 +- sqrt(2^2 - 4*1*2))/2 => x_(1,2) = (2 +- sqrt(4-8))/2 => x_(1,2) = (2 +- sqrt (-4))/2`

`x_(1,2) = (2 +- 2*i)/2 => x_(1,2) = 2(1+-i)/2 => x_(1,2) = 1 +- i`

Notice that using quadratic formula yields `x_1 = 1 + i ` and `x_2 = 1 - i` , the original roots.

Considering other pair of complex roots, `x_1 = 3 - 2i ` and `x_2 = 3 + 2i,`  you may form the second quadratic equation, such that:

`x_1 + x_2 = 3 - 2i + 3 + 2i = 2*3 = 6`

`x_1*x_2 = 3^2 - (2i)^2 = 9 + 4 = 13`

Hence, evaluating the second quadratic equation, whose roots are `x_(1,2) = 3 +- 2i` , yields `x^2 - 6x + 13 = 0` .

Considering the third pair of complex roots, `x_(1,2) =sqrt6+-sqrt6*i` , you may form the following quadratic equation, such that:

`x_1 + x_2 = 2sqrt6`

`x_1*x_2 = (sqrt6)^2 - (sqrt 6*i)^2 = 6 + 6 = 12`

`x^2 - 2sqrt6*x + 12 = 0`

Hence, evaluating the third quadratic equation, whose roots are `x_(1,2) = sqrt6+-sqrt6*i` , yields `x^2 - 2sqrt6*x + 12 = 0.`