How do you find the zeros, real and imaginary, of y= x^2-5x+16 using the quadratic formula?

1 Answer
Ken C.
Feb 3, 2016

x=(5+isqrt(39))/(2) and x=(5-isqrt(39))/(2)

Explanation:

The quadratic formula is: (-b+-sqrt(b^2-4ac))/(2a).

This is used to solve equations of the form y=ax^2+bx+c, which is the same type of equation as y=x^2-5x+16. We can see that for our example, a=1, b=-5, c=16. Plugging these guys into the formula,
(-(-5)+-sqrt((-5)^2-4(1)(16)))/(2(1))

And simplifying,
(5+-sqrt(25-64))/(2)

(5+-sqrt(-39))/(2)

We now see we have an issue - a negative in the square root. That means both of our solutions will be imaginary, so to represent that, we take the sqrt(-1) (i) out of the square root to make it positive:
(5+-isqrt(39))/(2)

The next thing we look for is any way to simplify the square root. But sqrt(39) can't be simplified further, which means our solutions are:
x=(5+isqrt(39))/(2)
x=(5-isqrt(39))/(2)

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