# what is one example that uses the quaratic equation to show that quadratic equation can not have one real root and one imaginary root? Start with:`ax^2+bx+c=0`Consider`b^2 - 4ac`this is either negative, or it is not.  If it is negative, then you have to take the square root of something negative, and you get `i` if it is not negative (so 0 or positive), then when you take the square root,...

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`ax^2+bx+c=0`

Consider

`b^2 - 4ac`

this is either negative, or it is not.  If it is negative, then you have to take the square root of something negative, and you get `i`
if it is not negative (so 0 or positive), then when you take the square root, you still have something real.

So, you either get:

real `+-` real

or you get

real `+-` imaginary

That is, you can't get one real and one complex:

if one of them is
real + imaginary,
then the other one is
real - imaginary

So, doing some concrete examples, we consider:

`x^2 + 0x + 1`

`x^2 + 0x -1`

`x^2 + 2x +1`

For the first:

`x = -0/(2*1) +- (sqrt(0^2 - 4*1*1))/(2*1) = 0 +- sqrt(-4)/2 = +- i`

For the second:

`x = -0/(2*1) +- (sqrt(0^2 - 4*-1*1))/(2*1) = 0 +- sqrt(4)/2 = +- 1`

For the third:

`x = -2/(2*1) +- (sqrt(2^2 - 4*1*1))/(2*1) = -1 +- sqrt(0)/2 = - 1`

The key is `b^2 - 4ac`

If it is negative, you get two complex numbers (numbers that include `i`)

If it is positive, you get two real numbers

If it is zero, you get one real number

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