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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Start with:



`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

Approved by eNotes Editorial Team

Posted on

Soaring plane image

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial