what is one example that uses the quaratic equation to show that quadratic equation can not have one real root and one imaginary root?
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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, 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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