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

1 Answer
Feb 3, 2016

Answer:

#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)#